## Learning R: The Ultimate Introduction (incl. Machine Learning!)

There are a million reasons to learn R (see e.g. Why R for Data Science – and not Python?), but where to start? I present to you the ultimate introduction to bring you up to speed! So read on…

I call it ultimate because it is the essence of many years of teaching R… or put differently: it is the kind of introduction I would have liked to have when I started out with R back in the days!

A word of warning though: this is a introduction to R and not to statistics, so I won’t explain the statistics terms used here. You do not need to know any other programming language but it does no harm either. Ok, now let us start!

First you need to install R (https://www.r-project.org) and preferably RStudio as a Graphical User Interface (GUI): https://www.rstudio.com/products/RStudio/#Desktop. Both are free and available for all common operating systems.

To get a quick overview of RStudio watch this video:

You can either type in the following commands in the console or open a new script tab (File -> New File -> R Script) and run the commands by pressing Ctrl + Enter/Return after having typed them.

First of all R is a very good calculator:

2 + 2
## [1] 4

sin(0.5)
## [1] 0.4794255

abs(-10) # absolute value
## [1] 10

pi
## [1] 3.141593

exp(1) # e
## [1] 2.718282

factorial(6)
## [1] 720


By the way: The hash is used for comments, everything after it will be ignored!

Of course you can define variables and use them in your calculations:

n1 <- 2
n2 <- 3
n1 # show content of variable by just typing the name
## [1] 2

n1 + n2
## [1] 5

n1 * n2
## [1] 6

n1^n2
## [1] 8


Part of R’s power stems from the fact that functions can handle several numbers at once, called vectors, and do calculations on them. When calling a function arguments are passed with round brackets:

n3 <- c(12, 5, 27) # concatenate (combine) elements into a vector
n3
## [1] 12  5 27

min(n3)
## [1] 5

max(n3)
## [1] 27

sum(n3)
## [1] 44

mean(n3)
## [1] 14.66667

sd(n3) # standard deviation
## [1] 11.23981

var(n3) # variance
## [1] 126.3333

median(n3)
## [1] 12

n3 / 12
## [1] 1.0000000 0.4166667 2.2500000


In the last example the 12 was recycled three times. R always tries to do that (when feasible), sometimes giving a warning when it might not be intended:

n3 / c(1, 2)
## Warning in n3/c(1, 2): longer object length is not a multiple of shorter
## object length
## [1] 12.0  2.5 27.0


In cases you only want parts of your vectors you can apply subsetting with square brackets:

n3[1]
## [1] 12

n3[c(2, 3)]
## [1]  5 27


Ranges can easily be created with the colon:

n4 <- 10:20
n4
##  [1] 10 11 12 13 14 15 16 17 18 19 20


When you test whether this vector is bigger than a certain number you will get logicals as a result. You can use those logicals for subsetting:

n4 > 15
##  [1] FALSE FALSE FALSE FALSE FALSE FALSE  TRUE  TRUE  TRUE  TRUE  TRUE

n4[n4 > 15]
## [1] 16 17 18 19 20


Perhaps you have heard the story of little Gauss where his teacher gave him the task to add all numbers from 1 to 100 to keep him busy for a while? Well, he found a mathematical trick to add them within seconds… for us normal people we can use R:

sum(1:100)
## [1] 5050


When we want to use some code several times we can define our own function (a user-defined function). We do that the same way we create a vector (or any other data structure) because R is a so called functional programming language and functions are so called first-class citizens (i.e. on the same level as other data structures like vectors). The code that is being executed is put in curly brackets:

gauss <- function(x) {
sum(1:x)
}
gauss(100)
## [1] 5050

gauss(1000)
## [1] 500500


Of course we also have other data types, e.g. matrices are basically two dimensional vectors:

M <- matrix(1:12, nrow = 3, byrow = TRUE) # create a matrix
M
##      [,1] [,2] [,3] [,4]
## [1,]    1    2    3    4
## [2,]    5    6    7    8
## [3,]    9   10   11   12

dim(M)
## [1] 3 4


Subsetting now has to provide two numbers, the first for the row, the second for the column (like in the game Battleship). If you leave one out, all data of the respective dimension will be shown:

M[2, 3]
## [1] 7

M[ , c(1, 3)]
##      [,1] [,2]
## [1,]    1    3
## [2,]    5    7
## [3,]    9   11


Another possibility to create matrices:

v1 <- 1:4
v2 <- 4:1
M1 <- rbind(v1, v2) # row bind
M1
##    [,1] [,2] [,3] [,4]
## v1    1    2    3    4
## v2    4    3    2    1

M2 <- cbind(v1, v2)  # column bind
M2
##      v1 v2
## [1,]  1  4
## [2,]  2  3
## [3,]  3  2
## [4,]  4  1


Naming rows, here with inbuilt datasets:

rownames(M2) <- LETTERS[1:4]
M2
##   v1 v2
## A  1  4
## B  2  3
## C  3  2
## D  4  1

LETTERS
##  [1] "A" "B" "C" "D" "E" "F" "G" "H" "I" "J" "K" "L" "M" "N" "O" "P" "Q"
## [18] "R" "S" "T" "U" "V" "W" "X" "Y" "Z"

letters
##  [1] "a" "b" "c" "d" "e" "f" "g" "h" "i" "j" "k" "l" "m" "n" "o" "p" "q"
## [18] "r" "s" "t" "u" "v" "w" "x" "y" "z"


When some result is Not Available:

LETTERS[50]
## [1] NA


Getting the structure of your variables:

str(LETTERS)
##  chr [1:26] "A" "B" "C" "D" "E" "F" "G" "H" "I" "J" "K" "L" "M" "N" ...

str(M2)
##  int [1:4, 1:2] 1 2 3 4 4 3 2 1
##  - attr(*, "dimnames")=List of 2
##   ..$: chr [1:4] "A" "B" "C" "D" ## ..$ : chr [1:2] "v1" "v2"


Another famous dataset (iris) that is also built into base R (to get help on any function or dataset just put the cursor in it and press F1):

iris
##     Sepal.Length Sepal.Width Petal.Length Petal.Width    Species
## 1            5.1         3.5          1.4         0.2     setosa
## 2            4.9         3.0          1.4         0.2     setosa
## 3            4.7         3.2          1.3         0.2     setosa
## 4            4.6         3.1          1.5         0.2     setosa
## 5            5.0         3.6          1.4         0.2     setosa
## 6            5.4         3.9          1.7         0.4     setosa
## 7            4.6         3.4          1.4         0.3     setosa
## 8            5.0         3.4          1.5         0.2     setosa
## 9            4.4         2.9          1.4         0.2     setosa
## 10           4.9         3.1          1.5         0.1     setosa
## 11           5.4         3.7          1.5         0.2     setosa
## 12           4.8         3.4          1.6         0.2     setosa
## 13           4.8         3.0          1.4         0.1     setosa
## 14           4.3         3.0          1.1         0.1     setosa
## 15           5.8         4.0          1.2         0.2     setosa
## 16           5.7         4.4          1.5         0.4     setosa
## 17           5.4         3.9          1.3         0.4     setosa
## 18           5.1         3.5          1.4         0.3     setosa
## 19           5.7         3.8          1.7         0.3     setosa
## 20           5.1         3.8          1.5         0.3     setosa
## 21           5.4         3.4          1.7         0.2     setosa
## 22           5.1         3.7          1.5         0.4     setosa
## 23           4.6         3.6          1.0         0.2     setosa
## 24           5.1         3.3          1.7         0.5     setosa
## 25           4.8         3.4          1.9         0.2     setosa
## 26           5.0         3.0          1.6         0.2     setosa
## 27           5.0         3.4          1.6         0.4     setosa
## 28           5.2         3.5          1.5         0.2     setosa
## 29           5.2         3.4          1.4         0.2     setosa
## 30           4.7         3.2          1.6         0.2     setosa
## 31           4.8         3.1          1.6         0.2     setosa
## 32           5.4         3.4          1.5         0.4     setosa
## 33           5.2         4.1          1.5         0.1     setosa
## 34           5.5         4.2          1.4         0.2     setosa
## 35           4.9         3.1          1.5         0.2     setosa
## 36           5.0         3.2          1.2         0.2     setosa
## 37           5.5         3.5          1.3         0.2     setosa
## 38           4.9         3.6          1.4         0.1     setosa
## 39           4.4         3.0          1.3         0.2     setosa
## 40           5.1         3.4          1.5         0.2     setosa
## 41           5.0         3.5          1.3         0.3     setosa
## 42           4.5         2.3          1.3         0.3     setosa
## 43           4.4         3.2          1.3         0.2     setosa
## 44           5.0         3.5          1.6         0.6     setosa
## 45           5.1         3.8          1.9         0.4     setosa
## 46           4.8         3.0          1.4         0.3     setosa
## 47           5.1         3.8          1.6         0.2     setosa
## 48           4.6         3.2          1.4         0.2     setosa
## 49           5.3         3.7          1.5         0.2     setosa
## 50           5.0         3.3          1.4         0.2     setosa
## 51           7.0         3.2          4.7         1.4 versicolor
## 52           6.4         3.2          4.5         1.5 versicolor
## 53           6.9         3.1          4.9         1.5 versicolor
## 54           5.5         2.3          4.0         1.3 versicolor
## 55           6.5         2.8          4.6         1.5 versicolor
## 56           5.7         2.8          4.5         1.3 versicolor
## 57           6.3         3.3          4.7         1.6 versicolor
## 58           4.9         2.4          3.3         1.0 versicolor
## 59           6.6         2.9          4.6         1.3 versicolor
## 60           5.2         2.7          3.9         1.4 versicolor
## 61           5.0         2.0          3.5         1.0 versicolor
## 62           5.9         3.0          4.2         1.5 versicolor
## 63           6.0         2.2          4.0         1.0 versicolor
## 64           6.1         2.9          4.7         1.4 versicolor
## 65           5.6         2.9          3.6         1.3 versicolor
## 66           6.7         3.1          4.4         1.4 versicolor
## 67           5.6         3.0          4.5         1.5 versicolor
## 68           5.8         2.7          4.1         1.0 versicolor
## 69           6.2         2.2          4.5         1.5 versicolor
## 70           5.6         2.5          3.9         1.1 versicolor
## 71           5.9         3.2          4.8         1.8 versicolor
## 72           6.1         2.8          4.0         1.3 versicolor
## 73           6.3         2.5          4.9         1.5 versicolor
## 74           6.1         2.8          4.7         1.2 versicolor
## 75           6.4         2.9          4.3         1.3 versicolor
## 76           6.6         3.0          4.4         1.4 versicolor
## 77           6.8         2.8          4.8         1.4 versicolor
## 78           6.7         3.0          5.0         1.7 versicolor
## 79           6.0         2.9          4.5         1.5 versicolor
## 80           5.7         2.6          3.5         1.0 versicolor
## 81           5.5         2.4          3.8         1.1 versicolor
## 82           5.5         2.4          3.7         1.0 versicolor
## 83           5.8         2.7          3.9         1.2 versicolor
## 84           6.0         2.7          5.1         1.6 versicolor
## 85           5.4         3.0          4.5         1.5 versicolor
## 86           6.0         3.4          4.5         1.6 versicolor
## 87           6.7         3.1          4.7         1.5 versicolor
## 88           6.3         2.3          4.4         1.3 versicolor
## 89           5.6         3.0          4.1         1.3 versicolor
## 90           5.5         2.5          4.0         1.3 versicolor
## 91           5.5         2.6          4.4         1.2 versicolor
## 92           6.1         3.0          4.6         1.4 versicolor
## 93           5.8         2.6          4.0         1.2 versicolor
## 94           5.0         2.3          3.3         1.0 versicolor
## 95           5.6         2.7          4.2         1.3 versicolor
## 96           5.7         3.0          4.2         1.2 versicolor
## 97           5.7         2.9          4.2         1.3 versicolor
## 98           6.2         2.9          4.3         1.3 versicolor
## 99           5.1         2.5          3.0         1.1 versicolor
## 100          5.7         2.8          4.1         1.3 versicolor
## 101          6.3         3.3          6.0         2.5  virginica
## 102          5.8         2.7          5.1         1.9  virginica
## 103          7.1         3.0          5.9         2.1  virginica
## 104          6.3         2.9          5.6         1.8  virginica
## 105          6.5         3.0          5.8         2.2  virginica
## 106          7.6         3.0          6.6         2.1  virginica
## 107          4.9         2.5          4.5         1.7  virginica
## 108          7.3         2.9          6.3         1.8  virginica
## 109          6.7         2.5          5.8         1.8  virginica
## 110          7.2         3.6          6.1         2.5  virginica
## 111          6.5         3.2          5.1         2.0  virginica
## 112          6.4         2.7          5.3         1.9  virginica
## 113          6.8         3.0          5.5         2.1  virginica
## 114          5.7         2.5          5.0         2.0  virginica
## 115          5.8         2.8          5.1         2.4  virginica
## 116          6.4         3.2          5.3         2.3  virginica
## 117          6.5         3.0          5.5         1.8  virginica
## 118          7.7         3.8          6.7         2.2  virginica
## 119          7.7         2.6          6.9         2.3  virginica
## 120          6.0         2.2          5.0         1.5  virginica
## 121          6.9         3.2          5.7         2.3  virginica
## 122          5.6         2.8          4.9         2.0  virginica
## 123          7.7         2.8          6.7         2.0  virginica
## 124          6.3         2.7          4.9         1.8  virginica
## 125          6.7         3.3          5.7         2.1  virginica
## 126          7.2         3.2          6.0         1.8  virginica
## 127          6.2         2.8          4.8         1.8  virginica
## 128          6.1         3.0          4.9         1.8  virginica
## 129          6.4         2.8          5.6         2.1  virginica
## 130          7.2         3.0          5.8         1.6  virginica
## 131          7.4         2.8          6.1         1.9  virginica
## 132          7.9         3.8          6.4         2.0  virginica
## 133          6.4         2.8          5.6         2.2  virginica
## 134          6.3         2.8          5.1         1.5  virginica
## 135          6.1         2.6          5.6         1.4  virginica
## 136          7.7         3.0          6.1         2.3  virginica
## 137          6.3         3.4          5.6         2.4  virginica
## 138          6.4         3.1          5.5         1.8  virginica
## 139          6.0         3.0          4.8         1.8  virginica
## 140          6.9         3.1          5.4         2.1  virginica
## 141          6.7         3.1          5.6         2.4  virginica
## 142          6.9         3.1          5.1         2.3  virginica
## 143          5.8         2.7          5.1         1.9  virginica
## 144          6.8         3.2          5.9         2.3  virginica
## 145          6.7         3.3          5.7         2.5  virginica
## 146          6.7         3.0          5.2         2.3  virginica
## 147          6.3         2.5          5.0         1.9  virginica
## 148          6.5         3.0          5.2         2.0  virginica
## 149          6.2         3.4          5.4         2.3  virginica
## 150          5.9         3.0          5.1         1.8  virginica


Oops, that is a bit long… if you only want to show the first or last rows do the following:

head(iris) # first 6 rows
##   Sepal.Length Sepal.Width Petal.Length Petal.Width Species
## 1          5.1         3.5          1.4         0.2  setosa
## 2          4.9         3.0          1.4         0.2  setosa
## 3          4.7         3.2          1.3         0.2  setosa
## 4          4.6         3.1          1.5         0.2  setosa
## 5          5.0         3.6          1.4         0.2  setosa
## 6          5.4         3.9          1.7         0.4  setosa

tail(iris, 10) # last 10 rows
##     Sepal.Length Sepal.Width Petal.Length Petal.Width   Species
## 141          6.7         3.1          5.6         2.4 virginica
## 142          6.9         3.1          5.1         2.3 virginica
## 143          5.8         2.7          5.1         1.9 virginica
## 144          6.8         3.2          5.9         2.3 virginica
## 145          6.7         3.3          5.7         2.5 virginica
## 146          6.7         3.0          5.2         2.3 virginica
## 147          6.3         2.5          5.0         1.9 virginica
## 148          6.5         3.0          5.2         2.0 virginica
## 149          6.2         3.4          5.4         2.3 virginica
## 150          5.9         3.0          5.1         1.8 virginica


Iris is a so called data frame, the working horse of R and data science (you will see how to create one below):

str(iris)
## 'data.frame':    150 obs. of  5 variables:
##  $Sepal.Length: num 5.1 4.9 4.7 4.6 5 5.4 4.6 5 4.4 4.9 ... ##$ Sepal.Width : num  3.5 3 3.2 3.1 3.6 3.9 3.4 3.4 2.9 3.1 ...
##  $Petal.Length: num 1.4 1.4 1.3 1.5 1.4 1.7 1.4 1.5 1.4 1.5 ... ##$ Petal.Width : num  0.2 0.2 0.2 0.2 0.2 0.4 0.3 0.2 0.2 0.1 ...
##  $Species : Factor w/ 3 levels "setosa","versicolor",..: 1 1 1 1 1 1 1 1 1 1 ...  As you can see, data frames can combine different data types. If you try to do that with e.g. vectors, which can only hold one data type, something called coercion happens, i.e. at least one data type is forced to become another one so that consistency is maintained: str(c(2, "Hello")) # 2 is coerced to become a character string too ## chr [1:2] "2" "Hello"  You can get a fast overview of your data like so: summary(iris[1:4]) ## Sepal.Length Sepal.Width Petal.Length Petal.Width ## Min. :4.300 Min. :2.000 Min. :1.000 Min. :0.100 ## 1st Qu.:5.100 1st Qu.:2.800 1st Qu.:1.600 1st Qu.:0.300 ## Median :5.800 Median :3.000 Median :4.350 Median :1.300 ## Mean :5.843 Mean :3.057 Mean :3.758 Mean :1.199 ## 3rd Qu.:6.400 3rd Qu.:3.300 3rd Qu.:5.100 3rd Qu.:1.800 ## Max. :7.900 Max. :4.400 Max. :6.900 Max. :2.500 boxplot(iris[1:4])  As you have seen, R often runs a function on all of the data simultaneously. This feature is called vectorization and in many other languages you would need a loop for that. In R you don’t use loops that often, but of course they are available: for (i in seq(5)) { print(1:i) } ## [1] 1 ## [1] 1 2 ## [1] 1 2 3 ## [1] 1 2 3 4 ## [1] 1 2 3 4 5  Speaking of control structures: of course conditional statements are available too: even <- function(x) ifelse(x %% 2 == 0, TRUE, FALSE) # %% gives remainder of division (= modulo operator) even(1:5) ## [1] FALSE TRUE FALSE TRUE FALSE  Linear modelling (e.g. correlation and linear regression) couldn’t be any easier, it is included in the core language: age <- c(21, 46, 55, 35, 28) income <- c(1850, 2500, 2560, 2230, 1800) df <- data.frame(age, income) # create a data frame df ## age income ## 1 21 1850 ## 2 46 2500 ## 3 55 2560 ## 4 35 2230 ## 5 28 1800 cor(df) # correlation ## age income ## age 1.0000000 0.9464183 ## income 0.9464183 1.0000000 LinReg <- lm(income ~ age, data = df) # income as a linear model of age LinReg ## ## Call: ## lm(formula = income ~ age, data = df) ## ## Coefficients: ## (Intercept) age ## 1279.37 24.56 summary(LinReg) ## ## Call: ## lm(formula = income ~ age, data = df) ## ## Residuals: ## 1 2 3 4 5 ## 54.92 90.98 -70.04 91.12 -166.98 ## ## Coefficients: ## Estimate Std. Error t value Pr(>|t|) ## (Intercept) 1279.367 188.510 6.787 0.00654 ** ## age 24.558 4.838 5.076 0.01477 * ## --- ## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 ## ## Residual standard error: 132.1 on 3 degrees of freedom ## Multiple R-squared: 0.8957, Adjusted R-squared: 0.8609 ## F-statistic: 25.77 on 1 and 3 DF, p-value: 0.01477 plot(df, pch = 16, main = "Linear model") abline(LinReg, col = "blue", lwd = 2) # adding the regression line  You could directly use the model to make predictions: pred_LinReg <- predict(LinReg, data.frame(age = seq(15, 70, 5))) names(pred_LinReg) <- seq(15, 70, 5) round(pred_LinReg, 2) ## 15 20 25 30 35 40 45 50 55 ## 1647.73 1770.52 1893.31 2016.10 2138.88 2261.67 2384.46 2507.25 2630.04 ## 60 65 70 ## 2752.83 2875.61 2998.40  If you want to know more about the modelling process you can find it here: Learning Data Science: Modelling Basics Another strength of R is the huge number of add-on packages for all kinds of specialized tasks. For the grand finale of this introduction, we’re gonna get a little taste of machine learning. For that matter we install the OneR package from CRAN (the official package repository of R): Tools -> Install packages… -> type in “OneR” -> click “Install”. After that we build a simple model on the iris dataset to predict the Species column: library(OneR) # load package data <- optbin(Species ~., data = iris) # find optimal bins for numeric predictors model <- OneR(data, verbose = TRUE) # build actual model ## ## Attribute Accuracy ## 1 * Petal.Width 96% ## 2 Petal.Length 95.33% ## 3 Sepal.Length 74.67% ## 4 Sepal.Width 55.33% ## --- ## Chosen attribute due to accuracy ## and ties method (if applicable): '*' summary(model) # show rules ## ## Call: ## OneR.data.frame(x = data, verbose = TRUE) ## ## Rules: ## If Petal.Width = (0.0976,0.791] then Species = setosa ## If Petal.Width = (0.791,1.63] then Species = versicolor ## If Petal.Width = (1.63,2.5] then Species = virginica ## ## Accuracy: ## 144 of 150 instances classified correctly (96%) ## ## Contingency table: ## Petal.Width ## Species (0.0976,0.791] (0.791,1.63] (1.63,2.5] Sum ## setosa * 50 0 0 50 ## versicolor 0 * 48 2 50 ## virginica 0 4 * 46 50 ## Sum 50 52 48 150 ## --- ## Maximum in each column: '*' ## ## Pearson's Chi-squared test: ## X-squared = 266.35, df = 4, p-value < 2.2e-16 plot(model)  We’ll now see how well the model is doing: prediction <- predict(model, data) eval_model(prediction, data) ## ## Confusion matrix (absolute): ## Actual ## Prediction setosa versicolor virginica Sum ## setosa 50 0 0 50 ## versicolor 0 48 4 52 ## virginica 0 2 46 48 ## Sum 50 50 50 150 ## ## Confusion matrix (relative): ## Actual ## Prediction setosa versicolor virginica Sum ## setosa 0.33 0.00 0.00 0.33 ## versicolor 0.00 0.32 0.03 0.35 ## virginica 0.00 0.01 0.31 0.32 ## Sum 0.33 0.33 0.33 1.00 ## ## Accuracy: ## 0.96 (144/150) ## ## Error rate: ## 0.04 (6/150) ## ## Error rate reduction (vs. base rate): ## 0.94 (p-value < 2.2e-16)  96% accuracy is not too bad, even for this simple dataset! If you want to know more about the OneR package you can read the vignette: OneR – Establishing a New Baseline for Machine Learning Classification Models. Well, and that’s it for the ultimate introduction to R – hopefully you liked it and you learned something! Please share your first experiences with R in the comments and also if you miss something (I might add it in the future!) – Thank you for reading and stay tuned for more to come! ## Was the Bavarian Abitur too hard this time? Bavaria is known for its famous Oktoberfest… and within Germany also for its presumably difficult Abitur, a qualification granted by university-preparatory schools in Germany. A mandatory part for all students is maths. This year many students protested that the maths part was way too hard, they even started an online petition with more than seventy thousand supporters at this time of writing! It is not clear yet whether their marks will be adjusted upwards, the ministry of education is investigating the case. As a professor in Bavaria who also teaches statistics I will take the opportunity to share with you an actual question from the original examination with solution, so read on… Let us have a look at the first (and easiest) question in the stochastics part: Every sixth visitor to the Oktoberfest wears a gingerbread heart around his neck. During the festival 25 visitors are chosen at random. Determine the probability that at most one of the selected visitors will have a gingerbread heart. Before you read on try to solve the task yourself… Of course students are not allowed to use R in the examination but in general the good thing about this kind of questions is that if you have no idea how to solve them analytically solving them by simulation is often much easier: set.seed(12) N <- 1e7 v <- matrix(sample(c(0L, 1L), size = 25 * N, replace = TRUE, prob = c(5/6, 1/6)), ncol = 25) sum(rowSums(v) <= 1) / N ## [1] 0.062936  The answer is about . Now for the analytical solution: “At least one” implies that we have to differentiate between two cases, “no gingerbread heart” and “exactly one gingerbread heart”. “No gingerbread heart” is just . “Exactly one gingerbread heart” is because there are possibilities of where the gingerbread heart could occur. We have to add both probabilities: (5/6)^25 + 25*1/6*(5/6)^24 ## [1] 0.06289558  If you know a little bit about probability distributions you will recognize the above as the binomial distribution: pbinom(q = 1, size = 25, prob = 1/6) ## [1] 0.06289558  Of course it is a little unfair to judge just on basis of the easiest task and without knowing the general maths level that is required. But still, I would like to hear your opinion on this. Also outsiders’ views from different countries and different educational systems are very welcome! So, what do you think: Was the Bavarian Abitur too hard this time? Please leave your reply in the comment section below! ## Backtest Trading Strategies like a real Quant R is one of the best choices when it comes to quantitative finance. Here we will show you how to load financial data, plot charts and give you a step-by-step template to backtest trading strategies. So, read on… We begin by just plotting a chart of the Standard & Poor’s 500 (S&P 500), an index of the 500 biggest companies in the US. To get the index data and plot the chart we use the powerful quantmod package (on CRAN). After that we add two popular indicators, the simple moving average (SMI) and the exponential moving average (EMA). Have a look at the code: library(quantmod) ## Loading required package: xts ## Loading required package: zoo ## ## Attaching package: 'zoo' ## The following objects are masked from 'package:base': ## ## as.Date, as.Date.numeric ## Loading required package: TTR ## Version 0.4-0 included new data defaults. See ?getSymbols. getSymbols("^GSPC", from = "2000-01-01") ## 'getSymbols' currently uses auto.assign=TRUE by default, but will ## use auto.assign=FALSE in 0.5-0. You will still be able to use ## 'loadSymbols' to automatically load data. getOption("getSymbols.env") ## and getOption("getSymbols.auto.assign") will still be checked for ## alternate defaults. ## ## This message is shown once per session and may be disabled by setting ## options("getSymbols.warning4.0"=FALSE). See ?getSymbols for details. ## [1] "^GSPC" head(GSPC) ## GSPC.Open GSPC.High GSPC.Low GSPC.Close GSPC.Volume ## 2000-01-03 1469.25 1478.00 1438.36 1455.22 931800000 ## 2000-01-04 1455.22 1455.22 1397.43 1399.42 1009000000 ## 2000-01-05 1399.42 1413.27 1377.68 1402.11 1085500000 ## 2000-01-06 1402.11 1411.90 1392.10 1403.45 1092300000 ## 2000-01-07 1403.45 1441.47 1400.73 1441.47 1225200000 ## 2000-01-10 1441.47 1464.36 1441.47 1457.60 1064800000 ## GSPC.Adjusted ## 2000-01-03 1455.22 ## 2000-01-04 1399.42 ## 2000-01-05 1402.11 ## 2000-01-06 1403.45 ## 2000-01-07 1441.47 ## 2000-01-10 1457.60 tail(GSPC) ## GSPC.Open GSPC.High GSPC.Low GSPC.Close GSPC.Volume ## 2019-04-24 2934.00 2936.83 2926.05 2927.25 3448960000 ## 2019-04-25 2928.99 2933.10 2912.84 2926.17 3425280000 ## 2019-04-26 2925.81 2939.88 2917.56 2939.88 3248500000 ## 2019-04-29 2940.58 2949.52 2939.35 2943.03 3118780000 ## 2019-04-30 2937.14 2948.22 2924.11 2945.83 3919330000 ## 2019-05-01 2952.33 2954.13 2923.36 2923.73 3645850000 ## GSPC.Adjusted ## 2019-04-24 2927.25 ## 2019-04-25 2926.17 ## 2019-04-26 2939.88 ## 2019-04-29 2943.03 ## 2019-04-30 2945.83 ## 2019-05-01 2923.73 chartSeries(GSPC, theme = chartTheme("white"), subset = "last 10 months", show.grid = TRUE)  addSMA(20)  addEMA(20)  As you can see the moving averages are basically smoothed out versions of the original data shifted by the given number of days. While with the SMA (red curve) all days are weighted equally with the EMA (blue curve) the more recent days are weighted stronger, so that the resulting indicator detects changes quicker. The idea is that by using those indicators investors might be able to detect longer term trends and act accordingly. For example a trading rule could be to buy the index whenever it crosses the MA from below and sell when it goes the other direction. Judge for yourself if this could have worked. Well, having said that it might not be that easy to find out the profitability of certain trading rules just by staring at a chart. We are looking for something more systematic! We would need a decent backtest! This can of course also be done with R, a great choice is the PerformanceAnalytics package (on CRAN). To backtest a trading strategy I provide you with a step-by-step template: 1. Load libraries and data 2. Create your indicator 3. Use indicator to create equity curve 4. Evaluate strategy performance As an example we want to test the idea that it might be profitable to sell the index when the financial markets exhibit significant stress. Interestingly enough “stress” can be measured by certain indicators that are freely available. One of them is the National Financial Conditions Index (NFCI) of the Federal Reserve Bank of Chicago (https://www.chicagofed.org/publications/nfci/index): The Chicago Fed’s National Financial Conditions Index (NFCI) provides a comprehensive weekly update on U.S. financial conditions in money markets, debt and equity markets and the traditional and “shadow” banking systems. […] The NFCI [is] constructed to have an average value of zero and a standard deviation of one over a sample period extending back to 1971. Positive values of the NFCI have been historically associated with tighter-than-average financial conditions, while negative values have been historically associated with looser-than-average financial conditions. To make it more concrete we want to create a buy signal when the index is above one standard deviation in negative territory and a sell signal otherwise. Have a look at the following code: # Step 1: Load libraries and data library(quantmod) library(PerformanceAnalytics) ## ## Attaching package: 'PerformanceAnalytics' ## The following object is masked from 'package:graphics': ## ## legend getSymbols('NFCI', src = 'FRED', , from = '2000-01-01') ## [1] "NFCI" NFCI <- na.omit(lag(NFCI)) # we can only act on the signal after release, i.e. the next day getSymbols("^GSPC", from = '2000-01-01') ## [1] "^GSPC" data <- na.omit(merge(NFCI, GSPC)) # merge before (!) calculating returns) data$GSPC <- na.omit(ROC(Cl(GSPC))) # calculate returns of closing prices

# Step 2: Create your indicator
data$sig <- ifelse(data$NFCI < 1, 1, 0)
data$sig <- na.locf(data$sig)

# Step 3: Use indicator to create equity curve
perf <- na.omit(merge(data$sig * data$GSPC, data$GSPC)) colnames(perf) <- c("Stress-based strategy", "SP500") # Step 4: Evaluate strategy performance table.DownsideRisk(perf) ## Stress-based strategy SP500 ## Semi Deviation 0.0075 0.0087 ## Gain Deviation 0.0071 0.0085 ## Loss Deviation 0.0079 0.0095 ## Downside Deviation (MAR=210%) 0.0125 0.0135 ## Downside Deviation (Rf=0%) 0.0074 0.0087 ## Downside Deviation (0%) 0.0074 0.0087 ## Maximum Drawdown 0.5243 0.6433 ## Historical VaR (95%) -0.0173 -0.0188 ## Historical ES (95%) -0.0250 -0.0293 ## Modified VaR (95%) -0.0166 -0.0182 ## Modified ES (95%) -0.0268 -0.0311 table.Stats(perf) ## Stress-based strategy SP500 ## Observations 4858.0000 4858.0000 ## NAs 0.0000 0.0000 ## Minimum -0.0690 -0.0947 ## Quartile 1 -0.0042 -0.0048 ## Median 0.0003 0.0005 ## Arithmetic Mean 0.0002 0.0002 ## Geometric Mean 0.0002 0.0001 ## Quartile 3 0.0053 0.0057 ## Maximum 0.0557 0.1096 ## SE Mean 0.0001 0.0002 ## LCL Mean (0.95) -0.0001 -0.0002 ## UCL Mean (0.95) 0.0005 0.0005 ## Variance 0.0001 0.0001 ## Stdev 0.0103 0.0120 ## Skewness -0.1881 -0.2144 ## Kurtosis 3.4430 8.5837 charts.PerformanceSummary(perf)  chart.RelativePerformance(perf[ , 1], perf[ , 2])  chart.RiskReturnScatter(perf)  The first chart shows that the stress-based strategy (black curve) clearly outperformed its benchmark, the S&P 500 (red curve). This can also be seen in the second chart, showing the relative performance. In the third chart we see that both return (more) and (!) risk (less) of our backtested strategy are more favourable compared to the benchmark. So, all in all this seems to be a viable strategy! But of course before investing real money many more tests have to be performed! You can use this framework for backtesting your own ideas. Here is not the place to explain all of the above tables and plots but as you can see both packages are very, very powerful and I have only shown you a small fraction of their capabilities. To use their full potential you should have a look at the extensive documentation that comes with it on CRAN. Disclaimer: This is no investment advice! No responsibility is taken whatsoever if you lose money! If you gain money though I would be happy if you could buy me a coffee… that is not too much to ask, is it? 😉 ## The Rich didn’t earn their Wealth, they just got Lucky Tomorrow, on the First of May, many countries celebrate the so called International Workers’ Day (or Labour Day): time to talk about the unequal distribution of wealth again! A few months ago I posted a piece with the title “If wealth had anything to do with intelligence…” where I argued that ability, e.g. intelligence, as an input has nothing to do with wealth as an output. It drew a lot of criticism (as expected), most of it unfounded in my opinion but one piece merits some discussion: the fact that the intelligence quotient (IQ) is normally distributed by construction. The argument goes that intelligence per se may be a distribution with fat tails too but by the way the IQ is constructed the metric is being transformed into a well formed gaussian distribution. To a degree this is certainly true, yet I would still argue that the distribution of intelligence and all other human abilities are far more well behaved than the extremely unequal distribution of wealth. I wrote in a comment: There are many aspects in your comment that are certainly true. Obviously there are huge problems in measuring “true” mental abilities, which is the exact reason why people came up with a somewhat artificial “intelligence quotient” with all its shortcomings. What would be interesting to see is (and I don’t know if you perhaps have a source about this) what the outcome of an intelligence test would look like without the “quotient” part, i.e. without subsequently normalizing the results. I guess the relationship wouldn’t be strictly linear but it wouldn’t be as extreme as the wealth distribution either. What I think is true in any case, independent of the distributions, is when you rank all people by intelligence and by wealth respectively you wouldn’t really see any stable connection – and that spirit was the intention of my post in the first place and I still stand by it, although some of the technicalities are obviously debatable. So, if you have a source, Dear Reader, you are more than welcome to share it in the comments – I am always eager to learn! I ended my post with: But if it is not ability/intelligence that determines the distribution of wealth what else could account for the extreme inequality we perceive in the world? In this post I will argue that luck is a good candidate, so read on… In 2014 there was a special issue of the renowned magazine Science titled “The science of inequality”. In one of the articles (Cho, A.: “Physicists say it’s simple”) the following thought experiment is being proposed: Suppose you randomly divide 500 million in income among 10,000 people. There’s only one way to give everyone an equal, 50,000 share. So if you’re doling out earnings randomly, equality is extremely unlikely. But there are countless ways to give a few people a lot of cash and many people a little or nothing. In fact, given all the ways you could divvy out income, most of them produce an exponential distribution of income. So, the basic idea is to randomly throw 9,999 darts at a scale ranging from zero to 500 million and study the resulting distribution of intervals: library(MASS) w <- 5e8 # wealth p <- 1e4 # no. of people set.seed(123) d <- diff(c(0, sort(runif(p-1, max = w)), w)) # wealth distribution h <- hist(d, col = "red", main = "Exponential decline", freq = FALSE, breaks = 45, xlim = c(0, quantile(d, 0.99))) fit <- fitdistr(d, "exponential") curve(dexp(x, rate = fit$estimate), col = "black", type = "p", pch = 16, add = TRUE)


The resulting distribution fits an exponential distribution very well. You can read some interesting discussions concerning this result on CrossValidated StackExchange: How can I analytically prove that randomly dividing an amount results in an exponential distribution (of e.g. income and wealth)?

Just to give you an idea of how unfair this distribution is: the richest six persons have more wealth than the poorest ten percent together:

sum(sort(d)[9994:10000]) - sum(sort(d)[0:1000])
## [1] 183670.8


If you think that this is ridiculous just look at the real global wealth distribution: here it is not six but three persons who own more than the poorest ten percent!

Now, what does that mean? Well, equality seems to be the exception and (extreme) inequality the rule. The intervals were found randomly, no interval had any special skills, just luck – and the result is (extreme) inequality – as in the real world!

If you can reproduce the wealth distribution of a society stochastically this could have the implication that it weren’t so much the extraordinary skills of the rich which made them rich but they just got lucky.

Some rich people are decent enough to admit this. In his impressive essay “Why Poverty Is Like a Disease” Christian H. Cooper, a hillbilly turned investment banker writes:

So how did I get out? By chance.

It’s easy to attach a post-facto narrative of talent and hard work to my story, because that’s what we’re fed by everything from Hollywood to political stump speeches. But it’s the wrong story. My escape was made up of a series of incredibly unlikely events, none of which I had real control over.

[…]

I am the exception that proves the rule—but that rule is that escape from poverty is a matter of chance, and not a matter of merit.

A consequence would be that you cannot really learn much from the rich. So throw away all of your self help books on how to become successful. I will end with a cartoon, which brings home this message, on a closely related concept, the so called survivorship bias (which is also important to keep in mind when backtesting trading strategies in quantitative finance, the topic of an upcoming post… so stay tuned!):

## Google’s Eigenvector… or how a Random Surfer finds the most relevant Webpages

Like most people you will have used a search engine lately, like Google. But have you ever thought about how it manages to give you the most fitting results? How does it order the results so that the best are on top? Read on to find out!

The earliest search engines either had human curated indices, like Yahoo! or used some simple heuristic like the more often the keyword you were looking for was mentioned on a page the better, like Altavista – which led to all kinds of crazy effects like certain keywords being repeated thousands of times on webpages to make them more “relevant”.

Now, most of those search engines are long gone because a new kid arrived on the block: Google! Google’s search engine results were much better than all of the competition and they became the dominant player in no time. How did they do that?

The big idea was in fact published by the two founders: Sergey Brin and Lawrence Page, it is called the pagerank algorithm (which is of course a pun because one of the authors was named Page too). The original paper can be found here: S. Brin, L. Page: The Anatomy of a Large-Scale Hypertextual Web Search Engine.

Let us start with another, related question: which properties are the best to own in Monopoly? Many would instinctively answer with the most expensive ones, i.e. Park Place and Boardwalk. But a second thought reveals that those might be the the ones where you get the biggest rent if somebody lands on them but that the last part is the caveat… “IF” somebody lands on them! The best streets are actually the ones where players land on the most. Those happen to be the orange streets, St. James Place, Tennessee Avenue and New York Avenue and therefore they are the key to winning the game.

How do find those properties? For example by simulation: you just simulate thousands of dice rolls and see where the players land.

A similar idea holds true for finding the best web pages: you just start from a random position and simulate a surfer who visits different web pages by chance. For each surfing session you tally the respective webpage where she ends up and after many runs we get a percentage for each page. The higher this percentage is the more relevant the webpage!

Let us do this with some R code. First we define a very small net and plot it (the actual example can be found in chapter 30 of the very good book “Chaotic Fishponds and Mirror Universes” by Richard Elwes):

library(igraph)
##
## Attaching package: 'igraph'
## The following objects are masked from 'package:stats':
##
##     decompose, spectrum
## The following object is masked from 'package:base':
##
##     union

M <- matrix(c(0, 0, 1, 1,
1, 0, 0, 0,
1, 0, 0, 0,
1, 1, 1, 0), nrow = 4)
colnames(M) <- rownames(M) <- c("A", "B", "C", "D")
M
##   A B C D
## A 0 1 1 1
## B 0 0 0 1
## C 1 0 0 1
## D 1 0 0 0

g <- graph_from_adjacency_matrix(t(M)) # careful with how the adjacency matrix is defined -> transpose of matrix
plot(g)


Now, we are running the actual simulation. We define two helper functions for that, next_page for getting a random but possible next page given the page our surfer is on at the moment and last_page which gives the final page after N clicks:

next_page <- function(page, graph) {
l <- sample(rownames(graph)[as.logical(graph[ , as.logical(page)])], 1)
as.numeric(rownames(graph) == l)
}

last_page <- function(page, graph, N = 100) {
for (i in seq(N)) {
page <- next_page(page, graph)
}
page
}

current_page <- c(1, 0, 0, 0) # random surfer starting from A
random_surfer <- replicate(2e4, last_page(current_page, M, 50))
round(rowSums(random_surfer) / sum(random_surfer), 2)
## [1] 0.43 0.07 0.28 0.22


So we see that page A is the most relevant one because our surfer ends up being there in more than 40% of all sessions, after that come the pages C, D and B. When you look at the net that makes sense, since all pages refer to A whereas B gets only one link, so it doesn’t seem to be that relevant.

As you have seen the simulation even for this small net took quite long so we need some clever mathematics to speed up the process. One idea is to transform our matrix which represents the network into a matrix which gives the probabilities of landing on the next pages and then multiply the probability matrix with the current position (and thereby transform the probabilities). Let us do this for the first step:

M_prob <- prop.table(M, 2) # create probability matrix
M_prob
##     A B C         D
## A 0.0 1 1 0.3333333
## B 0.0 0 0 0.3333333
## C 0.5 0 0 0.3333333
## D 0.5 0 0 0.0000000

M_prob %*% current_page
##   [,1]
## A  0.0
## B  0.0
## C  0.5
## D  0.5


The result says that there is a fifty-fifty chance of landing on C or D. When you look at the graph you see that this is correct since there are two links, one to C and one to D! For the next step you would have to multiply the matrix with the result again, or first multiply the matrix with itself before multiplying with the current position, which gives:

If we want to do this a hundred times we just have to raise this probability matrix to the one hundredth power:

We use the %^% operator in the expm package (on CRAN) for that:

library(expm)
##
## Attaching package: 'expm'
## The following object is masked from 'package:Matrix':
##
##     expm

r <- M_prob %^% 100 %*% current_page
r
##         [,1]
## A 0.42857143
## B 0.07142857
## C 0.28571429
## D 0.21428571


Again, we get the same result! You might ask: why ? The answer is that this is in most cases enough to get a stable result so that any further multiplication still results in the same result:

The last equations opens up still another possibility: we are obviously looking for a vector which goes unaffected when multiplied by the matrix . There is a mathematical name for that kind of behaviour: eigenvector! As you might have guessed the name is an import from the German language where it means something like “own vector”.

This hints at the problem we were solving all along (without consciously realizing perhaps): a page is the more relevant the more relevant a page is that links to it… now we have to know the importance of that page but that page two is the more relevant… and so on and so forth, we are going in circles here. The same is true when you look at the equation above: you define in terms of is the eigenvector of matrix !

There are very fast and powerful methods to find the eigenvectors of a matrix, and the corresponding eigen function is even a function in base R:

lr <- Re(eigen(M_prob)$vectors[ , 1]) # real parts of biggest eigenvector lr / sum(lr) # normalization ## [1] 0.42857143 0.07142857 0.28571429 0.21428571  Again, the same result! You can now understand the title of this post and titles of other articles about the pagerank algorithm and Google like “The$25,000,000,000 eigenvector”.

Yet, a word of warning is in order: there are cases where the probability matrix is not diagonalizable (we won’t get into the mathematical details here), which means that the eigenvector method won’t give sensible results. To check this the following code must evaluate to TRUE:

ev <- eigen(M_prob)$values length(unique(ev)) == length(ev) ## [1] TRUE  We now repeat the last two methods for a bigger network: set.seed(1415) n <- 10 g <- sample_gnp(n, p = 1/4, directed = TRUE) # create random graph g <- set_vertex_attr(g, "name", value = LETTERS[1:n]) plot(g)  M <- t(as_adjacency_matrix(g, sparse = FALSE)) M_prob <- prop.table(M, 2) # create probability matrix M_prob ## A B C D E F G H I J ## A 0.00 0 0 1 0.5 0.5 0.5 0.0000000 0.0000000 0.5 ## B 0.00 0 0 0 0.0 0.0 0.0 0.3333333 0.0000000 0.0 ## C 0.00 1 0 0 0.0 0.0 0.0 0.0000000 0.3333333 0.5 ## D 0.25 0 0 0 0.0 0.0 0.0 0.0000000 0.0000000 0.0 ## E 0.25 0 0 0 0.0 0.0 0.5 0.3333333 0.3333333 0.0 ## F 0.00 0 1 0 0.0 0.0 0.0 0.0000000 0.3333333 0.0 ## G 0.25 0 0 0 0.0 0.0 0.0 0.0000000 0.0000000 0.0 ## H 0.00 0 0 0 0.5 0.0 0.0 0.0000000 0.0000000 0.0 ## I 0.00 0 0 0 0.0 0.5 0.0 0.0000000 0.0000000 0.0 ## J 0.25 0 0 0 0.0 0.0 0.0 0.3333333 0.0000000 0.0 current_page <- c(1, rep(0, n-1)) r <- M_prob %^% 100 %*% current_page r ## [,1] ## A 0.27663574 ## B 0.02429905 ## C 0.08878509 ## D 0.06915881 ## E 0.14579434 ## F 0.10654199 ## G 0.06915881 ## H 0.07289723 ## I 0.05327107 ## J 0.09345787 lr <- Re(eigen(M_prob)$vectors[ , 1])
lr / sum(lr) # normalization of the real parts
##  [1] 0.27663551 0.02429907 0.08878505 0.06915888 0.14579439 0.10654206
##  [7] 0.06915888 0.07289720 0.05327103 0.09345794


We can now order the pages according to their importance – like the first 10 results of a google search:

search <- data.frame(Page = LETTERS[1:n], Rank = r)
search[order(search$Rank, decreasing = TRUE), ] ## Page Rank ## A A 0.27663574 ## E E 0.14579434 ## F F 0.10654199 ## J J 0.09345787 ## C C 0.08878509 ## H H 0.07289723 ## D D 0.06915881 ## G G 0.06915881 ## I I 0.05327107 ## B B 0.02429905  Looking at the net, does the resulting order make sense to you? Congratulations, you now understand the big idea behind one the greatest revolutions in information technology! ## Base Rate Fallacy – or why No One is justified to believe that Jesus rose In this post we are talking about one of the most unintuitive results in statistics: the so called false positive paradox which is an example of the so called base rate fallacy. It describes a situation where a positive test result of a very sensitive medical test shows that you have the respective disease… yet you are most probably healthy! The reason for this is that the disease itself is so rare that even with a very sensitive test the result is most probably false positive: it shows that you have the disease yet this result is false, you are healthy. The key to understanding this result is to understand the difference between two conditional probabilities: the probability that you have a positive test result when you are sick and the probability that you are sick in case you got a positive test result – you are interested in the last (am I really sick?) but you only know the first. Now for some notation (the vertical dash means “under the condition that”, P stands for probability): • : if you are sick () you will probably have a positive test result () – this is what we know • : if you have a positive test result () you are probably not sick () – this is what we want to know To calculate one conditional probability from the other we use the famous Bayes’ theorem: In the following example we assume a disease with an infection rate of 1 in 1000 and a test to detect this disease with a sensitivity of 99%. Have a look at the following code which illustrates the situation with Euler diagrams, first the big picture, then a zoomed-in version: library(eulerr) A <- 0.001 # prevalence of disease BlA <- 0.99 # sensitivity of test B <- A * BlA + (1 - A) * (1 - BlA) # positive test (specificity same as sensitivity) AnB <- BlA * A AlB <- BlA * A / B # Bayes's theorem #AnB / B # Bayes's theorem in different form C <- 1 # the whole population main <- paste0("P(B|A) = ", round(BlA, 2), ", but P(A|B) = ", round(AlB, 2)) set.seed(123) fit1 <- euler(c("A" = A, "B" = B, "C" = C, "A&B" = AnB, "A&C" = A, "B&C" = B, "A&B&C" = AnB), input = "union") plot(fit1, main = main, fill = c("red", "green", "gray90"))  fit2 <- euler(c("A" = A, "B" = B, "A&B" = AnB), input = "union") plot(fit2, main = main, fill = c("red", "green"))  As you can see although this test is very sensitive when you get a positive test result the probability of you being infected is only 9%! In the diagrams C is the whole population and A are the infected individuals. B shows the people with a positive test result and you can see in the second diagram that almost all of the infected A are also part of B (the brown area = true positive), but still most ob B are outside of A (the green area), so although they are not infected they have a positive test result! They are false positive. The red area shows the people that are infected (A) but get a negative test result, stating that they are healthy. This is called false negative. The grey area shows the people who are healthy and get a negative test result, they are true negative. Due to the occasion we are now coming to an even more extreme example: did Jesus rise from the dead? It is inspired by the very good essay “A drop in the sea”: Don’t believe in miracles until you’ve done the math. Let us assume that we had very, very reliable witnesses (as a side note what is strange though is that the gospels cannot even agree on how many men or angels appeared at the tomb: it is one angel in Matthew, a young man in Mark, two men in Luke and two angels in John… but anyway), yet the big problem is that not many people so far have been able to defy death. I have only heard of two cases: supposedly the King of Kings (Jesus) but also of course the King himself (Elvis!), whereby sightings of Elvis after his death are much more numerous than of Jesus (just saying… 😉 ) Have a look at the following code (source for the number of people who have ever lived: WolframAlpha) A <- 2/108500000000 # probability of coming back from the dead (The King = Elvis and the King of Kings = Jesus) BlA <- 0.9999999 # sensitivity of test -> very, very reliable witnesses (many more in case of Elvis 😉 B <- A * BlA + (1 - A) * (1 - BlA) # positive test = witnesses say He rose AnB <- BlA * A AlB <- BlA * A / B # Bayes's theorem C <- 1 # all people main <- paste0("P(B|A) = ", round(BlA, 2), ", but P(A|B) = ", round(AlB, 2)) fit1 <- euler(c("A" = A, "B" = B, "C" = C, "A&B" = AnB, "A&C" = A, "B&C" = B, "A&B&C" = AnB), input = "union") plot(fit1, main = main, fill = c("red", "green", "gray90"))  fit2 <- euler(c("A" = A, "B" = B, "A&B" = AnB), input = "union") plot(fit2, main = main, fill = c("red", "green"))  So, in this case C is the unfortunate group of people who have to go for good… it is us. 🙁 As you can see although the witnesses are super reliable when they claim that somebody rose it is almost certain that they are wrong: • : if Jesus rose () the very, very reliable witnesses would with a very high probability say so () • : if the very, very reliable witnesses said that Jesus rose () Jesus would still almost surely have stayed dead Or in the words of the above mentioned essay: No one is justified in believing in Jesus’s resurrection. The numbers simply don’t justify the conclusion. But this chimes well with a famous Christian saying “I believe because it is absurd” (or in Latin “Credo quia absurdum”) – you can find out more about that in another highly interesting essay: ‘I believe because it is absurd’: Christianity’s first meme Unfortunately this devastating conclusion is also true in the case of Elvis… ## Separating the Signal from the Noise: Robust Statistics for Pedestrians One of the problems of navigating an autonomous car through a city is to extract robust signals in the face of all the noise that is present in the different sensors. Just taking something like an arithmetic mean of all the data points could possibly end in a catastrophe: if a part of a wall looks similar to the street and the algorithm calculates an average trajectory of the two this would end in leaving the road and possibly crashing into pedestrians. So we need some robust algorithm to get rid of the noise. The area of statistics that especially deals with such problems is called robust statistics and the methods used therein robust estimation. Now, one of the problems is that one doesn’t know what is signal and what is noise. The big idea behind RANSAC (short for RAndom SAmple Consensus) is to get rid of outliers by basically taking as many points as possible which form a well-defined region and leaving out the others. It does that iteratively, similar to the famous k-means algorithm (the topic of one of the upcoming posts, so stay tuned…). To really understand how RANSAC works we will now build it with R. We will take a simple linear regression as an example and make it robust against outliers. For educational purposes we will do this step by step: 1. Understanding the general outline of the algorithm. 2. Looking at the steps in more detail. 3. Expressing the steps in pseudocode. 4. Translating this into R! Conveniently enough Wikipedia gives a good outline of the algorithm and even provides us with the very clear pseudocode which will serve as the basis for our own R implementation (the given Matlab code is not very good in my opinion and has nothing to do with the pseudocode): The RANSAC algorithm is essentially composed of two steps that are iteratively repeated: 1. In the first step, a sample subset containing minimal data items is randomly selected from the input dataset. A fitting model and the corresponding model parameters are computed using only the elements of this sample subset. The cardinality of the sample subset is the smallest sufficient to determine the model parameters. 2. In the second step, the algorithm checks which elements of the entire dataset are consistent with the model instantiated by the estimated model parameters obtained from the first step. A data element will be considered as an outlier if it does not fit the fitting model instantiated by the set of estimated model parameters within some error threshold that defines the maximum deviation attributable to the effect of noise. The set of inliers obtained for the fitting model is called consensus set. The RANSAC algorithm will iteratively repeat the above two steps until the obtained consensus set in certain iteration has enough inliers. In more detail: RANSAC achieves its goal by repeating the following steps: 1. Select a random subset of the original data. Call this subset the hypothetical inliers. 2. A model is fitted to the set of hypothetical inliers. 3. All other data are then tested against the fitted model. Those points that fit the estimated model well, according to some model-specific loss function, are considered as part of the consensus set. 4. The estimated model is reasonably good if sufficiently many points have been classified as part of the consensus set. 5. Afterwards, the model may be improved by reestimating it using all members of the consensus set. This procedure is repeated a fixed number of times, each time producing either a model which is rejected because too few points are part of the consensus set, or a refined model together with a corresponding consensus set size. In the latter case, we keep the refined model if its consensus set is larger than the previously saved model. Now, this can be expressed in pseudocode: Given: data - a set of observed data points model - a model that can be fitted to data points n - minimum number of data points required to fit the model k - maximum number of iterations allowed in the algorithm t - threshold value to determine when a data point fits a model d - number of close data points required to assert that a model fits well to data Return: bestfit - model parameters which best fit the data (or nul if no good model is found) iterations = 0 bestfit = nul besterr = something really large while iterations < k { maybeinliers = n randomly selected values from data maybemodel = model parameters fitted to maybeinliers alsoinliers = empty set for every point in data not in maybeinliers { if point fits maybemodel with an error smaller than t add point to alsoinliers } if the number of elements in alsoinliers is > d { % this implies that we may have found a good model % now test how good it is bettermodel = model parameters fitted to all points in maybeinliers and alsoinliers thiserr = a measure of how well model fits these points if thiserr < besterr { bestfit = bettermodel besterr = thiserr } } increment iterations } return bestfit  It is quite easy to convert this into valid R code (as a learner of R you should try it yourself before looking at my solution!): ransac <- function(data, n, k, t, d) { iterations <- 0 bestfit <- NULL besterr <- 1e5 while (iterations < k) { maybeinliers <- sample(nrow(data), n) maybemodel <- lm(y ~ x, data = data, subset = maybeinliers) alsoinliers <- NULL for (point in setdiff(1:nrow(data), maybeinliers)) { if (abs(maybemodel$coefficients[2]*data[point, 1] - data[point, 2] + maybemodel$coefficients[1])/(sqrt(maybemodel$coefficients[2] + 1)) < t)
alsoinliers <- c(alsoinliers, point)
}
if (length(alsoinliers) > d) {
bettermodel <- lm(y ~ x, data = data, subset = c(maybeinliers, alsoinliers))
thiserr <- summary(bettermodel)$sigma if (thiserr < besterr) { bestfit <- bettermodel besterr <- thiserr } } iterations <- iterations + 1 } bestfit }  We now test this with some sample data: data <- read.csv("data/RANSAC.csv") plot(data) abline(lm(y ~ x, data = data)) set.seed(3141) abline(ransac(data, n = 10, k = 10, t = 0.5, d = 10), col = "blue")  The black line is a plain vanilla linear regression, the blue line is the RANSAC-enhanced version. As you can see: no more crashing into innocent pedestrians. 🙂 ## Symbolic Regression, Genetic Programming… or if Kepler had R A few weeks ago we published a post about using the power of the evolutionary method for optimization (see Evolution works!). In this post we will go a step further, so read on… A problem researchers often face is that they have an amount of data and need to find some functional form, e.g. some kind of physical law, for it. The standard approach is to try different functional forms, like linear, quadratic or polynomial functions with higher order terms. Also possible is a fourier analysis with trigonometric functions. But all of those approaches are quite limited and it would be nice if we had a program to do this for us and come up with a functional form that is both accurate and parsimonious… well, your prayers were heard! This approach is called symbolic regression (also sometimes called free-form regression) – a special case of what is called genetic programming – and the idea is to give the algorithm a grammar which defines some basic functional building blocks (like addition, subtraction, multiplication, logarithms, trigonometric functions and so on) and then try different combinations in an evolutionary process which keeps the better terms and recombines them to even more fitting terms. And the end we want to have a nice formula which captures the dynamics of the system without overfitting the noise. The package with which you can do such magic is the gramEvol package (on CRAN). Let us start with a simple example where we first generate some data on the basis of a combination of trig functions: x <- seq(0, 4*pi, length.out = 201) y <- sin(x) + cos(x + x) plot(y)  We now try to find this functional form by just giving the program the generated data points. As a first step we load the package and define the grammar, i.e. the allowed functional building blocks (for details how to define your own grammar consult the package’s documentation): library("gramEvol") ruleDef <- list(expr = grule(op(expr, expr), func(expr), var), func = grule(sin, cos), op = grule('+', '-', '*'), var = grule(x)) grammarDef <- CreateGrammar(ruleDef) grammarDef ## <expr> ::= <op>(<expr>, <expr>) | <func>(<expr>) | <var> ## <func> ::= sin | cos ## <op> ::= "+" | "-" | "*" ## <var> ::= x  Just to give some examples of randomly created formulas from this grammar you could use the GrammarRandomExpression function: set.seed(123) GrammarRandomExpression(grammarDef, 6) ## [[1]] ## expression(sin(cos(x + x))) ## ## [[2]] ## expression(sin(cos(x * x)) + x) ## ## [[3]] ## expression((x - cos(x)) * x) ## ## [[4]] ## expression(x) ## ## [[5]] ## expression(sin(x)) ## ## [[6]] ## expression(x + x)  After that we have to define some cost function which is used to evaluate how good the respective formula is (again, for details please consult the documentation): SymRegFitFunc <- function(expr) { result <- eval(expr) if (any(is.nan(result))) return(Inf) return (mean(log(1 + abs(y - result)))) }  The last step is starting the search and see what the algorithm finds: set.seed(314) ge <- GrammaticalEvolution(grammarDef, SymRegFitFunc, terminationCost = 0.1, iterations = 2500, max.depth = 5) ge ## Grammatical Evolution Search Results: ## No. Generations: 2149 ## Best Expression: sin(x) + cos(x + x) ## Best Cost: 0 plot(y) points(eval(ge$best$expressions), col = "red", type = "l")  Quite impressive, isn’t it? It found the exact same form by just “looking at” the data and trying different combinations of functions, guided by evolution. Now, in a real world setting you normally don’t have perfect data but you always have some measurement errors (= noise), so we run the example again but this time with some added noise (we use the jitter function for that): x <- seq(0, 4*pi, length.out = 201) y <- jitter(sin(x) + cos(x + x), amount = 0.2) plot(y)  And now for the rest of the steps: ruleDef <- list(expr = grule(op(expr, expr), func(expr), var), func = grule(sin, cos), op = grule('+', '-', '*'), var = grule(x)) grammarDef <- CreateGrammar(ruleDef) grammarDef ## <expr> ::= <op>(<expr>, <expr>) | <func>(<expr>) | <var> ## <func> ::= sin | cos ## <op> ::= "+" | "-" | "*" ## <var> ::= x SymRegFitFunc <- function(expr) { result <- eval(expr) if (any(is.nan(result))) return(Inf) return (mean(log(1 + abs(y - result)))) } set.seed(314) ge <- GrammaticalEvolution(grammarDef, SymRegFitFunc, terminationCost = 0.1, iterations = 2500, max.depth = 5) ge ## Grammatical Evolution Search Results: ## No. Generations: 2149 ## Best Expression: sin(x) + cos(x + x) ## Best Cost: 0.0923240003917875 plot(y) points(eval(ge$best$expressions), col = "red", type = "l")  Although we added quite some noise the program was still successful in finding the original functional form! Now, we are ready to try something more useful: finding a real physical law of nature! We want to find the relationship between orbital periods and distances from the sun of our solar system. First we provide the distance and period data, normalised for the earth: planets <- c("Venus", "Earth", "Mars", "Jupiter", "Saturn", "Uranus") distance <- c(0.72, 1.00, 1.52, 5.20, 9.53, 19.10) period <- c(0.61, 1.00, 1.84, 11.90, 29.40, 83.50) data.frame(planets, distance, period) ## planets distance period ## 1 Venus 0.72 0.61 ## 2 Earth 1.00 1.00 ## 3 Mars 1.52 1.84 ## 4 Jupiter 5.20 11.90 ## 5 Saturn 9.53 29.40 ## 6 Uranus 19.10 83.50  And after that we perform just the same steps as above: ruleDef <- list(expr = grule(op(expr, expr), func(expr), var), func = grule(sin, cos, tan, log, sqrt), op = grule('+', '-', '*', '/', '^'), var = grule(distance, n), n = grule(1, 2, 3, 4, 5, 6, 7, 8, 9)) grammarDef <- CreateGrammar(ruleDef) grammarDef ## <expr> ::= <op>(<expr>, <expr>) | <func>(<expr>) | <var> ## <func> ::= sin | cos | tan | log | sqrt ## <op> ::= "+" | "-" | "*" | "/" | "^" ## <var> ::= distance | <n> ## <n> ::= 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 SymRegFitFunc <- function(expr) { result <- eval(expr) if (any(is.nan(result))) return(Inf) return (mean(log(1 + abs(period - result)))) } set.seed(2) suppressWarnings(ge <- GrammaticalEvolution(grammarDef, SymRegFitFunc, terminationCost = 0.05)) ge ## Grammatical Evolution Search Results: ## No. Generations: 42 ## Best Expression: sqrt(distance) * distance ## Best Cost: 0.0201895728693589  Wow, the algorithm just rediscovered the third law of Kepler in no time! The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit. If only Kepler could have used R! 😉 ## Inverse Statistics – and how to create Gain-Loss Asymmetry plots in R Asset returns have certain statistical properties, also called stylized facts. Important ones are: • Absence of autocorrelation: basically the direction of the return of one day doesn’t tell you anything useful about the direction of the next day. • Fat tails: returns are not normal, i.e. there are many more extreme events than there would be if returns were normal. • Volatility clustering: basically financial markets exhibit high-volatility and low-volatility regimes. • Leverage effect: high-volatility regimes tend to coincide with falling prices and vice versa. A good introduction and overview can be found in R. Cont: Empirical properties of asset returns: stylized facts and statistical issues. One especially fascinating statistical property is the so called gain-loss asymmetry: it basically states that upward movements tend to take a lot longer than downward movements which often come in the form of sudden hefty crashes. So, an abstract illustration of this property would be a sawtooth pattern: The same effect in real life: suppressWarnings(suppressMessages(library(quantmod))) suppressWarnings(suppressMessages(getSymbols("^GSPC", from = "1950-01-01"))) ## [1] "GSPC" plot.zoo(GSPC$GSPC.Close, xlim = c(as.Date("2000-01-01"), as.Date("2013-01-01")), ylim = c(600, 1700), ylab ="", main ="S&P from 2000 to 2013")


The practical implication for your investment horizon is that your losses often come much faster than your gains (life is just not fair…). To illustrate this authors often plot the investment horizon distribution. It illustrates how long you have to wait for a certain target return, negative as well as positive (for some examples see e.g. here, also the source of the following plot):

This is closely related to what statisticians call first passage time: when is a given threshold passed for the first time? To perform such an analysis you need something called inverse statistics. Normally you would plot the distribution of returns given a fixed time window (= forward statistics). Here we do it the other way around: you fix the return and want to find the shortest waiting time needed to obtain at least the respective return. To achieve that you have to test all possible time windows which can be quite time consuming.

Because I wanted to reproduce those plots I tried to find some code somewhere… to no avail. I then contacted some of the authors of the respective papers… no answer. I finally asked a question on Quantitative Finance StackExchange… and got no satisfying answer either. I therefore wrote the code myself and thereby answered my own question:

inv_stat <- function(symbol, name, target = 0.05) {
p <- coredata(Cl(symbol))
end <- length(p)
days_n <- days_p <- integer(end)

# go through all days and look when target is reached the first time from there
for (d in 1:end) {
ret <- cumsum(as.numeric(na.omit(ROC(p[d:end]))))
cond_n <- ret < -target
cond_p <- ret > target
suppressWarnings(days_n[d] <- min(which(cond_n)))
suppressWarnings(days_p[d] <- min(which(cond_p)))
}

days_n_norm <- prop.table(as.integer(table(days_n, exclude = "Inf")))
days_p_norm <- prop.table(as.integer(table(days_p, exclude = "Inf")))

plot(days_n_norm, log = "x", xlim = c(1, 1000), main = paste0(name, " gain-/loss-asymmetry with target ", target), xlab = "days", ylab = "density", col = "red")
points(days_p_norm, col = "blue")

c(which.max(days_n_norm), which.max(days_p_norm)) # mode of days to obtain (at least) neg. and pos. target return
}

inv_stat(GSPC, name = "S&P 500")


## [1] 10 24


So, here you see that for the S&P 500 since 1950 the mode (peak) of the days to obtain a loss of at least 5% has been 10 days and a gain of the same size 24 days! That is the gain-loss asymmetry in action!

Still two things are missing in the code:

• Detrending of the time series.
• Fitting a probability distribution (the generalized gamma distribution seems to work well).

If you want to add them or if you have ideas how to improve the code, please let me know in the comments! Thank you and stay tuned!

## Learning Data Science: Predicting Income Brackets

As promised in the post Learning Data Science: Modelling Basics we will now go a step further and try to predict income brackets with real world data and different modelling approaches. We will learn a thing or two along the way, e.g. about the so-called Accuracy-Interpretability Trade-Off, so read on…

The data we will use is from here: Marketing data set. The description reads:

This dataset contains questions from questionnaires that were filled out by shopping mall customers in the San Francisco Bay area. The goal is to predict the Annual Income of Household from the other 13 demographics attributes.

The following extra information (or metadata) is provided with the data:

cat(readLines('data/marketing-names.txt'), sep = '\n')
Marketing data set
1: Description.
This dataset contains questions from questionaries that were filled out by shopping mall customers in the San Francisco Bay area. The goal is to predict the Anual Income of Household from the other 13 demographics  attributes.
2: Type.            Classification
3: Origin.          Real world
4: Instances.       6876 (8993)
5: Features.        14
6: Classes          9
7: Missing values.  Yes
@relation marketing
@attribute Sex integer [1, 2]
@attribute MaritalStatus integer [1, 5]
@attribute Age integer [1, 7]
@attribute Education integer [1, 6]
@attribute Occupation integer [1, 9]
@attribute YearsInSf integer [1, 5]
@attribute DualIncome integer [1, 3]
@attribute HouseholdMembers integer [1, 9]
@attribute Under18 integer [0, 9]
@attribute HouseholdStatus integer [1, 3]
@attribute TypeOfHome integer [1, 5]
@attribute EthnicClass integer [1, 8]
@attribute Language integer [1, 3]
@attribute Income {1, 2, 3, 4, 5, 6, 7, 8, 9}
@inputs Sex, MaritalStatus, Age, Education, Occupation, YearsInSf, DualIncome, HouseholdMembers, Under18, HouseholdStatus, TypeOfHome, EthnicClass, Language
@outputs Income
DATA DICTIONARY
1    ANNUAL INCOME OF HOUSEHOLD (PERSONAL INCOME IF SINGLE)
1. Less than $10,000 2.$10,000 to $14,999 3.$15,000 to $19,999 4.$20,000 to $24,999 5.$25,000 to $29,999 6.$30,000 to $39,999 7.$40,000 to $49,999 8.$50,000 to $74,999 9.$75,000 or more

2    SEX
1. Male
2. Female
3    MARITAL STATUS
1. Married
2. Living together, not married
3. Divorced or separated
4. Widowed
5. Single, never married
4    AGE
1. 14 thru 17
2. 18 thru 24
3. 25 thru 34
4. 35 thru 44
5. 45 thru 54
6. 55 thru 64
7. 65 and Over
5    EDUCATION
4. 1 to 3 years of college
6    OCCUPATION
1. Professional/Managerial
2. Sales Worker
3. Factory Worker/Laborer/Driver
4. Clerical/Service Worker
5. Homemaker
6. Student, HS or College
7. Military
8. Retired
9. Unemployed
7    HOW LONG HAVE YOU LIVED IN THE SAN FRAN./OAKLAND/SAN JOSE AREA?
1. Less than one year
2. One to three years
3. Four to six years
4. Seven to ten years
5. More than ten years
8    DUAL INCOMES (IF MARRIED)
1. Not Married
2. Yes
3. No
1. One
2. Two
3. Three
4. Four
5. Five
6. Six
7. Seven
8. Eight
9. Nine or more

10    PERSONS IN HOUSEHOLD UNDER 18
0. None
1. One
2. Two
3. Three
4. Four
5. Five
6. Six
7. Seven
8. Eight
9. Nine or more
11    HOUSEHOLDER STATUS
1. Own
2. Rent
3. Live with Parents/Family
12    TYPE OF HOME
1. House
2. Condominium
3. Apartment
4. Mobile Home
5. Other
13    ETHNIC CLASSIFICATION
1. American Indian
2. Asian
3. Black
4. East Indian
5. Hispanic
6. Pacific Islander
7. White
8. Other
14    WHAT LANGUAGE IS SPOKEN MOST OFTEN IN YOUR HOME?
1. English
2. Spanish
3. Other


Our task is to predict the variable “Income”.

So, let us first load the data (you can find the correctly formatted csv-file here: marketing.csv), have a look at some of its characteristics and perform a little bit of additional formatting. After that we divide it into a training (80%) and a test set (20%) to account for potential overfitting (also see Learning Data Science: Modelling Basics):

data <- read.csv("data/marketing.csv")
dim(data)
## [1] 6876   14

str(data)
## 'data.frame':    6876 obs. of  14 variables:
##  $Sex : int 1 2 2 2 1 1 1 1 1 1 ... ##$ MaritalStatus   : int  1 1 5 5 1 5 3 1 1 5 ...
##  $Age : int 5 3 1 1 6 2 3 6 7 2 ... ##$ Education       : int  5 5 2 2 4 3 4 3 4 4 ...
##  $Occupation : int 5 1 6 6 8 9 3 8 8 9 ... ##$ YearsInSf       : int  5 5 5 3 5 4 5 5 4 5 ...
##  $DualIncome : int 3 2 1 1 3 1 1 3 3 1 ... ##$ HouseholdMembers: int  5 3 4 4 2 3 1 3 2 1 ...
##  $Under18 : int 2 1 2 2 0 1 0 0 0 0 ... ##$ HouseholdStatus : int  1 2 3 3 1 2 2 2 2 2 ...
##  $TypeOfHome : int 1 3 1 1 1 3 3 3 3 3 ... ##$ EthnicClass     : int  7 7 7 7 7 7 7 7 7 7 ...
##  $Language : int 1 1 1 1 1 1 1 1 1 1 ... ##$ Income          : int  9 9 1 1 8 1 6 2 4 1 ...

data_names <- names(data)
data <- cbind(data[-ncol(data)], factor(data\$Income)) # make variable Income (which should be predicted) a factor
names(data) <- data_names

set.seed(12)
random <- sample(1:nrow(data), 0.8 * nrow(data))
data_train <- data[random, ]
data_test <- data[-random, ]


We start with a simple but comprehensible model, OneR (on CRAN), as a benchmark:

library(OneR)
data <- optbin(data_train)
model <- OneR(data, verbose = TRUE)
##
##     Attribute        Accuracy
## 1 * Age              28.2%
## 2   MaritalStatus    28.11%
## 3   Occupation       28.07%
## 4   HouseholdStatus  27.56%
## 5   DualIncome       27.04%
## 6   Education        25.98%
## 7   HouseholdMembers 22.51%
## 8   Under18          20.69%
## 9   TypeOfHome       19.36%
## 10  EthnicClass      19.29%
## 11  Sex              18.07%
## 12  Language         17.82%
## 13  YearsInSf        17.75%
## ---
## Chosen attribute due to accuracy
## and ties method (if applicable): '*'

summary(model)
##
## Call:
## OneR.data.frame(x = data, verbose = TRUE)
##
## Rules:
## If Age = 1 then Income = 1
## If Age = 2 then Income = 1
## If Age = 3 then Income = 6
## If Age = 4 then Income = 8
## If Age = 5 then Income = 8
## If Age = 6 then Income = 8
## If Age = 7 then Income = 6
##
## Accuracy:
## 1551 of 5500 instances classified correctly (28.2%)
##
## Contingency table:
##       Age
## Income     1     2     3     4     5    6    7  Sum
##    1   * 421 * 352    99    43    21   15   25  976
##    2      16   204   107    39    13   22   33  434
##    3       9   147   122    49    12   21   35  395
##    4       5   121   188    71    39   29   42  495
##    5       3    77   179    81    29   23   34  426
##    6      10    93 * 234   156    70   56 * 47  666
##    7      12    92   185   155   107   66   33  650
##    8      12   111   211 * 251 * 160 * 86   44  875
##    9      11    76   114   187   104   69   22  583
##    Sum   499  1273  1439  1032   555  387  315 5500
## ---
## Maximum in each column: '*'
##
## Pearson's Chi-squared test:
## X-squared = 2671.2, df = 48, p-value < 2.2e-16

plot(model)


prediction <- predict(model, data_test)
eval_model(prediction, data_test)
##
## Confusion matrix (absolute):
##           Actual
## Prediction    1    2    3    4    5    6    7    8    9  Sum
##        1    232   45   46   32   33   27   19   27   24  485
##        2      0    0    0    0    0    0    0    0    0    0
##        3      0    0    0    0    0    0    0    0    0    0
##        4      0    0    0    0    0    0    0    0    0    0
##        5      0    0    0    0    0    0    0    0    0    0
##        6     31   30   44   44   41   66   44   57   50  407
##        7      0    0    0    0    0    0    0    0    0    0
##        8     16   20   20   47   27   87   71  110   86  484
##        9      0    0    0    0    0    0    0    0    0    0
##        Sum  279   95  110  123  101  180  134  194  160 1376
##
## Confusion matrix (relative):
##           Actual
## Prediction    1    2    3    4    5    6    7    8    9  Sum
##        1   0.17 0.03 0.03 0.02 0.02 0.02 0.01 0.02 0.02 0.35
##        2   0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00
##        3   0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00
##        4   0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00
##        5   0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00
##        6   0.02 0.02 0.03 0.03 0.03 0.05 0.03 0.04 0.04 0.30
##        7   0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00
##        8   0.01 0.01 0.01 0.03 0.02 0.06 0.05 0.08 0.06 0.35
##        9   0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00
##        Sum 0.20 0.07 0.08 0.09 0.07 0.13 0.10 0.14 0.12 1.00
##
## Accuracy:
## 0.2965 (408/1376)
##
## Error rate:
## 0.7035 (968/1376)
##
## Error rate reduction (vs. base rate):
## 0.1176 (p-value < 2.2e-16)


What can we conclude from this? First the most important feature is “Age” while “Marital Status”, “Occupation” and “Household Status” perform comparably well. The overall accuracy (i.e. the percentage of correctly predicted instances) is with about 30% not that great, on the other hand not that extraordinarily uncommon when dealing with real-world data. Looking at the model itself, in the form of rules and the diagnostic plot, we can see that we have non-linear relationship between Age and Income: the older one gets the higher the income bracket, except for people who are old enough to retire. This seems plausible.

OneR is basically a one step decision tree, so the natural choice for our next analysis would be to have a full decision tree built (all packages are on CRAN):

library(rpart)
library(rpart.plot)
model <- rpart(Income ~., data = data_train)
rpart.plot(model, type = 3, extra = 0, box.palette = "Grays")


prediction <- predict(model, data_test, type = "class")
eval_model(prediction, data_test)
##
## Confusion matrix (absolute):
##           Actual
## Prediction    1    2    3    4    5    6    7    8    9  Sum
##        1    201   36   22   13   16   12    8   15   12  335
##        2     43   25   32   22   17   12   10   14    6  181
##        3      0    0    0    0    0    0    0    0    0    0
##        4      0    0    0    0    0    0    0    0    0    0
##        5      0    0    0    0    0    0    0    0    0    0
##        6     18   24   40   50   42   68   32   33   22  329
##        7      0    0    0    0    0    0    0    0    0    0
##        8     17   10   16   38   26   88   84  132  120  531
##        9      0    0    0    0    0    0    0    0    0    0
##        Sum  279   95  110  123  101  180  134  194  160 1376
##
## Confusion matrix (relative):
##           Actual
## Prediction    1    2    3    4    5    6    7    8    9  Sum
##        1   0.15 0.03 0.02 0.01 0.01 0.01 0.01 0.01 0.01 0.24
##        2   0.03 0.02 0.02 0.02 0.01 0.01 0.01 0.01 0.00 0.13
##        3   0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00
##        4   0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00
##        5   0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00
##        6   0.01 0.02 0.03 0.04 0.03 0.05 0.02 0.02 0.02 0.24
##        7   0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00
##        8   0.01 0.01 0.01 0.03 0.02 0.06 0.06 0.10 0.09 0.39
##        9   0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00
##        Sum 0.20 0.07 0.08 0.09 0.07 0.13 0.10 0.14 0.12 1.00
##
## Accuracy:
## 0.3096 (426/1376)
##
## Error rate:
## 0.6904 (950/1376)
##
## Error rate reduction (vs. base rate):
## 0.134 (p-value < 2.2e-16)


The new model has an accuracy that is a little bit better but the interpretability is a little bit worse. You have to go through the different branches to see in which income bracket it ends. So for example when the age bracket is below (which means that it is ) it predicts income bracket , when it is bigger than and the Household Status bracket is it predicts income income bracket and so on. The most important variable, as you can see is again Age (which is reassuring that OneR was on the right track) but we also see Household Status and Occupation again.

What is better than one tree? Many trees! So the next natural step is to have many trees built, while varying the features and the examples that should be included in each tree. At the end it may be that different trees give different income brackets as their respective prediction, which we solve via voting as in a good democracy. This whole method is fittingly called Random Forests and fortunately there are many good packages available in R. We use the randomForest package (also on CRAN) here:

library(randomForest)
## randomForest 4.6-14
## Type rfNews() to see new features/changes/bug fixes.

set.seed(4543) # for reproducibility
model <- randomForest(Income ~., data = data_train, importance = TRUE)
varImpPlot(model)


prediction <- predict(model, data_test)
eval_model(prediction, data_test)
##
## Confusion matrix (absolute):
##           Actual
## Prediction    1    2    3    4    5    6    7    8    9  Sum
##        1    223   35   26   16   19   11    9   18   16  373
##        2     24   15   12   20    7   11    1    4    1   95
##        3      9   10   11   14    9    6    3    4    2   68
##        4      5   15   25   22   10   22    6    9    5  119
##        5      2    2    8    9    6   12    6    3    1   49
##        6      3    5   15   14   19   40   23   17   15  151
##        7      8    4    7   13   14   26   25   24    5  126
##        8      3    8    5   11   13   44   49   87   68  288
##        9      2    1    1    4    4    8   12   28   47  107
##        Sum  279   95  110  123  101  180  134  194  160 1376
##
## Confusion matrix (relative):
##           Actual
## Prediction    1    2    3    4    5    6    7    8    9  Sum
##        1   0.16 0.03 0.02 0.01 0.01 0.01 0.01 0.01 0.01 0.27
##        2   0.02 0.01 0.01 0.01 0.01 0.01 0.00 0.00 0.00 0.07
##        3   0.01 0.01 0.01 0.01 0.01 0.00 0.00 0.00 0.00 0.05
##        4   0.00 0.01 0.02 0.02 0.01 0.02 0.00 0.01 0.00 0.09
##        5   0.00 0.00 0.01 0.01 0.00 0.01 0.00 0.00 0.00 0.04
##        6   0.00 0.00 0.01 0.01 0.01 0.03 0.02 0.01 0.01 0.11
##        7   0.01 0.00 0.01 0.01 0.01 0.02 0.02 0.02 0.00 0.09
##        8   0.00 0.01 0.00 0.01 0.01 0.03 0.04 0.06 0.05 0.21
##        9   0.00 0.00 0.00 0.00 0.00 0.01 0.01 0.02 0.03 0.08
##        Sum 0.20 0.07 0.08 0.09 0.07 0.13 0.10 0.14 0.12 1.00
##
## Accuracy:
## 0.3459 (476/1376)
##
## Error rate:
## 0.6541 (900/1376)
##
## Error rate reduction (vs. base rate):
## 0.1796 (p-value < 2.2e-16)


As an aside you can see that the basic modelling workflow stayed the same – no matter what approach (OneR, decision tree or random forest) you chose. This standard is kept for most (modern) packages and one of the great strengths of R! With thousands of packages on CRAN alone there are of course sometimes differences but those are normally well documented (so the help function or the vignette are your friends!)

Now, back to the result of the last analysis: again, the overall accuracy is better but because we have hundreds of trees now the interpretability suffered a lot. This is also known under the name Accuracy-Interpretability Trade-Off. The best we can do in the case of random forests is to find out which features are the most important ones. Again Age, Occupation and Household Status show up, which is consistent with our analyses so far. Additionally, because of the many trees that had to be built, this analysis ran a lot longer than the other two.

Random forests are one of the best methods out of the box, so the accuracy of about 34% tells us that our first model (OneR) wasn’t doing too bad in the first place! Why are able to reach comparatively good results with just one feature. This is true for many real-world data sets. Sometimes simple models are not much worse than very complicated ones – you should keep that in mind!

If you play around with this dataset I would be interested in your results! Please post them in the comments – Thank you and stay tuned!