Regression and Classification Practice

Practise train/test splits, linear and logistic regression, regression and classification metrics, k-nearest neighbours and decision trees with scikit-learn.

Beginner7 problemsUpdated October 24, 2026

Review first: Features, Labels and Train/Test Splits, Linear Regression, Regression Metrics, Logistic Regression, Classification Metrics, k-Nearest Neighbours, Decision Trees

1.Stratified split
Beginner
Load the wine dataset with load_wine(return_X_y=True). Split it into 75% training and 25% test data with random_state=0, stratified by class. Print the number of rows in each part and the class counts in the test set.
Show hint
train_test_split(X, y, test_size=0.25, random_state=0, stratify=y), then np.bincount(y_test).
Show solution
Python
import numpy as np from sklearn.datasets import load_wine from sklearn.model_selection import train_test_split X, y = load_wine(return_X_y=True) X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.25, random_state=0, stratify=y) print(len(X_train), len(X_test)) print(np.bincount(y_test))
Output
133 45 [15 18 12]
2.Rent from flat size
Beginner
Flat sizes (sq ft) [450, 600, 750, 900, 1100, 1300] have monthly rents (₹) [9000, 11500, 14200, 16800, 20500, 24000]. Fit a linear regression, print the rent increase per extra square foot (2 decimals), and predict the rent for a 1,000 sq ft flat (rounded to the nearest rupee).
Show hint
Reshape sizes into a column with .reshape(-1, 1). The slope is model.coef_[0].
Show solution
Python
import numpy as np from sklearn.linear_model import LinearRegression size = np.array([450, 600, 750, 900, 1100, 1300]).reshape(-1, 1) rent = np.array([9000, 11500, 14200, 16800, 20500, 24000]) model = LinearRegression().fit(size, rent) print(round(model.coef_[0], 2)) print(round(model.predict([[1000]])[0]))
Output
17.74 18661
Each extra square foot adds about ₹17.74 to the monthly rent, so a 1,000 sq ft flat is predicted at about ₹18,661.

3.Regression errors

Beginner

Actual delivery times are [30, 45, 25, 60, 40] minutes and a model predicted [35, 40, 30, 50, 42]. Print the MAE, RMSE and R², each rounded to 2 decimals.

Show hint

mean_absolute_error, np.sqrt(mean_squared_error(...)) and r2_score from sklearn.metrics.

Show solution

Python

import numpy as np
from sklearn.metrics import mean_absolute_error, mean_squared_error, r2_score

actual = [30, 45, 25, 60, 40]
predicted = [35, 40, 30, 50, 42]
print(round(mean_absolute_error(actual, predicted), 2))
print(round(np.sqrt(mean_squared_error(actual, predicted)), 2))
print(round(r2_score(actual, predicted), 2))

Output

5.4
5.98
0.76

4.Logistic regression pipeline

Beginner

On the breast cancer dataset, split 75/25 (random_state=0, stratified), fit a pipeline of StandardScaler and LogisticRegression, and print the test accuracy (3 decimals) and the predicted probability of class 1 for the first test row (3 decimals).

Show hint

make_pipeline(StandardScaler(), LogisticRegression()); predict_proba(X_test[:1])[0, 1].

Show solution

Python

from sklearn.datasets import load_breast_cancer
from sklearn.linear_model import LogisticRegression
from sklearn.model_selection import train_test_split
from sklearn.pipeline import make_pipeline
from sklearn.preprocessing import StandardScaler

X, y = load_breast_cancer(return_X_y=True)
X_train, X_test, y_train, y_test = train_test_split(X, y, random_state=0, stratify=y)
model = make_pipeline(StandardScaler(), LogisticRegression()).fit(X_train, y_train)
print(round(model.score(X_test, y_test), 3))
print(round(model.predict_proba(X_test[:1])[0, 1], 3))

Output

0.958
0.996

5.Confusion matrix by hand
Beginner
True labels are [1, 0, 1, 1, 0, 0, 1, 0, 1, 0] and predictions are [1, 0, 0, 1, 0, 1, 1, 0, 0, 0]. Print the confusion matrix, then precision, recall and F1 rounded to 2 decimals.
Show hint
confusion_matrix, precision_score, recall_score and f1_score from sklearn.metrics.
Show solution
Python
from sklearn.metrics import confusion_matrix, f1_score, precision_score, recall_score y_true = [1, 0, 1, 1, 0, 0, 1, 0, 1, 0] y_pred = [1, 0, 0, 1, 0, 1, 1, 0, 0, 0] print(confusion_matrix(y_true, y_pred)) print(round(precision_score(y_true, y_pred), 2), round(recall_score(y_true, y_pred), 2), round(f1_score(y_true, y_pred), 2))
Output
[[4 1] [2 3]] 0.75 0.6 0.67
The matrix rows are actual classes and columns predicted: 3 true positives, 1 false positive and 2 false negatives. Precision is 3/4 = 0.75 and recall 3/5 = 0.6.
6.Choosing k for k-NN
Intermediate
On the scaled wine data, compute the 5-fold cross-validated accuracy of k-nearest neighbours for k = 1, 5, 15 and 45. Print each k with its accuracy (3 decimals).
Show hint
Put StandardScaler and KNeighborsClassifier(n_neighbors=k) in a pipeline and pass it to cross_val_score.
Show solution
Python
from sklearn.datasets import load_wine from sklearn.model_selection import cross_val_score from sklearn.neighbors import KNeighborsClassifier from sklearn.pipeline import make_pipeline from sklearn.preprocessing import StandardScaler X, y = load_wine(return_X_y=True) for k in [1, 5, 15, 45]: model = make_pipeline(StandardScaler(), KNeighborsClassifier(n_neighbors=k)) print(k, round(cross_val_score(model, X, y, cv=5).mean(), 3))
Output
1 0.95 5 0.949 15 0.955 45 0.961
Accuracy changes only a little across these values of k; here the largest k scores slightly best. Small k follows individual neighbours (more variance); large k smooths over more of them.

7.Tree depth and overfitting

Intermediate

Split the breast cancer data 75/25 (random_state=0). For decision trees with max_depth 1, 3, 5 and None (random_state=0), print the training and test accuracy (3 decimals). Which depth generalises best?

Show hint

Loop over the depths, fit each tree on the training data and call .score on both parts.

Show solution

Python

from sklearn.datasets import load_breast_cancer
from sklearn.model_selection import train_test_split
from sklearn.tree import DecisionTreeClassifier

X, y = load_breast_cancer(return_X_y=True)
X_train, X_test, y_train, y_test = train_test_split(X, y, random_state=0)
for depth in [1, 3, 5, None]:
    tree = DecisionTreeClassifier(max_depth=depth, random_state=0).fit(X_train, y_train)
    print(depth, round(tree.score(X_train, y_train), 3), round(tree.score(X_test, y_test), 3))

Output

1 0.923 0.881
3 0.977 0.937
5 0.993 0.895
None 1.0 0.881

Depth 3 gives the best test accuracy (0.937). Deeper trees fit the training data better and better (up to 1.0) but do worse on the test set: overfitting.

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