DA5030 kNN

Question 2

# URL of the CSV file
url <- "https://s3.us-east-2.amazonaws.com/artificium.us/datasets/Prostate_Cancer_Modified.csv"

# Read the CSV file into a data frame
df <- read.csv(url, stringsAsFactors = FALSE, header = T)

# Show the first five rows of df
head(df, 5)

Question 3

# Calculate the 10% trimmed mean of the smoothness column
smoothness_trimmed_mean <-  mean(df$smoothness, trim = 0.1, na.rm = TRUE)

# Replace NAs with the trimmed mean
df$smoothness[is.na(df$smoothness)] <- smoothness_trimmed_mean

Question 4

library(dplyr)

# Function to calculate z-score
z_score <- function(x) {
  (x - mean(x)) / sd(x)
}

# Select all numeric columns
numeric_columns <- df %>% select(-id, -diagnosis_result)

# Calculate z-score
df.norm <- df %>%
  mutate(across(all_of(names(numeric_columns)), z_score))

Question 5

library(caret)

# Use caret function createDataPartition to split data into training and test sets
# Example from kNN practice exercise
indxTrain <- createDataPartition(y = df.norm$diagnosis_result, p = 0.8, list = FALSE)
df.train <- df.norm[indxTrain,]
df.val <- df.norm[-indxTrain,]

Question 6

# Function to calculate class distribution
# Asked ChatGPT for help formatting the output of this function
class_distribution <- function(data) {
  distribution <- table(data$diagnosis_result)
  proportions <- prop.table(distribution)
  return(data.frame(Class = names(distribution), Count = as.vector(distribution), Proportion = as.vector(proportions)))
}

# Analyze distributions
original_distribution <- class_distribution(df.norm)
train_distribution <- class_distribution(df.train)
val_distribution <- class_distribution(df.val)

# Print distributions
print(original_distribution)

##   Class Count Proportion
## 1     B    39  0.3823529
## 2     M    63  0.6176471

print(train_distribution)

##   Class Count Proportion
## 1     B    32  0.3855422
## 2     M    51  0.6144578

print(val_distribution)

##   Class Count Proportion
## 1     B     7  0.3684211
## 2     M    12  0.6315789

• Class Imbalance:
  • The training set has a class distribution that is very similar to the original dataset
  • The validation set also retains a similar distribution, but the absolute counts are lower because of the smaller amount of data
• Proportionality Compared to the Original Dataset:
  • The proportions of Benign and Malignant classifications in both the training and validation sets are consistent with the original data
  • Class imbalance is not a major issue
• Addressing Imbalance:
  • Given that the proportions are relatively consistent across the datasets, the need for intervention is low
  • If further analysis reveals that model performance is lacking, oversampling minority class B may yield better results

Question 7:

# Calculate k as the square root of the number of observations in df.train
k <- round(sqrt(nrow(df.train)))

# Train the KNN model using caret
model <- train(diagnosis_result ~ ., data = df.train, method = "knn", tuneGrid = expand.grid(k = k))

# Output the model summary
print(model)

## k-Nearest Neighbors 
## 
## 83 samples
##  9 predictor
##  2 classes: 'B', 'M' 
## 
## No pre-processing
## Resampling: Bootstrapped (25 reps) 
## Summary of sample sizes: 83, 83, 83, 83, 83, 83, ... 
## Resampling results:
## 
##   Accuracy   Kappa    
##   0.6482241  0.2884335
## 
## Tuning parameter 'k' was held constant at a value of 9

Question 8:

library(gmodels)

# Make predictions on the validation set
predictions <- predict(model, newdata = df.val)

# Create and display the confusion matrix
CrossTable(x = df.val$diagnosis_result, y = predictions, prop.chisq = FALSE)

## 
##  
##    Cell Contents
## |-------------------------|
## |                       N |
## |           N / Row Total |
## |           N / Col Total |
## |         N / Table Total |
## |-------------------------|
## 
##  
## Total Observations in Table:  19 
## 
##  
##                         | predictions 
## df.val$diagnosis_result |         B |         M | Row Total | 
## ------------------------|-----------|-----------|-----------|
##                       B |         2 |         5 |         7 | 
##                         |     0.286 |     0.714 |     0.368 | 
##                         |     0.250 |     0.455 |           | 
##                         |     0.105 |     0.263 |           | 
## ------------------------|-----------|-----------|-----------|
##                       M |         6 |         6 |        12 | 
##                         |     0.500 |     0.500 |     0.632 | 
##                         |     0.750 |     0.545 |           | 
##                         |     0.316 |     0.316 |           | 
## ------------------------|-----------|-----------|-----------|
##            Column Total |         8 |        11 |        19 | 
##                         |     0.421 |     0.579 |           | 
## ------------------------|-----------|-----------|-----------|
## 
## 

Question 9:

kNN Model Performance Summary

• The kNN algorithm exhibits an overall accuracy of 42.11% with a true positive rate of 50% and a true negative rate of 28.57%.