# Chapter 3 R Lab 2 - 15/04/2021

In this lecture we will learn how to implement the logistic regression model, the linear and the quadratic discriminant analysis (LDA and QDA).

The following packages are required: `MASS`

, `pROC`

and `tidyverse`

.

```
##
## Attaching package: 'MASS'
```

```
## The following object is masked from 'package:dplyr':
##
## select
```

`## Type 'citation("pROC")' for a citation.`

```
##
## Attaching package: 'pROC'
```

```
## The following objects are masked from 'package:stats':
##
## cov, smooth, var
```

## 3.1 Diabetes data

The data we use for this lab are from the Kaggle platform link. The objective of the analysis is to predict whether or not a patient has diabetes, based on certain diagnostic measurements.

The datasets consists of several predictor variables:

`Pregnancies`

: Number of pregnancies`Glucose`

: Plasma glucose concentration a 2 hours in an oral glucose tolerance test`BloodPressure`

: Diastolic blood pressure (mm Hg)`SkinThickness`

: Triceps skin fold thickness (mm)`Insulin`

: 2-Hour serum insulin (mu U/ml)`BMI`

: Body mass index (weight in kg/(height in m)^2)`DiabetesPedigreeFunction`

: Diabetes pedigree function`Age`

: Age (years)

The response variable is named `Outcome`

and is a binary variable: `0`

means that the patient does not have diabetes, while `1`

means the patient does have diabetes.

We import the data

```
diabetes <- read.csv("~/Dropbox/UniBg/Didattica/Economia/2020-2021/MLFE_2021/RLabs/Lab2/diabetes.csv")
glimpse(diabetes)
```

```
## Rows: 768
## Columns: 9
## $ Pregnancies <int> 6, 1, 8, 1, 0, 5, 3, 10, 2, 8, 4, 10, 10, 1, …
## $ Glucose <int> 148, 85, 183, 89, 137, 116, 78, 115, 197, 125…
## $ BloodPressure <int> 72, 66, 64, 66, 40, 74, 50, 0, 70, 96, 92, 74…
## $ SkinThickness <int> 35, 29, 0, 23, 35, 0, 32, 0, 45, 0, 0, 0, 0, …
## $ Insulin <int> 0, 0, 0, 94, 168, 0, 88, 0, 543, 0, 0, 0, 0, …
## $ BMI <dbl> 33.6, 26.6, 23.3, 28.1, 43.1, 25.6, 31.0, 35.…
## $ DiabetesPedigreeFunction <dbl> 0.627, 0.351, 0.672, 0.167, 2.288, 0.201, 0.2…
## $ Age <int> 50, 31, 32, 21, 33, 30, 26, 29, 53, 54, 30, 3…
## $ Outcome <int> 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 1, …
```

We transform this 0/1 integer variable into a factor with categories “No” and “Yes”:

```
##
## No Yes
## 500 268
```

Some exploratory analysis can be done to study the distribution of the regressors conditionally on the diabetes condition. We consider here, as an example, `BMI`

or `Pregnancies`

:

## 3.2 Definition of a function for computing performance indexes

For assessing the performance of a classifier we compare predicted categories with observed categories. This can be done by using the **confusion matrix** which is a 2x2 matrix reporting the joint distribution (with absolute frequencies) of predicted (by row) and observed (by column) categories.

No (obs.) | Yes (obs.) | |||
---|---|---|---|---|

No (pred.) | TN | FN | ||

Yes (pred.) | FP | TP | ||

Total | N | P |

We consider in particular the following performance indexes:

**sensitivity**(true positive rate): TP/P**specificity**(true negative rate): TN/N**accuracy**(rate of correctly classified observations): (TN+TP)/(N+P)

We define a function named `perf_indexes`

which computes the above defined indexes. The function has only one argument which is a confusion 2x2 matrix named `cm`

. It returns a vector with the 3 indexes.

## 3.3 A new method for creating the training and testing set

To create the training (70%) and test (30%) dataset we use a new approach different from the one introduced in Section 2.1.1.

We first create in the dataframe a new column that contains an unique ID for each observation (this will be used later). The ID is simply given by the row number of each observation and can be obtained by usinf the `row_number`

function:

```
## Pregnancies Glucose BloodPressure SkinThickness Insulin BMI
## 1 6 148 72 35 0 33.6
## 2 1 85 66 29 0 26.6
## 3 8 183 64 0 0 23.3
## 4 1 89 66 23 94 28.1
## 5 0 137 40 35 168 43.1
## 6 5 116 74 0 0 25.6
## DiabetesPedigreeFunction Age Outcome id
## 1 0.627 50 Yes 1
## 2 0.351 31 No 2
## 3 0.672 32 Yes 3
## 4 0.167 21 No 4
## 5 2.288 33 Yes 5
## 6 0.201 30 No 6
```

We are now ready to sample from the `diabetes`

dataset 70% of the observations (rows) by using the `slice_sample`

function:

```
## Pregnancies Glucose BloodPressure SkinThickness Insulin BMI
## 1 9 170 74 31 0 44.0
## 2 6 92 92 0 0 19.9
## 3 7 142 90 24 480 30.4
## 4 10 108 66 0 0 32.4
## 5 1 89 76 34 37 31.2
## 6 1 164 82 43 67 32.8
## DiabetesPedigreeFunction Age Outcome id
## 1 0.403 43 Yes 762
## 2 0.188 28 No 34
## 3 0.128 43 Yes 696
## 4 0.272 42 Yes 144
## 5 0.192 23 No 113
## 6 0.341 50 No 549
```

To get now the test set we use the `anti_join`

function. As reported in the following fiugre

the `anti_join`

function returns the rows of a first dataframe (for us `diabetes`

) which are not cointained in a second dataframe (for us `tr`

). For this operation it is necessary to specify an ID variable which is available in both the dataframes: it’s the ID variable that we have created before!

```
## Pregnancies Glucose BloodPressure SkinThickness Insulin BMI
## 1 2 197 70 45 543 30.5
## 2 8 125 96 0 0 0.0
## 3 4 110 92 0 0 37.6
## 4 1 189 60 23 846 30.1
## 5 1 115 70 30 96 34.6
## 6 7 147 76 0 0 39.4
## DiabetesPedigreeFunction Age Outcome id
## 1 0.158 53 Yes 9
## 2 0.232 54 Yes 10
## 3 0.191 30 No 11
## 4 0.398 59 Yes 14
## 5 0.529 32 Yes 20
## 6 0.257 43 Yes 27
```

We now remove from `tr`

and `te`

the ID variable as it is not useful for the following analysis:

## 3.4 Logistic regression

For implementing the logistic regression the function `glm`

(generalized linear model) is used. It requires the specification of the formula, of the dataframe containing the data and of the distribution (`family`

) of the response variable (in this case `binomial`

distribution given that we are working with a binary response):

```
##
## Call:
## glm(formula = Outcome ~ ., family = "binomial", data = tr)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -2.5055 -0.7661 -0.4241 0.7973 2.7234
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -8.2035808 0.8562214 -9.581 < 2e-16 ***
## Pregnancies 0.1193692 0.0381312 3.130 0.00175 **
## Glucose 0.0317281 0.0041851 7.581 3.42e-14 ***
## BloodPressure -0.0151853 0.0066067 -2.298 0.02153 *
## SkinThickness 0.0008759 0.0079027 0.111 0.91175
## Insulin -0.0011698 0.0010986 -1.065 0.28694
## BMI 0.1032936 0.0185240 5.576 2.46e-08 ***
## DiabetesPedigreeFunction 0.8886256 0.3439073 2.584 0.00977 **
## Age 0.0153383 0.0112087 1.368 0.17118
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 703.68 on 536 degrees of freedom
## Residual deviance: 519.01 on 528 degrees of freedom
## AIC: 537.01
##
## Number of Fisher Scoring iterations: 5
```

Note that the notation `Outcome ~.`

means that all the covariates but `Outcome`

are included as regressors (this avoids to write the formula in the standard way: `Outcome ~ Pregnancies+Glucose+...`

).

The summary contains the parameter estimates and the corresponding p-values of the test checking \(H_0:\beta =0\) vs \(H_1: \beta\neq 0\). Consider for example the parameter of the `Pregnancies`

variable which is positive (i.e. higher risk of diabetes) and equal to 0.1193692: this means that for a one-unit increase in the number of pregnancies, the log-odds increases by 0.1193692. For a simpler interpretation, in the odds scale, we can take the exponential transformation of the parameter:

```
## Pregnancies
## 1.126786
```

This means that this means that for a one-unit increase in the number of pregnancies, we expect a 12.68 increase in the diabetes odds (remember that the odds is strictly connected to the diabetes probability as it is defined as \(p/(1-p)\)).

We are now interested in computing predictions for the test observations given the estimated logistic model. They can be obtained by using the `predict.glm`

function. It is important to specify that we are interested in the prediction on the outcome scale (otherwise we will get the predictions on the logit scale);

```
## 1 2 3 4 5 6
## 0.67493875 0.02398662 0.24613189 0.59857381 0.25929919 0.75019202
```

The object `logreg_pred`

contains the conditional probability of having diabetes given the covariate vector. For example for the first test observation the diabetes probability is equal to `logreg_pred[1]`

. In order to have a categorical prediction it is necessary to fix a threshold for the probability (default is 0.5) and transform correspondingly the probabilities:

```
## 1 2 3 4 5 6
## "Yes" "No" "No" "Yes" "No" "Yes"
```

As expected, the predicted category for the first observation is “Yes”.

We are now ready to compute the confusion matrix

```
##
## logreg_pred_cat No Yes
## No 140 27
## Yes 18 46
```

and the performance index by using the function we defined in Section 3.2

```
## sens spec acc
## 0.6301370 0.8860759 0.8051948
```

## 3.5 Linear and quadratic discriminant analysis

To run LDA we use the `lda`

function of the `MASS`

package. Similarly, to logistic regression we have to specify the formula and the data:

```
## Call:
## lda(Outcome ~ ., data = tr)
##
## Prior probabilities of groups:
## No Yes
## 0.6368715 0.3631285
##
## Group means:
## Pregnancies Glucose BloodPressure SkinThickness Insulin BMI
## No 3.304094 110.0409 68.45322 19.58187 66.75146 30.21228
## Yes 4.846154 140.4564 71.47179 22.69231 94.42564 35.15077
## DiabetesPedigreeFunction Age
## No 0.4269415 31.16667
## Yes 0.5597333 36.84615
##
## Coefficients of linear discriminants:
## LD1
## Pregnancies 0.0929390001
## Glucose 0.0247988603
## BloodPressure -0.0129578404
## SkinThickness 0.0012477089
## Insulin -0.0008439518
## BMI 0.0698160977
## DiabetesPedigreeFunction 0.6347798644
## Age 0.0125305304
```

The output reports the prior probabilities for the two categories and the conditional means of all the covariates.

We go straight to the calculation of the predictions:

`## [1] "class" "posterior" "x"`

Note that the object `lda_pred`

is a list containing 3 objects including the vector of posterior probabilities (`posterior`

) and the vector of predicted categories (`class):

```
## No Yes
## 1 0.2871729 0.71282712
## 2 0.9683246 0.03167541
## 3 0.7805098 0.21949024
## 4 0.3587903 0.64120969
## 5 0.7525345 0.24746553
## 6 0.2584910 0.74150900
```

```
## [1] Yes No No Yes No Yes
## Levels: No Yes
```

To compute the confusion matrix we will use the vector `lda_pred$class`

:

```
##
## No Yes
## No 143 28
## Yes 15 45
```

```
## sens spec acc
## 0.6164384 0.9050633 0.8138528
```

For the quadratic discriminant analysis everything is very similar except the use of the function `qda`

instead of `lda`

:

```
## Call:
## qda(Outcome ~ ., data = tr)
##
## Prior probabilities of groups:
## No Yes
## 0.6368715 0.3631285
##
## Group means:
## Pregnancies Glucose BloodPressure SkinThickness Insulin BMI
## No 3.304094 110.0409 68.45322 19.58187 66.75146 30.21228
## Yes 4.846154 140.4564 71.47179 22.69231 94.42564 35.15077
## DiabetesPedigreeFunction Age
## No 0.4269415 31.16667
## Yes 0.5597333 36.84615
```

`## [1] "class" "posterior"`

```
## [1] Yes No No Yes No Yes
## Levels: No Yes
```

```
## No Yes
## 1 1.132643e-04 0.99988674
## 2 9.828051e-01 0.01719485
## 3 7.994316e-01 0.20056835
## 4 3.567559e-10 1.00000000
## 5 8.683338e-01 0.13166621
## 6 3.345753e-01 0.66542467
```

```
## sens spec acc
## 0.6301370 0.8417722 0.7748918
```

## 3.6 Classifiers comparison in terms of performance indexes

By comparing the 3 classifiers with respect to accuracy, sensitivity and specificity

```
## sens spec acc
## 0.6301370 0.8860759 0.8051948
```

```
## sens spec acc
## 0.6164384 0.9050633 0.8138528
```

```
## sens spec acc
## 0.6301370 0.8417722 0.7748918
```

we see that the logistic regression and LDA perform very similarly. In particular, LDA has the highest specificity but the lowest sensitivity. In terms of accuracy it has the best performance. The QDA instead is the poorest between the three tested methods.

## 3.7 Construction and plotting of the ROC curve

To obtain the ROC curve we use the function `roc`

contained in the `pROC`

package. It is necessary to specify as arguments the vector of observed categories (`response`

) and the vector of the probabilities for the “Yes” category (`predictor`

).

`## Setting levels: control = No, case = Yes`

`## Setting direction: controls < cases`

Given the ROC curve it is possible to plot it and to compute the area under the curve (auc):

`## Area under the curve: 0.8407`

Note that the plot uses specificity on the x-axis (which goes from 1 on the left to zero on the right) and sensitivity on the y-axis.

We now compute all the ROC curves and plot them together using the `ggroc`

function:

```
roc_lda = roc(response = te$Outcome,
predictor = lda_pred$posterior[,2]) #the second col. contains the probabilities for the yes category
```

`## Setting levels: control = No, case = Yes`

`## Setting direction: controls < cases`

```
## Setting levels: control = No, case = Yes
## Setting direction: controls < cases
```

It can be observed that the ROC curve lies below the other two curves meaning that QDA performs worse. Let’s compute the auc for the three methods:

`## Area under the curve: 0.8407`

`## Area under the curve: 0.8419`

`## Area under the curve: 0.8111`

LDA shows the highest auc (even if very similar to the logistic regression auc). All in all we can conclude that LDA is the best classifier.

## 3.8 Exercises Lab 2

### 3.8.1 Exercise 1

The data contained in the files **titanic_tr.csv** (for training) and **titanic_te.csv** (and testing) are about the Titanic disaster (the files are available in the e-learning). In particular the following variables are available:

`pclass`

: ticket class (1 = 1st, 2 = 2nd, 3 = 3rd)`survived`

: survival (0 = No, 1 = Yes)`name`

of the passenger`sex`

`age`

in years`sibsp`

: number of siblings / spouses aboard the Titanic`parch`

number of parents / children aboard the Titanic`ticket`

: ticket number`fare`

: passenger fare`cabin`

: cabin number`embarked`

: port of Embarkation (C = Cherbourg, Q = Queenstown, S = Southampton)

Import the data in `R`

and explore them. Note that the response variable is `survived`

which is classified as a `int`

0/1 variable in the dataframe. Transform it in a factor type object using the `factor`

function with categories “No” and “Yes”.

Convert

`pclass`

(ticket class) to factor (1 = “1st”, 2 = “2nd”, 3 = “3rd”) for both the training and test data set. Plot`survived`

as a function of`pclass`

. In which class do you observe the highest proportion of survivors?Check how many missing values we have in the variables

`age`

and`fare`

separately for training and test data.

- Separately for the training and test data,
**substitute**the missing values with the average of`age`

and`fare`

. Note that for computing the mean when you have missing values you have to run`mean(...,na.rm=T)`

. After this, check that you don’t have any other missing values.

Consider the

**training data**and`survived`

as response variable. Compute the percentage of survived people. Moreover, represent graphically the relationship between`age`

and`survived`

. Comment the plot.Consider the training data set, estimate a logistic model for

`survived`

considering`age`

as the only predictor (`mod1`

). Provide the model summary.

Comment the

`age`

coefficient.Compute

**manually**(without using`predict`

) by means of the formula of the logistic function the probability of surviving for a person of 50 years old. Remember that estimated coefficients can be retrieved by`mod1$coefficients`

. Repeat the same for a person of 20 years old. Finally compute the odds and the odds-ratio.Compute the probability of surviving for all the people in the

**test**data set (use`predict`

).Plot the estimated probabilities as a function of age (impose that the probability axis is between 0 and 1). Use for each point a color corresponding to the

`survived`

outcome. Moreover, include in the plot an horizontal line corresponding to the probability threshold (0.5) and the corresponding age threshold (age boundary between categories). Comment the plot.Compute the confusion matrix for

`mod1`

. Moreover, compute the**test**classification error rate.

- Consider the training data set, estimate a logistic model for
`survived`

considering`pclass`

as predictor (`mod2`

). Note that`pclass`

is a categorical variable with 3 categories and will be included in the model as a dummy variable with 3-1 categories (one category is the baseline). Provide the model summary and comment the coefficients. The following function`contrasts`

helps you in understanding which is the baseline category:

```
contrasts(tr$pclass)
# on the rows you have the categories
# on the column the 2 dummy variables that will be included in the model (for the 2nd and 3rd category)
#the baseline is the category 1st
#in the model 2 coefficients will be estimated for category 2nd and 3rd
```

For more information about the management of dummy variables in R please read this short note available here. It refers to a linear regression model but it generalizes to any model.

Compute (manually or by using

`predict`

) the probability of surviving for a person with a 1st class ticket. Repeat also for the other 2 classes. Compare the three predicted probabilities with the corresponding surviving proportion computed for each class.Estimate the probability of surviving (and the corresponding categories) for all the observations in the

**test**data set (using`predict`

). Then compute the test overall classification error rate and compare it with the one obtained for`mod1`

.

For the training observations represent graphically

`fare`

(amount paid for the ticket) as a function of`pclass`

. Moreover, compute the average and median`fare`

amount separately for each class.Consider the training data set, estimate a logistic model for

`survived`

considering`age`

and`fare`

as predictors (`mod3`

). Provide the model summary.

Comment the coefficient of

`fare`

.Using

`mod3`

compute the predicted probability of surviving for all the**test**observations and the corresponding categories.Plot

`fare`

(y-axis) as a function of`age`

using the predicted categories to color the points. Moreover, include in the plot the linear boundary (it’s the solution of $_0+_1 X_1+_2X_2=0) which separates the two groups.Given the previous plot, what’s the prediction for a 40 years old individual who paid 200 dollars ?

Compute the

**test**error rate for`mod3`

and compare it with the other models.

- Consider the training data set, estimate a logistic model for
`survived`

considering`age`

,`fare`

and`pclass`

as predictors (`mod4`

). Provide the model summary. Compare`mod4`

with the others by using the test classification error. Which is the best model?

### 3.8.2 Exercise 2

We will use some simulated data available from the `mlbench`

library (don’t forget to install it) with \(p=2\) regressors and a binary response variable. Use the following code to generate the data and create the data frame.

```
library(mlbench)
?mlbench.circle
set.seed(3333,sample.kind="Rejection")
data = mlbench.circle(1000) #simulate 1000 data
datadf = data.frame(data$x,
Y=factor(data$classes,labels=c("No","Yes")))
head(datadf)
```

```
## X1 X2 Y
## 1 0.4705415 -0.52006500 No
## 2 -0.1253492 0.20921556 No
## 3 -0.4381452 0.52445868 No
## 4 0.5877131 -0.86523067 Yes
## 5 0.1631693 -0.01084204 No
## 6 -0.6190528 -0.98775598 Yes
```

Plot the data (use the response variable to color the points).

Split the dataset in two parts: 80% of the observations are used for training and 20% for testing. The split is random (use as seed for random number generation the number 456).

Estimate a logistic regression model using

`X1`

and`X2`

as regressors. Provide the model summary.Comment the model coefficients.

Plot the training data and add to the plot the linear decision boundary (see the slides for the formula) defined by the logistic model. Do you think that the model is working well?

Compute the predictions using the test observations. Provide the confusion matrix (use 0.5 as probability threshold) and the overall classification error rate.

Define a function named

`classification_perf`

which computes the overall classification error rate, the sensitivity and the specificity for a 2x2 confusion matrix which has in the rows the predicted categories and in the columns the observed categories.Use the

`classification_perf`

function for the logistic regression model output. Comment about the performance of the logistic regression model.Consider now a very simple classifier (

**null classifier**) which uses as prediction for all the test observations the majority class observed in the training dataset (regardless of the values of the predictors). Compute the overall error rate for the null classifier. Does it perform better than the logistic regression model?

Use linear discriminant analysis to classify your data. Compute the predictions using the test observations and provide the confusion matrix together with performance indexes. Comment the results.

- [
**Advanced and optional**] Using a grid of values for`X1`

and`X2`

together with the`outer`

function, plot the decision boundary together with the training data. Add to the plot also the boundary estimated with the logistic model.

- [
Use quadratic discriminant analysis to classify your data. Compute the predictions using the test observations and provide the confusion matrix together with performance indexes. Comment the results.

- [
**Advanced and optional**] Using a grid of values for`X1`

and`X2`

together with the`outer`

function, plot the decision boundary together with the training data. Add to the plot also the boundary estimated with the logistic model.

- [
Compute and plot the ROC curves for all the estimated models (lr, lda, qda). For which model do we obtain the highest area under the ROC curve?

### 3.8.3 Exercise 3

We will use some simulated data available from the `mlbench`

library with \(p=2\) regressors and a response variable with 4 categories. Use the following code to generate the data.

```
library(mlbench)
?mlbench.smiley
set.seed(3333,sample.kind="Rejection")
data = mlbench.smiley(1000) #simulate 1000 data
datadf = data.frame(data$x, factor(data$classes))
head(datadf)
```

```
## x4 V2 factor.data.classes.
## 1 -0.7371167 0.9282289 1
## 2 -0.8580089 1.0668791 1
## 3 -0.7794051 1.0191270 1
## 4 -0.9232296 0.9328375 1
## 5 -0.7723922 0.9318714 1
## 6 -0.9142181 0.8255049 1
```

```
## x1 x2 y
## 1 -0.7371167 0.9282289 1
## 2 -0.8580089 1.0668791 1
## 3 -0.7794051 1.0191270 1
## 4 -0.9232296 0.9328375 1
## 5 -0.7723922 0.9318714 1
## 6 -0.9142181 0.8255049 1
```

```
##
## 1 2 3 4
## 167 167 250 416
```

Plot the data (use the response variable to color the points).

Split the dataset in two parts: 70% of the observations are used for training and 20% for testing. The split is random (use as seed for random number generation the number 456).

Consider linear and quadratic discriminant analysis for your data. Compare the two classifiers.

Use the KNN method to classify your data. Choose the best value of \(k\) among a sequence of values between 1 and 100 (with step equal to 5). To tune the hyperparameter \(k\) split the training dataset into two sets: one for training (70%) and one for validation (30%). Use a seed equal to 4.

Which model (between lda, qda and knn) do you prefer? Explain why.