Efficiency_of_Fire_Extinguishers.Rmd

---
title: "Efficiency of Fire Extinguishers"
output:
  pdf_document: default
  html_notebook: default
  html_document:
    df_print: paged
  word_document: default
---

```{r}
library(readxl)
library(tidyverse)
library(moments)
library(car)
library(corrplot)
library(gridExtra)
library(Boruta)
library(caret)
library(e1071)
library(ROCR)
library(nortest)
library(psych)
library(vcd)
```

# Problem description

There is a dataset [publicly available](https://www.muratkoklu.com/datasets/vtdhnd07.php), which provides the result of several experiments carried out with sound wave fire extinguishing systems. The objective of this work is to build a machine learning model capable of classifying whether a fire extinguishing system using sound waves is capable or not of extinguishing a heat source.

## Loading the Data

```{r}
data <-read_excel("data/Acoustic_Extinguisher_Fire_Dataset.xlsx", sheet="A_E_Fire_Dataset")
```


# 01 - Data Description

## 1.1 Rename Columns


```{r}
colnames(data) <- c('size', 'fuel', 'distance', 'desibel', 'airflow', 'frequency', 'status')
colnames(data)
```

## 1.2 Visualizing the data

```{r}
head(data)
```


## 1.2 Data Dimension 

```{r}
print(paste("Number of rows ", nrow(data)))
print(paste("Number of columns ", ncol(data)))
```



## 1.3 Data Types

```{r}
str(data)
```

## 1.4 Change Types
```{r}
data$status <- as.factor(data$status)
data$fuel <- as.factor(data$fuel)
data$size <- factor(data$size, ordered = TRUE)
str(data)
```


## 1.5 Check NAs

```{r}
# Row count with complete data
sum(complete.cases(data))
```

```{r}
# Row count with incomplete data
sum(!complete.cases(data))
```

## 1.6 Check Balanced Data
```{r}
round(100*prop.table(table(data$status)),2)
```


## 1.7 descriptive Statistics

```{r}
# Check the number of unique values in each column
quantidade_valores_unicos <- sapply(data, function(col) length(unique(col)))
print(quantidade_valores_unicos)
rm(quantidade_valores_unicos)
```

### 1.7.1 Numerical Attributes

```{r}

get_descriptive_analysis <- function(dados){

mean <- apply(dados, 2, mean, na.rm = TRUE)
std <- apply(dados, 2, sd, na.rm = TRUE)
min <- apply(dados, 2, min, na.rm = TRUE)
max <- apply(dados, 2, max, na.rm = TRUE)
median <- apply(dados, 2, median, na.rm = TRUE)
quartis <- apply(dados, 2, quantile, probs = c(0.25, 0.75), na.rm = TRUE)
IQR <- apply(dados, 2, IQR, na.rm = TRUE)
skew <- apply(dados, 2, skewness, na.rm = TRUE)
kurtosis <- apply(dados, 2, kurtosis, na.rm = TRUE)

result <- data.frame(min, max, max-min, mean, median, std, skew, kurtosis)

names(result) <- c("min", "max", "range", "mean", "median", "std", "skew", "kurtosis")

return(result)
}

get_descriptive_analysis(data %>% select_if(is.numeric))
```
# 2 - Exploratory Data Analysis

## 2.1 Univariate Analysis

```{r}
# Plotting boxplots for each numeric variable on a single graph
data %>% select_if(is.numeric) %>%
  gather(variable, value ) %>%
  ggplot( aes(x=value)) +
  geom_boxplot(fill='steelblue') +
  facet_wrap(~ variable, ncol = 2, scales = "free")
```
There are no numerical data with outliers.

```{r}
# Plotting histograms for each numeric variable on a single graph
data %>% select_if(is.numeric) %>%
  gather(variable, value) %>%
  ggplot( aes(x=value)) +
  geom_histogram(fill='steelblue') +
  facet_wrap(~ variable, ncol = 2, scales = "free")
```

None of the numeric variables seems to follow a distribution close to normal.

```{r}
# Set up the figure structure
par(mar = c(2, 2, 2, 2))  # Set smaller margins
layout(matrix(1:20, 5, 4, byrow = TRUE))

# Initialize the list to save p-values
p_values <- list()

for (column in colnames(data)) {
  if (is.numeric(data[[column]])) {
    
    # Calculate the p-value with Anderson-Darling normality test
    p_value <- ad.test(data[[column]])$p.value
    
    # Add the p-value to the list
    p_values[[column]] <- p_value
  }
}

as.data.frame(p_values) %>%
  gather(variable, 'p_value' ) %>%
  mutate(rejected_H0 = p_value < 0.05)

rm(p_values, p_value, column)
```

There is statistically significant evidence to reject the null hypothesis that each of the numeric predictor variables follow a normal distribution.


## 2.2 Multivariate Analysis


### 2.2.1 Numerical Variables

```{r}
# Exploring relationships between numerical variables
pairs.panels(data %>% select_if(is.numeric), method = "spearman")
```

### 2.2.2 Categorical Variables

```{r}

cat_cols <- sapply(data, is.factor)

nCatAttributes <- length(which(cat_cols))

matriz_assoc <- matrix(0, ncol = nCatAttributes, 
                       nrow = nCatAttributes, 
                       dimnames = list(names(data)[which(cat_cols)], names(data)[which(cat_cols)]))

Cat_df <- as.data.frame(data[,which(cat_cols)])

for (i in 1:(nCatAttributes - 1)) {
  for (j in (i+1):nCatAttributes) {
    tabela <- table(Cat_df[, i], Cat_df[, j])
    matriz_assoc[i, j] <- assocstats(tabela)$cramer
    matriz_assoc[j, i] <- matriz_assoc[i, j]
  }
}

diag(matriz_assoc) <- 1

corrplot(matriz_assoc, method = "color", type = "upper", order = "hclust", tl.cex = 0.8, tl.srt = 45, addCoef.col = "black")

rm(Cat_df, matriz_assoc, nCatAttributes, cat_cols, i, j, tabela)
```
## 2.3 Bivariate Analysis


#### Hypothesis 1: The distance of the extinguisher from the heat source is related to its effectiveness (success in extinguishing the fire).

To verify this hypothesis, the t-Test for two independent samples will first be evaluated, which has the following premises:

- 1. Data are random and representative of the population.
- 2. The dependent variable is continuous.
- 3. Both groups are independent (i.e. exhaustive and exclusionary groups).
- 4. The residuals of the model are normally distributed.
- 5. The residual variance is homogeneous (principle of homoscedasticity).

For our example in this case study, we will assume as true the
assumptions 1 to 3 and we will validate assumptions 4 and 5. For assumption 4 we will use
the Shapiro-Wilk Test and for assumption 5 we will use the F Test, in case the data in each group is normally distributed.

```{r}

# Shapiro-Wilk Normality Test in each traction group
# H0: Data is normally distributed
# H1: Data is not normally distributed

# Subsampling of groups, as there is a maximum limitation of 5000 samples
group_0 <- data$distance[data$status == 0]
group_0_sample <- sample(group_0, min(5000, length(group_0)))

group_1 <- data$distance[data$status == 1]
group_1_sample <- sample(group_1, min(5000, length(group_1)))

# Shapiro-Wilk normality test for subsamples
shapiro_0 <- shapiro.test(group_0_sample)
shapiro_1 <- shapiro.test(group_1_sample)

print(shapiro_0)
print(shapiro_1)

rm(group_0, group_0_sample, group_1, group_1_sample, shapiro_0, shapiro_1)

```

As the p-value is less than 0.05, there is enough statistical evidence to reject the null hypothesis of normality of the data in each group. Thus, the assumption of normality of the residuals is discarded. As the assumption of normality is not met, it is recommended to use a non-parametric test to compare samples, such as the Mann-Whitney test.


```{r}
# Mann-Whitney Test
# H0 (Null Hypothesis): The distributions of the two independent samples are identical.
# HA (Alternative Hypothesis): The distributions of the two independent samples are not identical.

group_0 <- data$distance[data$status == 0]
group_1 <- data$distance[data$status == 1]

wilcox_test_result <- wilcox.test(group_0, group_1)
print(wilcox_test_result)

rm(group_0, group_1, wilcox_test_result)
```

Hypothesis 1 is true. As the p-value returned by the Mann-Whitney test is less than 0.05, it is concluded that there is statistical evidence to reject that the distribution of samples between groups is identical.

The graphs below were generated to visualize the relationship between the variables 'distance' and 'status'.

```{r}
p1 <- ggplot(data, aes(x = distance, fill = factor(status))) +
  geom_density(alpha = 0.5) +
  labs(fill = "Status") +
  theme(legend.position = "top")

p2 <- ggplot(data, aes(x = factor(status), y = distance, fill = factor(status))) +
  geom_boxplot() +
  labs(x = "Status", y = "Distance", fill = "Status") +
  theme(legend.position = "none")

p3 <- ggplot(data, aes(x = distance, fill = factor(status))) +
  geom_bar(position = "fill") +
  labs(x = "Distance", y = "Proportion", fill = "Status") +
  theme(legend.position = "none")

p4 <- ggplot(data, aes(x = distance, y = factor(status), color = factor(status))) +
  geom_point() +
  labs(x = "Distance", y = "Status", color = "Status") +
  theme(legend.position = "none")

# Arrange subplots side by side
grid.arrange(p1, p2, p3, p4, ncol = 2, top = guide_legend(title = "Status"))

rm(p1, p2, p3, p4)
```
From the graphs, it can be seen that as the distance increases, the proportion of cases of failure increase and the proportion of cases of success decrease.


#### Hypothesis 2: The sound frequency emitted by the extinguisher is related to its effectiveness (success in extinguishing the fire).

To verify this hypothesis, the t-Test for two independent samples (success=0 and success=1) will first be evaluated.

But before carrying out the t-test, the assumption of normality must be evaluated.

```{r}

# Shapiro-Wilk Normality Test in each traction group
# H0: Data is normally distributed
# H1: Data is not normally distributed

# Subsampling of groups, as there is a maximum limitation of 5000 samples
group_0 <- data$frequency[data$status == 0]
group_0_sample <- sample(group_0, min(5000, length(group_0)))

group_1 <- data$frequency[data$status == 1]
group_1_sample <- sample(group_1, min(5000, length(group_1)))

# Shapiro-Wilk normality test for subsamples
shapiro_0 <- shapiro.test(group_0_sample)
shapiro_1 <- shapiro.test(group_1_sample)

print(shapiro_0)
print(shapiro_1)

rm(group_0, group_0_sample, group_1, group_1_sample, shapiro_0, shapiro_1)

```
As the p-value is less than 0.05, there is enough statistical evidence to reject the null hypothesis of normality of the data in each group. Thus, the assumption of normality of the residuals is discarded. As the assumption of normality is not met, it is recommended to use a non-parametric test to compare samples, such as the Mann-Whitney test.

```{r}
# Mann-Whitney Test
# H0 (Null Hypothesis): The distributions of the two independent samples are identical.
# HA (Alternative Hypothesis): The distributions of the two independent samples are not identical.

group_0 <- data$frequency[data$status == 0]
group_1 <- data$frequency[data$status == 1]

wilcox_test_result <- wilcox.test(group_0, group_1)
print(wilcox_test_result)

rm(group_0, group_1, wilcox_test_result)
```

Hypothesis 2 is true. As the p-value returned by the Mann-Whitney test is less than 0.05, it is concluded that there is statistical evidence to reject that the distribution of samples between groups is identical.

The graphs below were generated to visualize the relationship between the variables 'frequency' and 'status'.

```{r}
p1 <- ggplot(data, aes(x = frequency, fill = factor(status))) +
  geom_density(alpha = 0.5) +
  labs(fill = "Status") +
  theme(legend.position = "top")

p2 <- ggplot(data, aes(x = factor(status), y = frequency, fill = factor(status))) +
  geom_boxplot() +
  labs(x = "Status", y = "Frequency", fill = "Status") +
  theme(legend.position = "none")

p3 <- ggplot(data, aes(x = frequency, fill = factor(status))) +
  geom_bar(position = "fill") +
  labs(x = "Frequency", y = "Proportion", fill = "Status") +
  theme(legend.position = "none")

p4 <- ggplot(data, aes(x = frequency, y = factor(status), color = factor(status))) +
  geom_point() +
  labs(x = "Frequency", y = "Status", color = "Status") +
  theme(legend.position = "none")

# Arrange subplots side by side
grid.arrange(p1, p2, p3, p4, ncol = 2, top = guide_legend(title = "Status"))

rm(p1, p2, p3, p4)
```
It is observed that there are cases of success and failure in the entire frequency range that was evaluated. However, it is observed that the proportion of successful cases have an increasing behavior between 0 and 20Hz and decreasing from 20 Hz.

#### Hypothesis 3: The amplitude of the sound waves, measured in db, emitted by the extinguisher, is related to its effectiveness.

To verify this hypothesis, the t-Test for two independent samples (success=0 and success=1) will first be evaluated.

But before carrying out the t-test, the assumption of normality must be evaluated.

```{r}
# Shapiro-Wilk Normality Test in each traction group
# H0: Data is normally distributed
# H1: Data is not normally distributed

# Subsampling of groups, as there is a maximum limitation of 5000 samples
group_0 <- data$desibel[data$status == 0]
group_0_sample <- sample(group_0, min(5000, length(group_0)))

group_1 <- data$desibel[data$status == 1]
group_1_sample <- sample(group_1, min(5000, length(group_1)))

# Shapiro-Wilk normality test for subsamples
shapiro_0 <- shapiro.test(group_0_sample)
shapiro_1 <- shapiro.test(group_1_sample)

print(shapiro_0)
print(shapiro_1)

rm(group_0, group_0_sample, group_1, group_1_sample, shapiro_0, shapiro_1)
```
As the p-value is less than 0.05, there is enough statistical evidence to reject the null hypothesis of normality of the data in each group. Thus, the assumption of normality of the residuals is discarded. As the assumption of normality is not met, it is recommended to use a non-parametric test to compare samples, such as the Mann-Whitney test.

```{r}
# Mann-Whitney Test
# H0 (Null Hypothesis): The distributions of the two independent samples are identical.
# HA (Alternative Hypothesis): The distributions of the two independent samples are not identical.
group_0 <- data$desibel[data$status == 0]
group_1 <- data$desibel[data$status == 1]

wilcox_test_result <- wilcox.test(group_0, group_1)
print(wilcox_test_result)

rm(group_0, group_1, wilcox_test_result)
```
Hypothesis 3 is true. As the p-value returned by the Mann-Whitney test is less than 0.05, it is concluded that there is statistical evidence to reject that the distribution of samples between groups is identical.

The graphs below were generated to visualize the relationship between the variables 'desibel' and 'status'.

```{r}
p1 <- ggplot(data, aes(x = desibel, fill = factor(status))) +
  geom_density(alpha = 0.5) +
  labs(fill = "Status") +
  theme(legend.position = "top")

p2 <- ggplot(data, aes(x = factor(status), y = desibel, fill = factor(status))) +
  geom_boxplot() +
  labs(x = "Status", y = "Amplitude (db)", fill = "Status") +
  theme(legend.position = "none")

p3 <- ggplot(data, aes(x = desibel, fill = factor(status))) +
  geom_bar(position = "fill") +
  labs(x = "Amplitude (db)", y = "Proportion", fill = "Status") +
  theme(legend.position = "none")

p4 <- ggplot(data, aes(x = desibel, y = factor(status), color = factor(status))) +
  geom_point() +
  labs(x = "Amplitude (db)", y = "Status", color = "Status") +
  theme(legend.position = "none")

# Arrange subplots side by side
grid.arrange(p1, p2, p3, p4, ncol = 2, top = guide_legend(title = "Status"))

rm(p1, p2, p3, p4)
```
It is observed that for amplitudes below 75 db there are no cases of success and for amplitudes above 110 db there are no cases of failure. The proportion of successful cases grows from 75 db and reaches a peak around 90 db. Between 90 and 98 db the success rate drops and after 98 db it grows.

#### Hypothesis 4: The air flow velocity, measured in m/s, is related to the effectiveness of the extinguisher.

To verify this hypothesis, the t-Test for two independent samples (success=0 and success=1) will first be evaluated.

But before carrying out the t-test, the assumption of normality must be evaluated.

```{r}
# Shapiro-Wilk Normality Test in each traction group
# H0: Data is normally distributed
# H1: Data is not normally distributed

# Subsampling of groups, as there is a maximum limitation of 5000 samples
group_0 <- data$airflow[data$status == 0]
group_0_sample <- sample(group_0, min(5000, length(group_0)))

group_1 <- data$airflow[data$status == 1]
group_1_sample <- sample(group_1, min(5000, length(group_1)))

# Shapiro-Wilk normality test for subsamples
shapiro_0 <- shapiro.test(group_0_sample)
shapiro_1 <- shapiro.test(group_1_sample)

print(shapiro_0)
print(shapiro_1)

rm(group_0, group_0_sample, group_1, group_1_sample, shapiro_0, shapiro_1)
```
As the p-value is less than 0.05, there is enough statistical evidence to reject the null hypothesis of normality of the data in each group. Thus, the assumption of normality of the residuals is discarded. As the assumption of normality is not met, it is recommended to use a non-parametric test to compare samples, such as the Mann-Whitney test.

```{r}
# Mann-Whitney Test
# H0 (Null Hypothesis): The distributions of the two independent samples are identical.
# HA (Alternative Hypothesis): The distributions of the two independent samples are not identical.
group_0 <- data$airflow[data$status == 0]
group_1 <- data$airflow[data$status == 1]

wilcox_test_result <- wilcox.test(group_0, group_1)
print(wilcox_test_result)

rm(group_0, group_1, wilcox_test_result)
```

Hypothesis 4 is true. As the p-value returned by the Mann-Whitney test is less than 0.05, it is concluded that there is statistical evidence to reject that the distribution of samples between groups is identical.

The graphs below were generated to visualize the relationship between the variables 'airflow' and 'status'.

```{r}
p1 <- ggplot(data, aes(x = airflow, fill = factor(status))) +
  geom_density(alpha = 0.5) +
  labs(fill = "Status") +
  theme(legend.position = "top")

p2 <- ggplot(data, aes(x = factor(status), y = airflow, fill = factor(status))) +
  geom_boxplot() +
  labs(x = "Status", y = "Airflow (m/s)", fill = "Status") +
  theme(legend.position = "none")

p3 <- ggplot(data, aes(x = airflow, fill = factor(status))) +
  geom_bar(position = "fill") +
  labs(x = "Airflow (m/s)", y = "Proportion", fill = "Status") +
  theme(legend.position = "none")

p4 <- ggplot(data, aes(x = airflow, y = factor(status), color = factor(status))) +
  geom_point() +
  labs(x = "Airflow (m/s)", y = "Status", color = "Status") +
  theme(legend.position = "none")

# Arrange subplots side by side
grid.arrange(p1, p2, p3, p4, ncol = 2, top = guide_legend(title = "Status"))

rm(p1, p2, p3, p4)
```
It is observed that as the airflow velocity increases, the proportion of successful in extinguishing the fire increases.


#### Hypothesis 5: There is no relatioship between the fuel used to create the flame and the effectiveness of the extinguisher.

To validate a hypothesis, the chi-squared test will be performed.


```{r}
# chi-squared Test
# H0 - There is no relationship between x and y
# HA - x and y are related

# Create a contingency table with the observed frequencies
contingency_table <- table(data$fuel, data$status)

# Perform the chi-squares test
result <- chisq.test(contingency_table)

# Show the result of test
print(result)

rm(contingency_table, result)
```

Hypothesis 5 is false. As the p-value returned by the chi-squared test is less than 0.05, it is concluded that there is statistical evidence to confirm that 'status' and 'fuel' are related.

The graphs below were generated to visualize the relationship between the variables 'fuel' and 'status'.
```{r}
p1 <- ggplot(data, aes(x = factor(fuel), fill = factor(status))) +
  geom_bar(position = "dodge") +
  labs(x = "Fuel", y = "Count", fill = "Status") +
  theme(legend.position = "top")

p2 <- ggplot(data, aes(x = factor(fuel), fill = factor(status))) +
  geom_bar(position = "fill") +
  labs(x = "Fuel", y = "Proportion", fill = "Status") +
  theme(legend.position = "top")

# Arrange subplots side by side
grid.arrange(p1, p2, ncol = 2, top = guide_legend(title = "Status"))

rm(p1, p2)
```
- Gasoline and lpg have more success than failures. Kerosene and Thinner present more cases of failure than success in extinguishing fires.

#### Hypothesis 6: There is no relatioship between the size of liquid fuel cans used to create the flame and the effectiveness of the extinguisher.

To validate a hypothesis, the chi-squared test will be performed.

```{r}

# chi-squared Test
# H0 - There is no relationship between x and y
# HA - x and y are related

# Create a contingency table with the observed frequencies
contingency_table <- table(data$size, data$status)

# Perform the chi-squares test
result <- chisq.test(contingency_table)

# Show the result of test
print(result)

rm(contingency_table, result)
```
Hypothesis 6 is false. As the p-value returned by the chi-squared test is less than 0.05, it is concluded that there is statistical evidence to confirm that 'status' and 'size' are related.

The graphs below were generated to visualize the relationship between the variables 'size' and 'status'.

```{r}
p1 <- ggplot(data, aes(x = factor(size), fill = factor(status))) +
  geom_bar(position = "dodge") +
  labs(x = "Size", y = "Count", fill = "Status") +
  theme(legend.position = "top")

p2 <- ggplot(data, aes(x = factor(size), fill = factor(status))) +
  geom_bar(position = "fill") +
  labs(x = "Size", y = "Proportion", fill = "Status") +
  theme(legend.position = "top")

# Arrange subplots side by side
grid.arrange(p1, p2, ncol = 2, top = guide_legend(title = "Status"))

rm(p1, p2)
```

- It is observed that as the size of the fuel cans increases, there is a tendency for the failure rate to increase and the success rate to decrease in extinguishing the fire. However, this pattern isn't observed for sizes 6 and 7, which have 1026 records each, while for sizes 1 to 5 there are 3078 records each.

# 3 - Data Preparation

## 3.1 Division of Data into Training and Test Sets

```{r}
# Set a seed for reproducibility
set.seed(123)  
indexes <- createDataPartition(data$status, p = 0.8, list = FALSE)
train_data <- data[indexes, ]
test_data <- data[-indexes, ]

rm(indexes)
```


## 3.2 Rescaling

According to the univariate analysis of numerical variables (section 3.1), no variable has a normal or close to normal distribution. Min-Max normalization is more suitable when the data do not follow a normal distribution, thus preventing the data distribution from being distorted.

```{r}

# Gets all numeric column names
num_cols <- names(train_data %>% select_if(is.numeric))

# Create the pre process object
preprocess_obj <- preProcess(train_data[, num_cols], method = "range")

# Apply the MinMax transformation to the training data
num_train_data_normalized <- predict(preprocess_obj, train_data[, num_cols])

# Apply the MinMax transformation to the test data
num_test_data_normalized <- predict(preprocess_obj, test_data[, num_cols])

rm(num_cols, preprocess_obj)

```

## 3.3 Encoding


```{r}
# Create an object dummyVars (fuel col)
dummy_obj <- dummyVars(~ fuel, data = train_data)

# Aplly one-hot encoding to train data
fuel_train_encoded <- as.data.frame(predict(dummy_obj, newdata = train_data))

# Aplly one-hot encoding to test data
fuel_test_encoded <- as.data.frame(predict(dummy_obj, newdata = test_data))

rm(dummy_obj)
```

## 3.4 Concatenation

```{r}
# Concatenates the training data
train_data <- cbind(num_train_data_normalized,
                    fuel_train_encoded,
                    train_data[, c('size', 'status')])

# Concatenates the test data
test_data <- cbind(num_test_data_normalized,
                    fuel_test_encoded,
                    test_data[, c('size', 'status')])

rm(num_train_data_normalized, fuel_train_encoded)
rm(num_test_data_normalized, fuel_test_encoded)

# Check col names
names(train_data)
```

# 4 - Dimensionality reduction

## 4.1 PCA

```{r}
# Remove the dependent variable from the training dataset
train_data_pca <- subset(train_data, select = -status)
test_data_pca <- subset(test_data, select = -status)

# Convert variable "size" to numeric with ordinal encoding
train_data_pca$size <- as.numeric(train_data_pca$size)
test_data_pca$size <- as.numeric(test_data_pca$size)

# Application of the PCA
pca_model <- prcomp(train_data_pca, center = FALSE, scale = FALSE)

# Analysis of results
summary(pca_model)
plot(pca_model, type = "l", main = "Variance Explained")

```
```{r}
# Get the cumulative variance ratio
cumulative_variance <- cumsum(pca_model$sdev^2) / sum(pca_model$sdev^2)

# Find the minimum number of components that explain 99.9% of the variance
num_components <- min(which(cumulative_variance >= 0.999))

# Get the transformed data
train_data_pca <- as.data.frame(predict(pca_model, newdata = train_data_pca)[, 1:num_components])
test_data_pca <- as.data.frame(predict(pca_model, newdata = test_data_pca)[, 1:num_components])

train_data_pca <- cbind(train_data_pca, train_data$status)
test_data_pca <- cbind(test_data_pca, test_data$status)

names(train_data_pca) <- c("PC1", "PC2", "PC3", "PC4", "PC5", "PC6", "PC7", "status")
names(test_data_pca) <- c("PC1", "PC2", "PC3", "PC4", "PC5", "PC6", "PC7", "status")

print(paste('The dimensionality reduction of the data was from', ncol(train_data)-1, 'to', ncol(train_data_pca)-1, 'columns'))
```


# 5 - Machine Learning Modeling

## 5.1 - Random Model - Baseline

```{r}
# Generating random predictions
y_hat <- as.factor(sample(c(0,1), size = nrow(test_data_pca), replace = TRUE))

# Computing confusion matrix
cm_baseline <- confusionMatrix(test_data_pca$status, y_hat, mode = "prec_recall", positive = "1")

cm_baseline
```

## 5.2 - Logistic Regression

```{r}
#  Cross validation with 5 folds
ctrl <- trainControl(method = "cv", number = 5)

# Train model
model_lr  <- train(status ~ ., data = train_data_pca, method = "glm", trControl = ctrl)

# prediction
yhat_lr = predict(model_lr, newdata = test_data_pca)

# Computing confusion matrix
cm_glm <- confusionMatrix(test_data_pca$status, yhat_lr, mode = "prec_recall", positive = "1")
cm_glm
```

## 5.3 - Decision Tree

```{r}
#  Cross validation with 5 folds
ctrl <- trainControl(method = "cv", number = 5)

# Treinar o modelo usando o método rpart, avalaiando 3 valores diferentes de
# custo de complexidade,
model_dt  <- train(status ~ ., data = train_data_pca, method = "rpart", trControl = ctrl, tuneLength = 10)

print(model_dt)
plot(model_dt)
```


```{r}
# prediction
yhat_dt = predict(model_dt, newdata = test_data_pca)

# Computing confusion matrix
cm_rpart <- confusionMatrix(test_data_pca$status, yhat_dt, mode = "prec_recall", positive = "1")
cm_rpart
```

## 5.4 - K-Nearest Neighbors 

```{r}
#  Cross validation with 5 folds
ctrl <- trainControl(method = "cv", number = 5)

#Train the model using 10 different values of os neighbors, chosen automatically
model_knn  <- train(status ~ ., data = train_data_pca, method = "knn", trControl = ctrl, tuneLength = 10)

print(model_knn$finalModel)
plot(model_knn)

```

```{r}
# prediction
yhat_knn = predict(model_knn, newdata = test_data_pca)

# Computing confusion matrix
cm_knn <- confusionMatrix(test_data_pca$status, yhat_knn, mode = "prec_recall", positive = "1")
cm_knn
```

## 5.5 - Random Forest 

```{r}
# Cross validation with 5 folds
ctrl <- trainControl(method = "cv", number = 5)

# Train the model using 10 different neighborhood count values
model_rf  <- train(status ~ ., data = train_data_pca, method = "rf", trControl = ctrl, tuneLength = 10)

print(model_knn$finalModel)
plot(model_knn)
```

```{r}
# prediction
yhat_rf = predict(model_rf, newdata = test_data_pca)

# Computing confusion matrix
cm_rf <- confusionMatrix(test_data_pca$status, yhat_rf, mode = "prec_recall", positive = "1")
cm_rf
```

## 5.6 - Neural Network - 1 Hidden Layer
```{r}
#  Cross validation with 5 folds
ctrl <- trainControl(method = "cv", number = 5, verbose = FALSE)

# Train the model using 5 different values for the regularization factor and 5 values
# different neurons in the hidden layer, chosen automatically.
model_nnet <- train(status ~ ., data = train_data_pca, method = "nnet", trControl = ctrl, tuneLength=5)

plot(model_nnet)
print(model_nnet)
```

```{r}
# prediction
yhat_nnet = predict(model_nnet, newdata = test_data_pca)

# Computing confusion matrix
cm_nnet <- confusionMatrix(test_data_pca$status, yhat_nnet, mode = "prec_recall", positive = "1")
cm_nnet
```


## 5.7 - XGBoost

```{r}
#  Cross validation with 5 folds
ctrl <- trainControl(method = "cv", number = 5, verbose = FALSE)

# Train the model using 5 different values for each hyperparameter. The 5 values are chosen automatically
model_xgbTree <- train(status ~ ., data = train_data_pca, method = "xgbTree", trControl = ctrl, tuneLength=5)

plot(model_xgbTree)
print(model_xgbTree)
```
```{r}
# prediction
yhat_xgbTree = predict(model_xgbTree, newdata = test_data_pca)

# Computing confusion matrix
cm_xgbTree <- confusionMatrix(test_data_pca$status, yhat_xgbTree, mode = "prec_recall", positive = "1")
cm_xgbTree
```

## 7.8 Compare Model's Performance

```{r}
# Calculate predicted probabilities for each model in a test dataset
pred_model1 <- predict(model_lr, newdata = test_data_pca, type = "prob")
pred_model2 <- predict(model_dt, newdata = test_data_pca, type = "prob")
pred_model3 <- predict(model_knn, newdata = test_data_pca, type = "prob")
pred_model4 <- predict(model_rf, newdata = test_data_pca, type = "prob")
pred_model5 <- predict(model_nnet, newdata = test_data_pca, type = "prob")
pred_model6 <- predict(model_xgbTree, newdata = test_data_pca, type = "prob")

# Create prediction objects for each model
pred_model1 <- prediction(pred_model1[, "1"], test_data$status)
pred_model2 <- prediction(pred_model2[, "1"], test_data$status)
pred_model3 <- prediction(pred_model3[, "1"], test_data$status)
pred_model4 <- prediction(pred_model4[, "1"], test_data$status)
pred_model5 <- prediction(pred_model5[, "1"], test_data$status)
pred_model6 <- prediction(pred_model6[, "1"], test_data$status)

# Calculate performance metrics for each model
perf_model1 <- performance(pred_model1, "tpr", "fpr")
perf_model2 <- performance(pred_model2, "tpr", "fpr")
perf_model3 <- performance(pred_model3, "tpr", "fpr")
perf_model4 <- performance(pred_model4, "tpr", "fpr")
perf_model5 <- performance(pred_model5, "tpr", "fpr")
perf_model6 <- performance(pred_model6, "tpr", "fpr")


# Plot the overlapping ROC curves
plot(perf_model1, col = "blue", main = "ROC Curve")
plot(perf_model2, col = "red", add = TRUE)
plot(perf_model3, col = "green", add = TRUE)
plot(perf_model4, col = "orange", add = TRUE)
plot(perf_model5, col = "cyan", add = TRUE)
plot(perf_model6, col = "violet", add = TRUE)

# Calculate AUC values
auc_model1 <- performance(pred_model1, "auc")@y.values[[1]]
auc_model2 <- performance(pred_model2, "auc")@y.values[[1]]
auc_model3 <- performance(pred_model3, "auc")@y.values[[1]]
auc_model4 <- performance(pred_model4, "auc")@y.values[[1]]
auc_model5 <- performance(pred_model5, "auc")@y.values[[1]]
auc_model6 <- performance(pred_model6, "auc")@y.values[[1]]

# Add legend with AUC values
legend("bottomright", legend = c(paste("Logistic Regression' (AUC =", round(auc_model1, 3), ")"),
                                 paste("Decision Tree (AUC =", round(auc_model2, 3), ")"),
                                 paste("k-NN (AUC =", round(auc_model3, 3), ")"),
                                 paste("Random Forest 4 (AUC =", round(auc_model4, 3), ")"),
                                 paste("Neural Network (AUC =", round(auc_model5, 3), ")"),
                                 paste("XGBoost (AUC =", round(auc_model6, 3), ")")),
       col = c("blue", "red", "green", "orange", "cyan", "violet"), lwd = 2)

```



```{r}
# Set the number of decimal places to 3
options(digits = 3)

modelNames <- c('Logistic Regression', 'Decision Tree', 'K-Nearest Neighbors', 'Random Forest ', 'Neural Network - 1 Hidden Layer', 'XGBoost')
accuracies <- c(cm_glm$overall[1], cm_rpart$overall[1], cm_knn$overall[1], cm_rf$overall[1], cm_nnet$overall[1], cm_xgbTree$overall[1])
f1_scores <- c(cm_glm$byClass[7], cm_rpart$byClass[7], cm_knn$byClass[7], cm_rf$byClass[7], cm_nnet$byClass[7], cm_xgbTree$byClass[7])
precisions <- c(cm_glm$byClass[5], cm_rpart$byClass[5], cm_knn$byClass[5], cm_rf$byClass[5], cm_nnet$byClass[5], cm_xgbTree$byClass[5])
recalls <- c(cm_glm$byClass[6], cm_rpart$byClass[6], cm_knn$byClass[6], cm_rf$byClass[6], cm_nnet$byClass[6], cm_xgbTree$byClass[6])
AUCs <- c(auc_model1, auc_model2, auc_model3, auc_model4, auc_model5, auc_model6)
data.frame('Model Name' = modelNames,
           'Accuracy' = accuracies,
           'F1-Score' = f1_scores,
           'Precision' = precisions,
           'Recall' = recalls,
           'AUC' = AUCs)
```