Chhabra_Ridhi_Module8_Lab

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install.packages("tidymodels")

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install.packages("kknn")

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install.packages("rsample")

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library(kknn)
library(tidymodels)

── Attaching packages ─────────────
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✖ dplyr::lag()     masks stats::lag()
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• Use tidymodels_prefer() to resolve common conflicts.

library(rsample)

Modeling tasks:

1. Is this a supervised or unsupervised learning problem? Why?

This is a supervised learning problem because we have a labeled target variable (cmedv) that we want to predict based on input features.

2. There are 16 variables in this data set. Which variable is the response variable and which variables

are the predictor variables (aka features)?

- Response Variable (Target): cmedv

- Predictor Variables (Features): lon, lat, crim, zn, indus, chas, nox, rm, age, dis, rad, tax, ptratio, lstat

3. Given the type of variable cmedv is, is this a regression or classification problem?

This is a regression problem because cmedv is a continuous numerical variable representing median house prices.

# 4. Fill in the blanks to import the Boston housing data set (boston.csv). Are there any missing values?
# What is the minimum and maximum values of cmedv? What is the average cmedv value?

boston <- readr::read_csv("boston.csv")

Rows: 506 Columns: 16── Column specification ───────────
Delimiter: ","
dbl (16): lon, lat, cmedv, crim...
ℹ Use `spec()` to retrieve the full column specification for this data.
ℹ Specify the column types or set `show_col_types = FALSE` to quiet this message.

str(boston)

spc_tbl_ [506 × 16] (S3: spec_tbl_df/tbl_df/tbl/data.frame)
 $ lon    : num [1:506] -71 -71 -70.9 -70.9 -70.9 ...
 $ lat    : num [1:506] 42.3 42.3 42.3 42.3 42.3 ...
 $ cmedv  : num [1:506] 24 21.6 34.7 33.4 36.2 28.7 22.9 22.1 16.5 18.9 ...
 $ crim   : num [1:506] 0.00632 0.02731 0.02729 0.03237 0.06905 ...
 $ zn     : num [1:506] 18 0 0 0 0 0 12.5 12.5 12.5 12.5 ...
 $ indus  : num [1:506] 2.31 7.07 7.07 2.18 2.18 2.18 7.87 7.87 7.87 7.87 ...
 $ chas   : num [1:506] 0 0 0 0 0 0 0 0 0 0 ...
 $ nox    : num [1:506] 0.538 0.469 0.469 0.458 0.458 0.458 0.524 0.524 0.524 0.524 ...
 $ rm     : num [1:506] 6.58 6.42 7.18 7 7.15 ...
 $ age    : num [1:506] 65.2 78.9 61.1 45.8 54.2 58.7 66.6 96.1 100 85.9 ...
 $ dis    : num [1:506] 4.09 4.97 4.97 6.06 6.06 ...
 $ rad    : num [1:506] 1 2 2 3 3 3 5 5 5 5 ...
 $ tax    : num [1:506] 296 242 242 222 222 222 311 311 311 311 ...
 $ ptratio: num [1:506] 15.3 17.8 17.8 18.7 18.7 18.7 15.2 15.2 15.2 15.2 ...
 $ b      : num [1:506] 397 397 393 395 397 ...
 $ lstat  : num [1:506] 4.98 9.14 4.03 2.94 5.33 ...
 - attr(*, "spec")=
  .. cols(
  ..   lon = col_double(),
  ..   lat = col_double(),
  ..   cmedv = col_double(),
  ..   crim = col_double(),
  ..   zn = col_double(),
  ..   indus = col_double(),
  ..   chas = col_double(),
  ..   nox = col_double(),
  ..   rm = col_double(),
  ..   age = col_double(),
  ..   dis = col_double(),
  ..   rad = col_double(),
  ..   tax = col_double(),
  ..   ptratio = col_double(),
  ..   b = col_double(),
  ..   lstat = col_double()
  .. )
 - attr(*, "problems")=<externalptr> 

# Check for missing values in the dataset
any_missing <- anyNA(boston) || any(boston == "")

# Print the result
print(paste("Any missing values in the dataset? ", any_missing))

[1] "Any missing values in the dataset?  FALSE"

# Minimum and maximum values of cmedv
min(boston$cmedv)

[1] 5

max(boston$cmedv)

[1] 50

# Average cmedv value
mean(boston$cmedv)

[1] 22.52885

# Calculate the median value for cmedv
median_cmedv <- median(boston$cmedv)

# Print the result
print(paste("Median value for cmedv: ", median_cmedv))

[1] "Median value for cmedv:  21.2"

# 5. Fill in the blanks to split the data into a training set and test set using a 70-30% split.
# Be sure to include the set.seed(123) so that your train and test sets are the same size as mine.

# Set seed for reproducibility
set.seed(123)

# Split the data into training and test sets using a 70-30% split

split <- initial_split(boston, prop = 0.7, strata = cmedv)
train <- training(split)
test <- testing(split)

# 6. How many observations are in the training set and test set?

# Number of observations in the training set
nrow(train)

[1] 352

# Number of observations in the test set
nrow(test)

[1] 154

# 7. Compare the distribution of cmedv between the training set and test set.
# Do they appear to have the same distribution or do they differ significantly?

# Ensure cmedv is numeric
train$cmedv <- as.numeric(train$cmedv)
test$cmedv <- as.numeric(test$cmedv)

# Check the class of cmedv in both training and test sets
class(train$cmedv)

[1] "numeric"

class(test$cmedv)

[1] "numeric"

library(ggplot2)

ggplot() +
  geom_histogram(data = train, aes(x = cmedv), fill = "blue", alpha = 0.5, bins = 30) +
  geom_histogram(data = test, aes(x = cmedv), fill = "green", alpha = 0.5, bins = 30) +
  labs(title = "Distribution of cmedv in Training and Test Sets",
       x = "cmedv",
       y = "Frequency") +
  scale_fill_manual(values = c("blue", "green"), name = "Dataset", labels = c("Training", "Test"))

# 8. Fill in the blanks to fit a linear regression model using the rm feature variable to predict cmedv
# and compute the RMSE on the test data. What is the test set RMSE?

# fit model
lm1 <- linear_reg() %>%
  fit(cmedv ~ rm, data = train)

# compute the RMSE on the test data
lm1 %>%
  predict(new_data = test) %>%
  bind_cols(test %>% select(cmedv)) %>%
  rmse(truth = cmedv, estimate = .pred)

NA

# 9. Fill in the blanks to fit a linear regression model using all available features to predict cmedv
# and compute the RMSE on the test data. What is the test set RMSE? Is this better than the previous model’s performance?

# fit model
lm2 <- linear_reg() %>%
  fit(cmedv ~ ., data = train)

# compute the RMSE on the test data
lm2 %>%
  predict(new_data = test) %>%
  bind_cols(test %>% select(cmedv)) %>%
  rmse(truth = cmedv, estimate = .pred)

NA
NA

# Assuming you have your training data in 'train' and test data in 'test' data frames

# Fit a K-nearest neighbor model using all available features to predict cmedv

install.packages("caret")

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library(caret)

Loading required package: lattice

Attaching package: ‘caret’

The following objects are masked from ‘package:yardstick’:

    precision, recall,
    sensitivity, specificity

The following object is masked from ‘package:purrr’:

    lift

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    contr.dummy

# Specify the formula for prediction, assuming 'cmedv' is the target variable
formula <- cmedv ~ .

# Fit the K-nearest neighbor model
knn_model <- train(formula, data = train, method = "kknn")

# Make predictions on the test data
predictions <- predict(knn_model, newdata = test)

# Compute RMSE on the test data
rmse_value <- sqrt(mean((predictions - test$cmedv)^2))

# Print the RMSE value
print(rmse_value)

[1] 3.251742