# Chapter 3 R Lab 2 - 31/03/2022

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`

.

`library(MASS) #for LDA and QDA`

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

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

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

`library(pROC) #for ROC curve`

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

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

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

```
library(tidyverse)
library(plotly)
```

## 3.1 Satisfaction data

The data we use for this lab are from the Kaggle platform link. You can download the dataset from Moodle, or use the following code for recreating it:

```
# # Data management -> recreate the original dataset from train / test datasets
# satisf_test <- read.csv("~/Downloads/archive/test.csv")
# satisf_test <- satisf_test %>% dplyr::select(satisfaction, Age, Flight.Distance, Departure.Delay.in.Minutes, Arrival.Delay.in.Minutes)
# satisf_test$satisfaction <- ifelse(satisf_test$satisfaction=="satisfied",1,0)
# table(satisf_test$satisfaction) #check!
#
# satisf_train <- read.csv("~/Downloads/archive/train.csv")
# satisf_train$satisfaction <- ifelse(satisf_train$satisfaction=="satisfied",1,0)
# satisf_train <- satisf_train %>% dplyr::select(satisfaction, Age, Flight.Distance, Departure.Delay.in.Minutes, Arrival.Delay.in.Minutes)
# table(satisf_train$satisfaction) #check!
# satisf <- rbind(satisf_test, satisf_train)
# satisf %>% filter(is.na(satisfaction)==T |
# is.na(Age)==T |
# is.na(Flight.Distance)==T |
# is.na(Departure.Delay.in.Minutes)==T |
# is.na(Arrival.Delay.in.Minutes)==T) #check!
#
# summary(satisf) #check!
# satisf <- satisf %>% filter(is.na(Arrival.Delay.in.Minutes)==F)
# colnames(satisf) <- c("outcome","age","dist","dep_delay","arr_delay")
#
# #export
# write_csv(satisf, "~/Downloads/archive/satisf.csv")
```

The objective of the analysis is to predict whether or not a customer is overall satisfied with the company service, based on other quality measurements.The datasets consists of several predictor variables:

`outcome`

: Customer’s overall satisfaction about the airline (0 = Neutral or dissatisfied, 1 = Satisfied)`age`

: Age of passenger (years)`dist`

: Flight distance of the evaluated journey (not indicated :-( )`dep_delay`

: Delay when departure (minutes)`arr_delay`

: Delay delayed at the arrival (minutes)

The response variable is named `outcome`

and is a binary variable: `0`

means that the customer was not satisfied about the overall company service, while `1`

means he/she was pleased by the experience.

We import the data

```
<- read.csv("/Users/marco/Google Drive Personale/Formazione/3_Unibg/2021-22/ML_tutor/RLabs/Lab2/archive/satisf.csv")
satisf glimpse(satisf)
```

```
## Rows: 129,487
## Columns: 5
## $ outcome <int> 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 0, 0, 1, 0, 1, …
## $ age <int> 52, 36, 20, 44, 49, 16, 77, 43, 47, 46, 47, 33, 46, 60, 52, …
## $ dist <int> 160, 2863, 192, 3377, 1182, 311, 3987, 2556, 556, 1744, 1235…
## $ dep_delay <int> 50, 0, 0, 0, 0, 0, 0, 77, 1, 28, 29, 18, 0, 117, 10, 5, 2, 0…
## $ arr_delay <int> 44, 0, 0, 6, 20, 0, 0, 65, 0, 14, 19, 7, 0, 113, 0, 2, 0, 0,…
```

We transform this 0/1 integer variable into a factor with categories “Neutral or dissatisfied” and “Satisfied”:

```
$outcome= factor(satisf$outcome)
satisflevels(satisf$outcome) = c("Neutral-Dissatisfied","Satisfied")
table(satisf$outcome)
```

```
##
## Neutral-Dissatisfied Satisfied
## 73225 56262
```

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

or `Distance`

:

```
%>%
satisf ggplot() +
geom_boxplot(aes(age,fill=outcome))
```

```
# Seems that the most satisfied customers are also the elder one
%>%
satisf ggplot() +
geom_density(aes(dist,fill=outcome),alpha=0.3)
```

```
# Satisfaction seems to be related also to travel distance. Passenger travelling on longer flights are the one
# that face less dissatisfaction/neutrality.
```

## 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.

```
= function(cm){
perf_indexes = cm[2,2] / (cm[1,2] + cm[2,2])
sensitivity = cm[1,1] / (cm[1,1] + cm[2,1])
specificity = sum(diag(cm)) / sum(cm)
accuracy return(c(sens=sensitivity,spec=specificity,acc=accuracy))
}
```

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

To create the training (80%) and test (20%) dataset we use a new approach different from the one introduced in Section 2.1.1 and Section 2.2.

We first create a vector with the indexes we will use for the training dataset. In this case we must set `replace`

to false, since each index will indicate an observation belonging to just train or test. Recall also to set a seed, in order to be able to replicate your results:

```
set.seed(11, sample.kind="Rejection")
<- sample(1:nrow(satisf),0.8*nrow(satisf), replace=F) # notice that here resample is not allowed!
training_index head(training_index)
```

`## [1] 65570 66457 128113 19004 73612 28886`

We are now ready to sample from the `satisf`

dataset 80% of the observations (rows) by using the `training_index`

vector, and leave the non-sampled indexes for test. Just use a `minus (-)`

before the row selection to exclude them:

```
# recall: dataset [ row , columns ]
<- satisf[training_index,]
s_train <- satisf[-training_index,]
s_test
# check datasets
head(s_train)
```

```
## outcome age dist dep_delay arr_delay
## 65570 Neutral-Dissatisfied 50 541 35 39
## 66457 Neutral-Dissatisfied 39 1013 17 57
## 128113 Satisfied 42 3330 9 2
## 19004 Neutral-Dissatisfied 41 341 0 0
## 73612 Neutral-Dissatisfied 24 1084 45 42
## 28886 Neutral-Dissatisfied 22 258 18 19
```

`head(s_test)`

```
## outcome age dist dep_delay arr_delay
## 5 Satisfied 49 1182 0 20
## 12 Neutral-Dissatisfied 33 325 18 7
## 19 Neutral-Dissatisfied 43 1927 0 0
## 29 Satisfied 44 1543 0 0
## 30 Neutral-Dissatisfied 51 235 0 0
## 34 Neutral-Dissatisfied 22 1846 40 68
```

Note that this procedure could be the easiest choice when we don’t have a third dataset (validation).

We are now ready to train our models.

## 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):

```
= glm(outcome ~ .,
logreg data=s_train,
family="binomial") # binomial use as default "logit" link function
summary(logreg)
```

```
##
## Call:
## glm(formula = outcome ~ ., family = "binomial", data = s_train)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -2.2647 -0.9716 -0.7758 1.1563 2.5771
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -1.556e+00 2.037e-02 -76.361 < 2e-16 ***
## age 1.522e-02 4.403e-04 34.572 < 2e-16 ***
## dist 6.214e-04 7.038e-06 88.294 < 2e-16 ***
## dep_delay 3.118e-03 6.908e-04 4.514 6.37e-06 ***
## arr_delay -6.788e-03 6.842e-04 -9.921 < 2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 141820 on 103588 degrees of freedom
## Residual deviance: 130815 on 103584 degrees of freedom
## AIC: 130825
##
## Number of Fisher Scoring iterations: 4
```

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 ~ age+dist+...`

).

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 `Age`

variable which is positive (i.e. higher possibility of being satisfied) and equal to 0.0152229: this means that for a one-unit increase in age, the **log-odds** increases by 0.0152229. For a simpler interpretation, in the odds scale, we can take the exponential transformation of the parameter:

```
# odds ratio. Not of our interest
exp(logreg$coefficients[2])
```

```
## age
## 1.015339
```

```
# odds interpretation
exp(coef(logreg)[2]) -1)*100 (
```

```
## age
## 1.533937
```

This means that this means that for a one-unit increase in the customer’s age, we expect a 1.53 increase in the satisfaction **odds**. Remember that the odds is strictly connected to the satisfaction 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);

```
= predict.glm(logreg, #also predict is fine
logreg_pred newdata = s_test,
type="response")
head(logreg_pred)
```

```
## 5 12 19 29 30 34
## 0.4474067 0.3009241 0.5735149 0.5181946 0.3467344 0.3987565
```

The object `logreg_pred`

contains the conditional probability of being satisfied given the covariate vector. For example for the first test observation the satisfaction 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:

```
= ifelse(logreg_pred > 0.5, "Satisfied", "Neutral-Dissatisfied")
logreg_pred_class head(logreg_pred_class)
```

```
## 5 12 19
## "Neutral-Dissatisfied" "Neutral-Dissatisfied" "Satisfied"
## 29 30 34
## "Satisfied" "Neutral-Dissatisfied" "Neutral-Dissatisfied"
```

As expected, the predicted category for the first observation is **“Neutral - Dissatisfied”**.

We are now ready to compute the confusion matrix

`table(logreg_pred_class, s_test$outcome)`

```
##
## logreg_pred_class Neutral-Dissatisfied Satisfied
## Neutral-Dissatisfied 12117 6259
## Satisfied 2524 4998
```

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

`perf_indexes(table(logreg_pred_class, s_test$outcome))`

```
## sens spec acc
## 0.4439904 0.8276074 0.6608618
```

What happen to our performance indexes if we change prediction threshold?

```
rm(logreg_pred_class) # to erase a specific object
= ifelse(logreg_pred > 0.8, "Yes", "No")
logreg_pred_class perf_indexes(table(logreg_pred_class, s_test$outcome))
```

```
## sens spec acc
## 0.04557164 0.99112083 0.58012202
```

As you can see, increasing the satisfaction threshold can improve a lot the algorithm specificity (= our algorithm is more able to detect an unsatisfied customer). At the same time, the decreased in the sensitivity and overall accuracy should be considered. In general, the decision of lowering / increasing the decision boundary depends on the context we are applying our knowledge. Think about COVID. Should be “better” to have a diagnostic test with more false positive (= lower sensitivity) or false negative (= lower specificity)?

## 3.5 Linear 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:

```
= lda(outcome ~ ., data=s_train)
ldamod ldamod
```

```
## Call:
## lda(outcome ~ ., data = s_train)
##
## Prior probabilities of groups:
## Neutral-Dissatisfied Satisfied
## 0.5655427 0.4344573
##
## Group means:
## age dist dep_delay arr_delay
## Neutral-Dissatisfied 37.66313 928.5023 16.29445 17.02332
## Satisfied 41.73423 1526.8704 12.22791 12.36638
##
## Coefficients of linear discriminants:
## LD1
## age 0.0226054119
## dist 0.0009458021
## dep_delay 0.0045190445
## arr_delay -0.0093103673
```

The output reports the prior probabilities for the two categories and the conditional means of all the covariates. Recall that the prior probability is the probability that a randomly chosen observation comes from the \(k^{th}\) class.

We go straight to the calculation of the predictions:

```
= predict(ldamod,newdata = s_test)
lda_pred names(lda_pred)
```

`## [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):

`head(lda_pred$posterior)`

```
## Neutral-Dissatisfied Satisfied
## 5 0.5565935 0.4434065
## 12 0.7088309 0.2911691
## 19 0.4283927 0.5716073
## 29 0.4860507 0.5139493
## 30 0.6639620 0.3360380
## 34 0.5979763 0.4020237
```

`head(lda_pred$class)`

```
## [1] Neutral-Dissatisfied Neutral-Dissatisfied Satisfied
## [4] Satisfied Neutral-Dissatisfied Neutral-Dissatisfied
## Levels: Neutral-Dissatisfied Satisfied
```

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

:

`table(lda_pred$class, s_test$outcome)`

```
##
## Neutral-Dissatisfied Satisfied
## Neutral-Dissatisfied 12179 6306
## Satisfied 2462 4951
```

`perf_indexes(table(lda_pred$class, s_test$outcome))`

```
## sens spec acc
## 0.4398152 0.8318421 0.6614410
```

As before, we can choose a different threshold for our decision. In this case, the used functions do not allow to change directly the threshold. So, we need to use the same methodology applied to logistic regression classification method to generate new classes Recall that LDA assign the the observation to the majority class (= class with highest posterior probability). So, there is an implicit probability threshold with two classes (\(0.5\)). Check if by yourself here above:

```
= ifelse(lda_pred$posterior[,"Satisfied"] > 0.5, "Satisfied", "Neutral-Dissatisfied")
lda_pred_class # same results than previous LDA
table(lda_pred_class, s_test$outcome)
```

```
##
## lda_pred_class Neutral-Dissatisfied Satisfied
## Neutral-Dissatisfied 12179 6306
## Satisfied 2462 4951
```

```
# same results than previous LDA
perf_indexes(table(lda_pred_class, s_test$outcome))
```

```
## sens spec acc
## 0.4398152 0.8318421 0.6614410
```

We can now set an higher probability threshold, with the aim of being able to better select the “real” satisfied customers. The probability threshold will be set **on the “Satisfied” class** at \(0.8\) :

```
= ifelse(lda_pred$posterior[,"Satisfied"] > 0.8, "Satisfied", "Neutral-Dissatisfied")
lda_pred_class
table(lda_pred_class, s_test$outcome) #objects are changed w.r.t. before
```

```
##
## lda_pred_class Neutral-Dissatisfied Satisfied
## Neutral-Dissatisfied 14499 10673
## Satisfied 142 584
```

`perf_indexes(table(lda_pred_class, s_test$outcome)) #objects are changed w.r.t. before`

```
## sens spec acc
## 0.05187883 0.99030121 0.58240019
```

We reduced a lot the false positive observation (from \(2462\) to \(142\)). We were able to correctly detect more real unsatisfied customers (from \(12179\) to \(14499\)), increasing the specificity of our algorithm. However, we dramatically predict with a wrong class lots of truly satisfied customers (\(6306\) to \(10673\)), being able to predict very few true “positive” (Satisfied) travellers (\(584\)).

## 3.6 Classifiers comparison in terms of performance indexes

By comparing the 2 classifiers with respect to accuracy, sensitivity and specificity

`perf_indexes(table(logreg_pred_class, s_test$outcome))`

```
## sens spec acc
## 0.04557164 0.99112083 0.58012202
```

`perf_indexes(table(lda_pred$class, s_test$outcome))`

```
## sens spec acc
## 0.4398152 0.8318421 0.6614410
```

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.

## 3.7 Exercises Lab 2

### 3.7.1 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 \(\beta_0+\beta_1 X_1+\beta_2X_2=0\)) which separates the two groups. HINT: derive it considering \(X_2 = Y\) and \(\beta_1=X\).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?

### 3.7.2 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.circleset.seed(3333,sample.kind="Rejection")
= mlbench.circle(1000) #simulate 1000 data
data = data.frame(data$x,
datadf Y=factor(data$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.

- [
Compare the three previous classifier (LR, null classifier, LDA) using the confusion matrix (with the

`classification_perf`

formula) and the error rate. Are you able to compute the performance indexes with the null classifier? Which of the proposed model seems to perform better?

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

```
colnames(datadf) = c("x1", "x2", "y")
head(datadf)
```

```
## 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
```

`table(datadf$y)`

```
##
## 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 discriminant analysis for your data. Analyse your results computing the test error rate.

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 and knn) do you prefer? Explain why.