## What are Decision Trees?

** DECISION TREES** are versatile Machine Learning algorithm that can perform both classification and regression tasks. They are very powerful algorithms, capable of fitting complex datasets. Besides, decision trees are fundamental components of random forests, which are among the most potent Machine Learning algorithms available today.

### Training and Visualizing a decision trees

To build your first decision trees, we will proceed as follow:

- Step 1: Import the data
- Step 2: Clean the dataset
- Step 3: Create train/test set
- Step 4: Build the model
- Step 5: Make prediction
- Step 6: Measure performance
- Step 7: Tune the hyper-parameters

**Step 1) **Import the data

If you are curious about the fate of the titanic, you can watch this video on Youtube. The purpose of this dataset is to predict which people are more likely to survive after the collision with the iceberg. The dataset contains 13 variables and 1309 observations. The dataset is ordered by the variable X.

set.seed(678) path <- 'https://raw.githubusercontent.com/guru99-edu/R-Programming/master/titanic_data.csv' titanic <-read.csv(path) head(titanic)

**Output:**

## X pclass survived name sex ## 1 1 1 1 Allen, Miss. Elisabeth Walton female ## 2 2 1 1 Allison, Master. Hudson Trevor male ## 3 3 1 0 Allison, Miss. Helen Loraine female ## 4 4 1 0 Allison, Mr. Hudson Joshua Creighton male ## 5 5 1 0 Allison, Mrs. Hudson J C (Bessie Waldo Daniels) female ## 6 6 1 1 Anderson, Mr. Harry male ## age sibsp parch ticket fare cabin embarked ## 1 29.0000 0 0 24160 211.3375 B5 S ## 2 0.9167 1 2 113781 151.5500 C22 C26 S ## 3 2.0000 1 2 113781 151.5500 C22 C26 S ## 4 30.0000 1 2 113781 151.5500 C22 C26 S ## 5 25.0000 1 2 113781 151.5500 C22 C26 S ## 6 48.0000 0 0 19952 26.5500 E12 S ## home.dest ## 1 St Louis, MO ## 2 Montreal, PQ / Chesterville, ON ## 3 Montreal, PQ / Chesterville, ON ## 4 Montreal, PQ / Chesterville, ON ## 5 Montreal, PQ / Chesterville, ON ## 6 New York, NY

tail(titanic)

**Output:**

## X pclass survived name sex age sibsp ## 1304 1304 3 0 Yousseff, Mr. Gerious male NA 0 ## 1305 1305 3 0 Zabour, Miss. Hileni female 14.5 1 ## 1306 1306 3 0 Zabour, Miss. Thamine female NA 1 ## 1307 1307 3 0 Zakarian, Mr. Mapriededer male 26.5 0 ## 1308 1308 3 0 Zakarian, Mr. Ortin male 27.0 0 ## 1309 1309 3 0 Zimmerman, Mr. Leo male 29.0 0 ## parch ticket fare cabin embarked home.dest ## 1304 0 2627 14.4583 C ## 1305 0 2665 14.4542 C ## 1306 0 2665 14.4542 C ## 1307 0 2656 7.2250 C ## 1308 0 2670 7.2250 C ## 1309 0 315082 7.8750 S

From the head and tail output, you can notice the data is not shuffled. This is a big issue! When you will split your data between a train set and test set, you will select **only** the passenger from class 1 and 2 (No passenger from class 3 are in the top 80 percent of the observations), which means the algorithm will never see the features of passenger of class 3. This mistake will lead to poor prediction.

To overcome this issue, you can use the function sample().

shuffle_index <- sample(1:nrow(titanic)) head(shuffle_index)

Code Explanation

- sample(1:nrow(titanic)): Generate a random list of index from 1 to 1309 (i.e. the maximum number of rows).

**Output:**

## [1] 288 874 1078 633 887 992

You will use this index to shuffle the titanic dataset.

titanic <- titanic[shuffle_index, ] head(titanic)

**Output:**

## X pclass survived ## 288 288 1 0 ## 874 874 3 0 ## 1078 1078 3 1 ## 633 633 3 0 ## 887 887 3 1 ## 992 992 3 1 ## name sex age ## 288 Sutton, Mr. Frederick male 61 ## 874 Humblen, Mr. Adolf Mathias Nicolai Olsen male 42 ## 1078 O'Driscoll, Miss. Bridget female NA ## 633 Andersson, Mrs. Anders Johan (Alfrida Konstantia Brogren) female 39 ## 887 Jermyn, Miss. Annie female NA ## 992 Mamee, Mr. Hanna male NA ## sibsp parch ticket fare cabin embarked home.dest## 288 0 0 36963 32.3208 D50 S Haddenfield, NJ ## 874 0 0 348121 7.6500 F G63 S ## 1078 0 0 14311 7.7500 Q ## 633 1 5 347082 31.2750 S Sweden Winnipeg, MN ## 887 0 0 14313 7.7500 Q ## 992 0 0 2677 7.2292 C

**Step 2) **Clean the dataset

The structure of the data shows some variables have NA's. Data clean up to be done as follows

- Drop variables home.dest,cabin, name, X and ticket
- Create factor variables for pclass and survived
- Drop the NA

library(dplyr) # Drop variables clean_titanic <- titanic % > % select(-c(home.dest, cabin, name, X, ticket)) % > % #Convert to factor level mutate(pclass = factor(pclass, levels = c(1, 2, 3), labels = c('Upper', 'Middle', 'Lower')), survived = factor(survived, levels = c(0, 1), labels = c('No', 'Yes'))) % > % na.omit() glimpse(clean_titanic)

Code Explanation

- select(-c(home.dest, cabin, name, X, ticket)): Drop unnecessary variables
- pclass = factor(pclass, levels = c(1,2,3), labels= c('Upper', 'Middle', 'Lower')): Add label to the variable pclass. 1 becomes Upper, 2 becomes MIddle and 3 becomes lower
- factor(survived, levels = c(0,1), labels = c('No', 'Yes')): Add label to the variable survived. 1 Becomes No and 2 becomes Yes
- na.omit(): Remove the NA observations

**Output:**

## Observations: 1,045 ## Variables: 8 ## $ pclass <fctr> Upper, Lower, Lower, Upper, Middle, Upper, Middle, U... ## $ survived <fctr> No, No, No, Yes, No, Yes, Yes, No, No, No, No, No, Y... ## $ sex <fctr> male, male, female, female, male, male, female, male... ## $ age <dbl> 61.0, 42.0, 39.0, 49.0, 29.0, 37.0, 20.0, 54.0, 2.0, ... ## $ sibsp <int> 0, 0, 1, 0, 0, 1, 0, 0, 4, 0, 0, 1, 1, 0, 0, 0, 1, 1,... ## $ parch <int> 0, 0, 5, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 2, 0, 4, 0,... ## $ fare <dbl> 32.3208, 7.6500, 31.2750, 25.9292, 10.5000, 52.5542, ... ## $ embarked <fctr> S, S, S, S, S, S, S, S, S, C, S, S, S, Q, C, S, S, C...

**Step 3)** Create train/test set

Before you train your model, you need to perform two steps:

- Create a train and test set: You train the model on the train set and test the prediction on the test set (i.e. unseen data)
- Install rpart.plot from the console

The common practice is to split the data 80/20, 80 percent of the data serves to train the model, and 20 percent to make predictions. You need to create two separate data frames. You don't want to touch the test set until you finish building your model. You can create a function name create_train_test() that takes three arguments.

create_train_test(df, size = 0.8, train = TRUE) arguments: -df: Dataset used to train the model. -size: Size of the split. By default, 0.8. Numerical value -train: If set to `TRUE`, the function creates the train set, otherwise the test set. Default value sets to `TRUE`. Boolean value.You need to add a Boolean parameter because R does not allow to return two data frames simultaneously.

create_train_test <- function(data, size = 0.8, train = TRUE) { n_row = nrow(data) total_row = size * n_row train_sample < - 1: total_row if (train == TRUE) { return (data[train_sample, ]) } else { return (data[-train_sample, ]) } }

Code Explanation

- function(data, size=0.8, train = TRUE): Add the arguments in the function
- n_row = nrow(data): Count number of rows in the dataset
- total_row = size*n_row: Return the nth row to construct the train set
- train_sample <- 1:total_row: Select the first row to the nth rows
- if (train ==TRUE){ } else { }: If condition sets to true, return the train set, else the test set.

You can test your function and check the dimension.

data_train <- create_train_test(clean_titanic, 0.8, train = TRUE) data_test <- create_train_test(clean_titanic, 0.8, train = FALSE) dim(data_train)

**Output:**

## [1] 836 8

dim(data_test)

**Output:**

## [1] 209 8

The train dataset has 1046 rows while the test dataset has 262 rows.

You use the function prop.table() combined with table() to verify if the randomization process is correct.

prop.table(table(data_train$survived))

**Output:**

## ## No Yes ## 0.5944976 0.4055024

prop.table(table(data_test$survived))

**Output:**

## ## No Yes ## 0.5789474 0.4210526

In both dataset, the amount of survivors is the same, about 40 percent.

### Install rpart.plot

rpart.plot is not available from conda libraries. You can install it from the console:

install.packages("rpart.plot")

**Step 4) **Build the model

You are ready to build the model. The syntax for Rpart() function is:

rpart(formula, data=, method='') arguments: - formula: The function to predict - data: Specifies the data frame- method: - "class" for a classification tree - "anova" for a regression tree

You use the class method because you predict a class.

library(rpart) library(rpart.plot) fit <- rpart(survived~., data = data_train, method = 'class') rpart.plot(fit, extra = 106

Code Explanation

- rpart(): Function to fit the model. The arguments are:
- survived ~.: Formula of the Decision Trees
- data = data_train: Dataset
- method = 'class': Fit a binary model

- rpart.plot(fit, extra= 106): Plot the tree. The extra features are set to 101 to display the probability of the 2nd class (useful for binary responses). You can refer to the vignette for more information about the other choices.

**Output:**

You start at the root node (depth 0 over 3, the top of the graph):

- At the top, it is the overall probability of survival. It shows the proportion of passenger that survived the crash. 41 percent of passenger survived.
- This node asks whether the gender of the passenger is male. If yes, then you go down to the root's left child node (depth 2). 63 percent are males with a survival probability of 21 percent.
- In the second node, you ask if the male passenger is above 3.5 years old. If yes, then the chance of survival is 19 percent.
- You keep on going like that to understand what features impact the likelihood of survival.

Note that, one of the many qualities of Decision Trees is that they require very little data preparation. In particular, they don't require feature scaling or centering.

By default, rpart() function uses the **Gini** impurity measure to split the note. The higher the Gini coefficient, the more different instances within the node.

**Step 5)** Make a prediction

You can predict your test dataset. To make a prediction, you can use the predict() function. The basic syntax of predict for decision trees is:

predict(fitted_model, df, type = 'class') arguments: - fitted_model: This is the object stored after model estimation. - df: Data frame used to make the prediction - type: Type of prediction - 'class': for classification - 'prob': to compute the probability of each class - 'vector': Predict the mean response at the node level

You want to predict which passengers are more likely to survive after the collision from the test set. It means, you will know among those 209 passengers, which one will survive or not.

predict_unseen <-predict(fit, data_test, type = 'class')

Code Explanation

- predict(fit, data_test, type = 'class'): Predict the class (0/1) of the test set

Testing the passenger who didn't make it and those who did.

table_mat <- table(data_test$survived, predict_unseen) table_mat

Code Explanation

- table(data_test$survived, predict_unseen): Create a table to count how many passengers are classified as survivors and passed away compare to the correct classification

**Output:**

## predict_unseen ## No Yes ## No 106 15 ## Yes 30 58

The model correctly predicted 106 dead passengers but classified 15 survivors as dead. By analogy, the model misclassified 30 passengers as survivors while they turned out to be dead.

**Step 6)** Measure performance

You can compute an accuracy measure for classification task with the **confusion matrix**:

The **confusion matrix** is a better choice to evaluate the classification performance. The general idea is to count the number of times True instances are classified are False.

Each row in a confusion matrix represents an actual target, while each column represents a predicted target. The first row of this matrix considers dead passengers (the False class): 106 were correctly classified as dead (**True negative**), while the remaining one was wrongly classified as a survivor (**False positive**). The second row considers the survivors, the positive class were 58 (**True positive**), while the **True negative** was 30.

You can compute the **accuracy test** from the confusion matrix:

It is the proportion of true positive and true negative over the sum of the matrix. With R, you can code as follow:

accuracy_Test <- sum(diag(table_mat)) / sum(table_mat)

Code Explanation

- sum(diag(table_mat)): Sum of the diagonal
- sum(table_mat): Sum of the matrix.

You can print the accuracy of the test set:

print(paste('Accuracy for test', accuracy_Test))

**Output:**

## [1] "Accuracy for test 0.784688995215311"

You have a score of 78 percent for the test set. You can replicate the same exercise with the training dataset.

**Step 7) **Tune the hyper-parameters

Decision tree has various parameters that control aspects of the fit. In rpart library, you can control the parameters using the rpart.control() function. In the following code, you introduce the parameters you will tune. You can refer to the vignette for other parameters.

rpart.control(minsplit = 20, minbucket = round(minsplit/3), maxdepth = 30) Arguments: -minsplit: Set the minimum number of observations in the node before the algorithm perform a split -minbucket: Set the minimum number of observations in the final note i.e. the leaf -maxdepth: Set the maximum depth of any node of the final tree. The root node is treated a depth 0

We will proceed as follow:

- Construct function to return accuracy
- Tune the maximum depth
- Tune the minimum number of sample a node must have before it can split
- Tune the minimum number of sample a leaf node must have

You can write a function to display the accuracy. You simply wrap the code you used before:

- predict: predict_unseen <- predict(fit, data_test, type = 'class')
- Produce table: table_mat <- table(data_test$survived, predict_unseen)
- Compute accuracy: accuracy_Test <- sum(diag(table_mat))/sum(table_mat)

accuracy_tune <- function(fit) { predict_unseen <- predict(fit, data_test, type = 'class') table_mat <- table(data_test$survived, predict_unseen) accuracy_Test <- sum(diag(table_mat)) / sum(table_mat) accuracy_Test }

You can try to tune the parameters and see if you can improve the model over the default value. As a reminder, you need to get an accuracy higher than 0.78

control <- rpart.control(minsplit = 4, minbucket = round(5 / 3), maxdepth = 3, cp = 0) tune_fit <- rpart(survived~., data = data_train, method = 'class', control = control) accuracy_tune(tune_fit)

**Output:**

## [1] 0.7990431

With the following parameter:

minsplit = 4 minbucket= round(5/3) maxdepth = 3cp=0

You get a higher performance than the previous model. Congratulation!

### Summary

We can summarize the functions to train a decision trees algorithm.

Library | Objective | function | class | parameters | details |
---|---|---|---|---|---|

rpart | Train classification trees | rpart() | class | formula, df, method | |

rpart | Train regression tree | rpart() | anova | formula, df, method | |

rpart | Plot the trees | rpart.plot() | fitted model | ||

base | predict | predict() | class | fitted model, type | |

base | predict | predict() | prob | fitted model, type | |

base | predict | predict() | vector | fitted model, type | |

rpart | Control parameters | rpart.control() | minsplit | Set the minimum number of observations in the node before the algorithm perform a split | |

minbucket | Set the minimum number of observations in the final note i.e. the leaf | ||||

maxdepth | Set the maximum depth of any node of the final tree. The root node is treated a depth 0 | ||||

rpart | Train model with control parameter | rpart() | formula, df, method, control |

Note : Train the model on a training data and test the performance on an unseen dataset, i.e. test set.