Machine Learning Exercises in R: 50 Real Practice Problems

Fifty practice problems on machine learning in R: workflow basics, classification, regression, tuning, feature engineering, and end-to-end pipelines. Hidden solutions, runnable code.

By Selva Prabhakaran · Published May 11, 2026 · Last updated May 11, 2026

RRun this once before any exercise

library(dplyr) library(rpart) library(randomForest) library(class) library(caret) library(tibble) library(ggplot2) library(e1071) library(gbm) library(glmnet) library(recipes) library(xgboost)

Section 1. Workflow basics (8 problems)

Exercise 1.1: Train-test split

Scenario: Split iris 70/30 into training and test sets reproducibly.

Difficulty: Beginner

RYour turn

set.seed(1) # your code here

Click to reveal solution

RSolution

set.seed(1) n <- nrow(iris) idx <- sample(seq_len(n), size = floor(0.7 * n)) train <- iris[idx, ]; test <- iris[-idx, ] c(train = nrow(train), test = nrow(test))

Explanation: sample without replacement gives a random index. set.seed makes it reproducible.

Exercise 1.2: Stratified split

Scenario: Same split but ensure class proportions are preserved per fold.

Difficulty: Intermediate

RYour turn

set.seed(1) # your code here

Click to reveal solution

RSolution

set.seed(1) idx <- caret::createDataPartition(iris$Species, p = 0.7, list = FALSE) train <- iris[idx, ]; test <- iris[-idx, ] prop.table(table(train$Species))

Explanation: createDataPartition stratifies on the y argument. Critical for imbalanced classes; random split could leave some classes absent in a fold.

Exercise 1.3: First k-NN classifier

Scenario: k-NN with k=5 for iris Species.

Difficulty: Intermediate

RYour turn

set.seed(1) idx <- sample(seq_len(nrow(iris)), 100) tr <- iris[idx, 1:4]; te <- iris[-idx, 1:4] y_tr <- iris$Species[idx]; y_te <- iris$Species[-idx] ex_1_3 <- # your code here mean(ex_1_3 == y_te)

Click to reveal solution

RSolution

set.seed(1) idx <- sample(seq_len(nrow(iris)), 100) tr <- iris[idx, 1:4]; te <- iris[-idx, 1:4] y_tr <- iris$Species[idx]; y_te <- iris$Species[-idx] ex_1_3 <- class::knn(tr, te, y_tr, k = 5) mean(ex_1_3 == y_te)

Explanation: Class probabilities are decided by k nearest neighbors. Critical preprocessing: scale features so distances are comparable.

Exercise 1.4: First decision tree

Scenario: Fit a decision tree for iris Species using all 4 predictors.

Difficulty: Intermediate

RYour turn

ex_1_4 <- # your code here print(ex_1_4)

Click to reveal solution

RSolution

ex_1_4 <- rpart::rpart(Species ~ ., data = iris) print(ex_1_4)

Explanation: rpart uses CART. Default stops based on cp (complexity). Inspect splits via the printed tree or rpart.plot::rpart.plot.

Exercise 1.5: Compute accuracy

Scenario: Given predictions and truth, compute accuracy.

Difficulty: Beginner

RYour turn

truth <- c("a","b","a","c","b","a") pred <- c("a","b","b","c","b","c") ex_1_5 <- # your code here ex_1_5

Click to reveal solution

RSolution

ex_1_5 <- mean(pred == truth) ex_1_5

Explanation: Mean of TRUE/FALSE comparisons gives the proportion correct.

Exercise 1.6: Confusion matrix

Scenario: Build a confusion matrix from the same predictions.

Difficulty: Intermediate

RYour turn

truth <- factor(c("a","b","a","c","b","a")) pred <- factor(c("a","b","b","c","b","c")) ex_1_6 <- # your code here ex_1_6

Click to reveal solution

RSolution

truth <- factor(c("a","b","a","c","b","a")) pred <- factor(c("a","b","b","c","b","c")) ex_1_6 <- table(truth = truth, pred = pred) ex_1_6

Explanation: table cross-tabulates truth vs pred. Diagonal = correct. caret::confusionMatrix adds per-class metrics.

Exercise 1.7: Per-class accuracy from confusion matrix

Scenario: From the confusion matrix in 1.6, compute precision per class.

Difficulty: Intermediate

RYour turn

cm <- table(truth = factor(c("a","b","a","c","b","a")), pred = factor(c("a","b","b","c","b","c"))) ex_1_7 <- # your code here ex_1_7

Click to reveal solution

RSolution

cm <- table(truth = factor(c("a","b","a","c","b","a")), pred = factor(c("a","b","b","c","b","c"))) ex_1_7 <- diag(cm) / colSums(cm) ex_1_7

Explanation: Precision = TP / (TP + FP) per class = diag / column sum. Recall = diag / row sum. F1 = 2PR / (P+R).

Exercise 1.8: Save and load a model

Scenario: Save an rpart model to disk and load it back.

Difficulty: Intermediate

RYour turn

fit <- rpart::rpart(Species ~ ., data = iris) # your code here

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RSolution

fit <- rpart::rpart(Species ~ ., data = iris) saveRDS(fit, "model.rds") loaded <- readRDS("model.rds") identical(predict(fit), predict(loaded))

Explanation: saveRDS for one R object; readRDS to restore. Standard for serializing models.

Section 2. Classification (10 problems)

Exercise 2.1: Logistic regression

Scenario: Predict am (0/1) from mpg in mtcars.

Difficulty: Intermediate

RYour turn

ex_2_1 <- # your code here coef(ex_2_1)

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RSolution

ex_2_1 <- glm(am ~ mpg, data = mtcars, family = binomial) coef(ex_2_1)

Explanation: glm with family = binomial fits logistic regression. Coefficient is on log-odds scale; exp() gives odds ratio.

Exercise 2.2: Predict probabilities

Scenario: Get predicted probabilities from the model in 2.1.

Difficulty: Intermediate

RYour turn

fit <- glm(am ~ mpg, data = mtcars, family = binomial) ex_2_2 <- # your code here head(ex_2_2)

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RSolution

fit <- glm(am ~ mpg, data = mtcars, family = binomial) ex_2_2 <- predict(fit, type = "response") head(ex_2_2)

Explanation: type = "response" returns probabilities; type = "link" returns log-odds; default is "link" for glm.

Exercise 2.3: Threshold to classify

Scenario: Convert probabilities to 0/1 predictions at threshold 0.5.

Difficulty: Beginner

RYour turn

fit <- glm(am ~ mpg, data = mtcars, family = binomial) prob <- predict(fit, type = "response") ex_2_3 <- # your code here table(ex_2_3, mtcars$am)

Click to reveal solution

RSolution

fit <- glm(am ~ mpg, data = mtcars, family = binomial) prob <- predict(fit, type = "response") ex_2_3 <- as.integer(prob > 0.5) table(ex_2_3, mtcars$am)

Explanation: Default threshold 0.5; tune for imbalanced classes or asymmetric costs.

Exercise 2.4: ROC curve and AUC

Scenario: ROC curve and AUC for the logistic model.

Difficulty: Advanced

RYour turn

fit <- glm(am ~ mpg + hp, data = mtcars, family = binomial) prob <- predict(fit, type = "response") ex_2_4 <- # your code here ex_2_4

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RSolution

fit <- glm(am ~ mpg + hp, data = mtcars, family = binomial) prob <- predict(fit, type = "response") roc_obj <- pROC::roc(mtcars$am, prob) ex_2_4 <- pROC::auc(roc_obj) ex_2_4

Explanation: ROC plots TPR vs FPR across thresholds. AUC is the area: 1 = perfect, 0.5 = random. Threshold-independent metric.

Exercise 2.5: Random forest classifier

Scenario: Random forest for iris Species.

Difficulty: Intermediate

RYour turn

set.seed(1) ex_2_5 <- # your code here ex_2_5

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RSolution

set.seed(1) ex_2_5 <- randomForest::randomForest(Species ~ ., data = iris) ex_2_5

Explanation: Bootstrap aggregating with random feature subsets per split. ntree default 500; mtry sqrt(p) for classification. Out-of-bag error printed.

Exercise 2.6: Variable importance

Scenario: Extract variable importance from the random forest.

Difficulty: Intermediate

RYour turn

set.seed(1) fit <- randomForest::randomForest(Species ~ ., data = iris) ex_2_6 <- # your code here ex_2_6

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RSolution

set.seed(1) fit <- randomForest::randomForest(Species ~ ., data = iris) ex_2_6 <- randomForest::importance(fit) ex_2_6

Explanation: MeanDecreaseGini measures average impurity reduction by each feature. Higher = more useful.

Exercise 2.7: Naive Bayes

Scenario: Naive Bayes classifier for iris.

Difficulty: Intermediate

RYour turn

ex_2_7 <- # your code here ex_2_7

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RSolution

ex_2_7 <- e1071::naiveBayes(Species ~ ., data = iris) ex_2_7

Explanation: Assumes feature independence given class. Fast, simple, surprisingly competitive on text and high-dimensional data.

Exercise 2.8: Imbalanced class handling

Scenario: Up-sample the minority class to balance.

Difficulty: Advanced

RYour turn

df <- tibble(x = rnorm(100), y = factor(c(rep("A", 90), rep("B", 10)))) set.seed(1) ex_2_8 <- # your code here table(ex_2_8$y)

Click to reveal solution

RSolution

df <- tibble(x = rnorm(100), y = factor(c(rep("A", 90), rep("B", 10)))) set.seed(1) ex_2_8 <- caret::upSample(df["x"], df$y, yname = "y") table(ex_2_8$y)

Explanation: Replicates minority rows to match majority count. Alternatives: downSample (cut majority), SMOTE (synthetic), class weights in the model.

Exercise 2.9: SVM with caret

Scenario: Train an SVM with linear kernel on iris.

Difficulty: Advanced

RYour turn

set.seed(1) ex_2_9 <- # your code here ex_2_9

Click to reveal solution

RSolution

set.seed(1) ex_2_9 <- caret::train(Species ~ ., data = iris, method = "svmLinear") ex_2_9

Explanation: caret::train wraps many ML methods with consistent interface. Auto-tunes when method allows. tidymodels is the modern equivalent.

Exercise 2.10: Gradient boosting

Scenario: Fit a gradient-boosted classifier (gbm).

Difficulty: Advanced

RYour turn

ir <- iris ir$Species_int <- as.integer(ir$Species == "virginica") set.seed(1) ex_2_10 <- # your code here ex_2_10

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RSolution

ir <- iris ir$Species_int <- as.integer(ir$Species == "virginica") set.seed(1) ex_2_10 <- gbm::gbm(Species_int ~ Sepal.Length + Sepal.Width + Petal.Length + Petal.Width, data = ir, distribution = "bernoulli", n.trees = 500, interaction.depth = 3, verbose = FALSE) ex_2_10

Explanation: GBM trains many shallow trees, each correcting predecessors. interaction.depth controls tree complexity. xgboost::xgboost is the production-grade alternative.

Section 3. Regression ML (8 problems)

Exercise 3.1: Regression tree

Scenario: Predict mpg from mtcars predictors with rpart.

Difficulty: Intermediate

RYour turn

ex_3_1 <- # your code here ex_3_1

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RSolution

ex_3_1 <- rpart::rpart(mpg ~ ., data = mtcars) ex_3_1

Explanation: Same rpart, regression mode (continuous y). Splits minimize within-node variance.

Exercise 3.2: Random forest regression

Scenario: Random forest to predict mpg.

Difficulty: Intermediate

RYour turn

set.seed(1) ex_3_2 <- # your code here ex_3_2

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RSolution

set.seed(1) ex_3_2 <- randomForest::randomForest(mpg ~ ., data = mtcars) ex_3_2

Explanation: Same RF; default mtry = p/3 for regression. % Var explained reported via OOB.

Exercise 3.3: Compute RMSE

Scenario: Compute RMSE on a holdout for the RF in 3.2.

Difficulty: Intermediate

RYour turn

set.seed(1) idx <- sample(seq_len(nrow(mtcars)), 22) tr <- mtcars[idx, ]; te <- mtcars[-idx, ] fit <- randomForest::randomForest(mpg ~ ., data = tr) ex_3_3 <- # your code here ex_3_3

Click to reveal solution

RSolution

set.seed(1) idx <- sample(seq_len(nrow(mtcars)), 22) tr <- mtcars[idx, ]; te <- mtcars[-idx, ] fit <- randomForest::randomForest(mpg ~ ., data = tr) preds <- predict(fit, te) ex_3_3 <- sqrt(mean((te$mpg - preds)^2)) ex_3_3

Explanation: RMSE = sqrt(mean(squared errors)). Same units as y; standard regression metric.

Exercise 3.4: MAE

Scenario: Mean absolute error for the same predictions.

Difficulty: Beginner

RYour turn

preds <- c(20, 25, 30); actual <- c(22, 24, 31) ex_3_4 <- # your code here ex_3_4

Click to reveal solution

RSolution

preds <- c(20, 25, 30); actual <- c(22, 24, 31) ex_3_4 <- mean(abs(preds - actual)) ex_3_4

Explanation: MAE is more robust to outliers than RMSE. Each unit error contributes linearly, not squared.

Exercise 3.5: R-squared on test set

Scenario: Compute test-set R-squared for the RF.

Difficulty: Intermediate

RYour turn

Click to reveal solution

RSolution

Explanation: R-squared on test = 1 - residual SS / total SS. Tells you how much variance the model explains beyond the test mean.

Exercise 3.6: Ridge regression

Scenario: Ridge regression to predict mpg.

Difficulty: Advanced

RYour turn

x <- as.matrix(mtcars[, -1]) y <- mtcars$mpg ex_3_6 <- # your code here coef(ex_3_6, s = 0.1)

Click to reveal solution

RSolution

x <- as.matrix(mtcars[, -1]) y <- mtcars$mpg ex_3_6 <- glmnet::glmnet(x, y, alpha = 0) coef(ex_3_6, s = 0.1)

Explanation: alpha=0 is ridge (L2); alpha=1 is lasso (L1); 0

Exercise 3.7: Lasso for variable selection

Scenario: Lasso to predict mpg, identify zeroed coefficients.

Difficulty: Advanced

RYour turn

x <- as.matrix(mtcars[, -1]) y <- mtcars$mpg fit <- glmnet::glmnet(x, y, alpha = 1) ex_3_7 <- # your code here ex_3_7

Click to reveal solution

RSolution

x <- as.matrix(mtcars[, -1]) y <- mtcars$mpg fit <- glmnet::glmnet(x, y, alpha = 1) ex_3_7 <- coef(fit, s = 0.5) ex_3_7

Explanation: Lasso shrinks some coefficients to exactly 0, performing automatic variable selection. Increase s to drop more.

Exercise 3.8: Cross-validate to pick lambda

Scenario: Use cv.glmnet to find the best lambda for ridge.

Difficulty: Advanced

RYour turn

set.seed(1) x <- as.matrix(mtcars[, -1]) y <- mtcars$mpg ex_3_8 <- # your code here ex_3_8$lambda.min

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RSolution

set.seed(1) x <- as.matrix(mtcars[, -1]) y <- mtcars$mpg ex_3_8 <- glmnet::cv.glmnet(x, y, alpha = 0) ex_3_8$lambda.min

Explanation: cv.glmnet runs k-fold CV across a lambda grid. lambda.min minimizes CV error; lambda.1se gives a more parsimonious choice within 1 SE.

Section 4. Tuning and validation (8 problems)

Exercise 4.1: 5-fold CV manually

Scenario: 5-fold CV for a linear model on mtcars.

Difficulty: Intermediate

RYour turn

set.seed(1) folds <- sample(rep(1:5, length.out = nrow(mtcars))) # your code here

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RSolution

set.seed(1) folds <- sample(rep(1:5, length.out = nrow(mtcars))) rmses <- sapply(1:5, function(i) { tr <- mtcars[folds != i, ]; te <- mtcars[folds == i, ] fit <- lm(mpg ~ wt + hp, data = tr) sqrt(mean((te$mpg - predict(fit, te))^2)) }) mean(rmses)

Explanation: Each fold tests once. Average RMSE is the CV estimate.

Exercise 4.2: caret::train with CV

Scenario: Use caret to train a random forest with 5-fold CV.

Difficulty: Intermediate

RYour turn

set.seed(1) ex_4_2 <- # your code here ex_4_2

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RSolution

set.seed(1) ex_4_2 <- caret::train(mpg ~ ., data = mtcars, method = "rf", trControl = caret::trainControl(method = "cv", number = 5)) ex_4_2

Explanation: caret::train abstracts the CV loop. trainControl configures resampling. Method names: "lm", "rf", "glmnet", "xgbTree", etc.

Exercise 4.3: Hyperparameter grid search

Scenario: Tune RF mtry over c(2, 4, 6) using caret.

Difficulty: Advanced

RYour turn

set.seed(1) ex_4_3 <- # your code here ex_4_3$bestTune

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RSolution

set.seed(1) ex_4_3 <- caret::train(mpg ~ ., data = mtcars, method = "rf", trControl = caret::trainControl(method = "cv", number = 5), tuneGrid = expand.grid(mtry = c(2, 4, 6))) ex_4_3$bestTune

Explanation: tuneGrid passes candidate hyperparameter values. caret picks the best by CV metric. expand.grid supports multi-parameter grids.

Exercise 4.4: Repeated CV

Scenario: Repeated 5-fold CV (3 repeats) to reduce variance of the estimate.

Difficulty: Advanced

RYour turn

set.seed(1) ex_4_4 <- # your code here ex_4_4

Click to reveal solution

RSolution

set.seed(1) ex_4_4 <- caret::train(mpg ~ ., data = mtcars, method = "lm", trControl = caret::trainControl(method = "repeatedcv", number = 5, repeats = 3)) ex_4_4

Explanation: Repeated CV reshuffles folds and repeats. Smoother estimate of CV error at the cost of compute.

Exercise 4.5: Bootstrap validation

Scenario: Use 100 bootstrap resamples for model evaluation.

Difficulty: Advanced

RYour turn

set.seed(1) ex_4_5 <- # your code here ex_4_5

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RSolution

set.seed(1) ex_4_5 <- caret::train(mpg ~ ., data = mtcars, method = "lm", trControl = caret::trainControl(method = "boot", number = 100)) ex_4_5

Explanation: Bootstrap resamples with replacement; out-of-bag observations form the test fold. Common alternative to CV when n is small.

Exercise 4.6: Holdout for true generalization

Scenario: Split off a 20% test set BEFORE any CV, then evaluate at the end.

Difficulty: Intermediate

RYour turn

set.seed(1) n <- nrow(mtcars) test_idx <- sample(seq_len(n), size = floor(0.2 * n)) test <- mtcars[test_idx, ]; train <- mtcars[-test_idx, ] # your code here

Click to reveal solution

RSolution

set.seed(1) n <- nrow(mtcars) test_idx <- sample(seq_len(n), size = floor(0.2 * n)) test <- mtcars[test_idx, ]; train <- mtcars[-test_idx, ] fit <- caret::train(mpg ~ ., data = train, method = "rf", trControl = caret::trainControl(method = "cv", number = 5)) preds <- predict(fit, test) sqrt(mean((test$mpg - preds)^2))

Explanation: Two-level split: test never touches CV/tuning. Final test RMSE is the most honest estimate of deployment performance.

Exercise 4.7: Compare models on CV RMSE

Scenario: Compare lm, rf, gbm on mpg using caret resamples.

Difficulty: Advanced

RYour turn

set.seed(1) ctrl <- caret::trainControl(method = "cv", number = 5) # your code here

Click to reveal solution

RSolution

set.seed(1) ctrl <- caret::trainControl(method = "cv", number = 5) m_lm <- caret::train(mpg ~ ., data = mtcars, method = "lm", trControl = ctrl) m_rf <- caret::train(mpg ~ ., data = mtcars, method = "rf", trControl = ctrl) caret::resamples(list(lm = m_lm, rf = m_rf)) |> summary()

Explanation: caret::resamples bundles CV results from multiple models for direct comparison. Always evaluate at the same resampling seed for fair comparison.

Exercise 4.8: Early stopping

Scenario: xgboost with early stopping on a validation fold.

Difficulty: Advanced

RYour turn

set.seed(1) n <- nrow(mtcars) idx <- sample(seq_len(n), 22) x_tr <- as.matrix(mtcars[idx, -1]) y_tr <- mtcars$mpg[idx] x_te <- as.matrix(mtcars[-idx, -1]) y_te <- mtcars$mpg[-idx] # your code here

Click to reveal solution

RSolution

dtrain <- xgboost::xgb.DMatrix(x_tr, label = y_tr) dtest <- xgboost::xgb.DMatrix(x_te, label = y_te) fit <- xgboost::xgb.train( params = list(objective = "reg:squarederror"), data = dtrain, nrounds = 200, watchlist = list(test = dtest), early_stopping_rounds = 10, verbose = 0 ) fit$best_iteration

Explanation: Watch test loss per round; stop when no improvement for early_stopping_rounds. Prevents overfitting and saves compute.

Section 5. Feature engineering (8 problems)

Exercise 5.1: One-hot encode a factor

Scenario: Convert factor to dummy variables.

Difficulty: Beginner

RYour turn

df <- tibble(color = c("red","blue","green","red","blue")) ex_5_1 <- # your code here ex_5_1

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RSolution

df <- tibble(color = c("red","blue","green","red","blue")) ex_5_1 <- model.matrix(~ color - 1, data = df) ex_5_1

Explanation: model.matrix expands factors to dummy columns. -1 removes the intercept so all levels appear. lm/glm do this automatically; for ML methods you may need to do it manually.

Exercise 5.2: Min-max scale numeric features

Scenario: Rescale all numeric columns of iris to [0, 1].

Difficulty: Intermediate

RYour turn

ex_5_2 <- iris |> # your code here

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RSolution

ex_5_2 <- iris |> mutate(across(where(is.numeric), ~ (.x - min(.x)) / (max(.x) - min(.x)))) summary(ex_5_2[, 1:4])

Explanation: Critical preprocessing for distance-based methods (kNN, SVM, k-means). Z-scoring (mean 0, sd 1) is the alternative.

Exercise 5.3: Z-score standardization

Scenario: Standardize numeric columns to mean 0, sd 1.

Difficulty: Intermediate

RYour turn

ex_5_3 <- iris |> # your code here

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RSolution

ex_5_3 <- iris |> mutate(across(where(is.numeric), scale)) head(ex_5_3)

Explanation: scale() centers and scales by sd. across applies it to all numeric. Always fit the scaler on train only and apply to test.

Exercise 5.4: Log-transform skewed feature

Scenario: Log-transform diamonds carat (right-skewed).

Difficulty: Beginner

RYour turn

ex_5_4 <- diamonds |> # your code here

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RSolution

ex_5_4 <- diamonds |> mutate(log_carat = log(carat)) head(ex_5_4 |> select(carat, log_carat))

Explanation: log compresses right tail, often improving linear model fit. log1p (log(1+x)) handles 0 values.

Exercise 5.5: Bin a continuous variable

Scenario: Bin mpg into low/mid/high terciles.

Difficulty: Intermediate

RYour turn

ex_5_5 <- mtcars |> # your code here

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RSolution

ex_5_5 <- mtcars |> mutate(mpg_bin = cut(mpg, breaks = quantile(mpg, c(0, 1/3, 2/3, 1)), labels = c("low","mid","high"), include.lowest = TRUE)) count(ex_5_5, mpg_bin)

Explanation: cut creates bins from breaks. include.lowest brings the minimum value into the first bin. ntile() is the dplyr alternative for equal-frequency bins.

Exercise 5.6: Interaction features

Scenario: Add a wt:hp interaction column manually.

Difficulty: Intermediate

RYour turn

ex_5_6 <- mtcars |> # your code here

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RSolution

ex_5_6 <- mtcars |> mutate(wt_hp = wt * hp) head(ex_5_6 |> select(wt, hp, wt_hp))

Explanation: For methods that don't handle interactions natively (RF, k-NN), engineer them manually. lm/glm support * in formula directly.

Exercise 5.7: Impute missing with median

Scenario: Fill NAs in airquality with column median.

Difficulty: Intermediate

RYour turn

ex_5_7 <- airquality |> # your code here

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RSolution

ex_5_7 <- airquality |> mutate(across(where(is.numeric), ~ if_else(is.na(.x), median(.x, na.rm = TRUE), .x))) sum(is.na(ex_5_7))

Explanation: Median is more robust to outliers than mean. For tree-based models, you can often skip imputation; rpart and xgboost handle NAs.

Exercise 5.8: Recipe-based preprocessing

Scenario: Use recipes to define a reusable preprocessing pipeline.

Difficulty: Advanced

RYour turn

ex_5_8 <- recipes::recipe(mpg ~ ., data = mtcars) |> # your code here

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RSolution

ex_5_8 <- recipes::recipe(mpg ~ ., data = mtcars) |> recipes::step_normalize(recipes::all_numeric_predictors()) |> recipes::step_corr(recipes::all_numeric_predictors(), threshold = 0.9) |> recipes::prep() recipes::bake(ex_5_8, mtcars) |> head()

Explanation: recipes is the tidymodels preprocessing engine. Each step is named and parameterizable. prep() learns from data; bake() applies. Essential for production workflows.

Section 6. End-to-end ML pipelines (8 problems)

Exercise 6.1: Iris classifier mini-pipeline

Scenario: Split, scale, fit kNN, evaluate.

Difficulty: Intermediate

RYour turn

set.seed(1) # your code here

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RSolution

set.seed(1) idx <- caret::createDataPartition(iris$Species, p = 0.7, list = FALSE) tr <- iris[idx, ]; te <- iris[-idx, ] # Scale based on training only mu <- sapply(tr[, 1:4], mean); sg <- sapply(tr[, 1:4], sd) tr_x <- sweep(sweep(tr[, 1:4], 2, mu), 2, sg, "/") te_x <- sweep(sweep(te[, 1:4], 2, mu), 2, sg, "/") pred <- class::knn(tr_x, te_x, tr$Species, k = 5) mean(pred == te$Species)

Explanation: Critical: fit scaling on train only, apply same to test. sweep applies a row/column operation.

Exercise 6.2: Cross-validated tidymodels workflow

Scenario: tidymodels workflow with recipe + lm + 5-fold CV.

Difficulty: Advanced

RYour turn

library(tidymodels) set.seed(1) # your code here

Click to reveal solution

RSolution

library(tidymodels) set.seed(1) split <- initial_split(mtcars, prop = 0.7) tr <- training(split); te <- testing(split) rec <- recipe(mpg ~ ., data = tr) |> step_normalize(all_numeric_predictors()) mod <- linear_reg() |> set_engine("lm") wf <- workflow() |> add_recipe(rec) |> add_model(mod) cv_folds <- vfold_cv(tr, v = 5) results <- fit_resamples(wf, cv_folds, metrics = metric_set(rmse, rsq)) collect_metrics(results)

Explanation: tidymodels stack: rsample (split), recipes (preprocess), parsnip (model), workflows (combine), tune/yardstick (tune/eval). Modern, composable.

Exercise 6.3: Feature importance ranking

Scenario: Random forest, then plot variable importance.

Difficulty: Intermediate

RYour turn

set.seed(1) fit <- randomForest::randomForest(mpg ~ ., data = mtcars, importance = TRUE) # your code here

Click to reveal solution

RSolution

set.seed(1) fit <- randomForest::randomForest(mpg ~ ., data = mtcars, importance = TRUE) imp <- as.data.frame(randomForest::importance(fit)) imp$var <- rownames(imp) ggplot(imp, aes(x = reorder(var, `%IncMSE`), y = `%IncMSE`)) + geom_col() + coord_flip() + labs(title = "RF variable importance", x = NULL)

Explanation: importance = TRUE enables permutation importance. %IncMSE measures how much CV-MSE rises when a feature is shuffled.

Exercise 6.4: Confusion matrix viz

Scenario: Visualize confusion matrix as a heatmap.

Difficulty: Advanced

RYour turn

set.seed(1) fit <- randomForest::randomForest(Species ~ ., data = iris) pred <- predict(fit) cm <- table(truth = iris$Species, pred = pred) # your code here

Click to reveal solution

RSolution

set.seed(1) fit <- randomForest::randomForest(Species ~ ., data = iris) pred <- predict(fit) cm <- table(truth = iris$Species, pred = pred) cm_df <- as.data.frame(cm) ggplot(cm_df, aes(pred, truth, fill = Freq)) + geom_tile() + geom_text(aes(label = Freq), color = "white", size = 5) + scale_fill_gradient(low = "lightblue", high = "navy")

Explanation: Tile + text labels. Easy to scan vs raw table.

Exercise 6.5: Calibration check for probabilities

Scenario: Plot predicted probability vs actual rate in deciles.

Difficulty: Advanced

RYour turn

set.seed(1) fit <- glm(am ~ mpg + hp + wt, data = mtcars, family = binomial) prob <- predict(fit, type = "response") df <- tibble(prob = prob, actual = mtcars$am) # your code here

Click to reveal solution

RSolution

set.seed(1) fit <- glm(am ~ mpg + hp + wt, data = mtcars, family = binomial) prob <- predict(fit, type = "response") df <- tibble(prob = prob, actual = mtcars$am) calib <- df |> mutate(decile = ntile(prob, 10)) |> group_by(decile) |> summarise(mean_pred = mean(prob), actual_rate = mean(actual)) ggplot(calib, aes(mean_pred, actual_rate)) + geom_point() + geom_abline(slope = 1, linetype = "dashed")

Explanation: Well-calibrated model has predicted probability close to actual frequency. Off-diagonal points reveal miscalibration.

Exercise 6.6: Compare models with ROC curves

Scenario: Plot ROC curves for two models on the same axes.

Difficulty: Advanced

RYour turn

fit1 <- glm(am ~ mpg, data = mtcars, family = binomial) fit2 <- glm(am ~ mpg + hp + wt, data = mtcars, family = binomial) # your code here

Click to reveal solution

RSolution

fit1 <- glm(am ~ mpg, data = mtcars, family = binomial) fit2 <- glm(am ~ mpg + hp + wt, data = mtcars, family = binomial) r1 <- pROC::roc(mtcars$am, predict(fit1, type = "response")) r2 <- pROC::roc(mtcars$am, predict(fit2, type = "response")) plot(r1, col = "blue") plot(r2, col = "red", add = TRUE) legend("bottomright", legend = c("Simple","Full"), col = c("blue","red"), lty = 1)

Explanation: ROC curves on same plot reveal which model dominates across thresholds.

Exercise 6.7: SHAP-style local explanation (caret)

Scenario: Explain a single prediction with feature attribution.

Difficulty: Advanced

RYour turn

set.seed(1) # your code here

Click to reveal solution

RSolution

set.seed(1) fit <- randomForest::randomForest(mpg ~ wt + hp, data = mtcars) # DALEX/iml provide model-agnostic SHAP-like explanations explainer <- DALEX::explain(fit, data = mtcars[, c("wt","hp")], y = mtcars$mpg, verbose = FALSE) single_obs <- mtcars["Mazda RX4", c("wt","hp")] DALEX::predict_parts(explainer, single_obs)

Explanation: DALEX offers model-agnostic local explanations: feature contribution to a single prediction. iml is an alternative.

Exercise 6.8: Save and serve a tidymodels workflow

Scenario: Train, save, load, predict.

Difficulty: Intermediate

RYour turn

library(tidymodels) set.seed(1) # your code here

Click to reveal solution

RSolution

library(tidymodels) set.seed(1) rec <- recipe(mpg ~ ., data = mtcars) |> step_normalize(all_numeric_predictors()) mod <- linear_reg() |> set_engine("lm") wf <- workflow() |> add_recipe(rec) |> add_model(mod) fit <- fit(wf, mtcars) saveRDS(fit, "wf.rds") loaded <- readRDS("wf.rds") predict(loaded, new_data = mtcars[1:3, ])

Explanation: Workflow object is portable (preprocessing + model bundled). Save + load + predict on new data. Production deployment patterns.

What to do next

After 50 problems, the ML workflow loop should be second nature. Natural follow-ups:

Linear-Regression-Exercises (shipped), go deeper on regression diagnostics.
Hypothesis-Testing-Exercises (coming), statistical foundation.
Random-Forest-Exercises, XGBoost-Exercises, Clustering-Exercises (coming), algorithm-specific drills.
tidymodels-Exercises (coming), modern R ML stack end-to-end.