Problems

1 / 1

Gaussian Naive Bayes

Classic ML

Hard

Gaussian Naive Bayes is a classification algorithm based on Bayes' theorem with a "naive" assumption that features are conditionally independent given the class. Each feature's likelihood is modeled as a Gaussian distribution, making it fast and effective for many real-world problems.

Given labeled training data and unlabeled test data, predict the class for each test sample by computing the posterior probability for each class.

Algorithm

For each class c, compute the prior probability:

P(c) = \frac{n_c}{n}

For each class c and feature j, compute the mean and population variance:

\mu_{cj} = \frac{1}{n_c} \sum_{i \in c} x_{ij}, \quad \sigma^2_{cj} = \frac{1}{n_c} \sum_{i \in c} (x_{ij} - \mu_{cj})^2

For each test sample, compute the log posterior for each class:

\log P(c | x) \propto \log P(c) + \sum_j \left[ -\frac{1}{2} \log(2\pi\sigma^2_{cj}) - \frac{(x_j - \mu_{cj})^2}{2\sigma^2_{cj}} \right]

Predict the class with the highest log posterior. Add a small epsilon (1e-9) to variances to avoid division by zero.

Loading visualization...

Examples

Input:

X_train = [[1], [2], [3], [10], [11], [12]], y_train = [0, 0, 0, 1, 1, 1], X_test = [[2], [11], [6]]

Output:

[0, 1, 0]

Class 0 has mean 2 and class 1 has mean 11. Test points near class 0's mean are classified as 0, near class 1's mean as 1. The midpoint sample x=6 is closer to class 0 in this Gaussian model.

Input:

X_train = [[0, 0], [1, 0], [0, 1], [10, 10], [11, 10], [10, 11]], y_train = [0, 0, 0, 1, 1, 1], X_test = [[0.5, 0.5], [10.5, 10.5]]

Output:

[0, 1]

With 2D features, each class has a cluster of points. The test points fall clearly within each cluster's distribution.

Hint 1

Group training data by class. For each class, compute the mean and variance for each feature. Then for each test point, compute log_prior + sum of log-likelihoods for each class, and pick the class with the highest total.

Hint 2

The Gaussian log-likelihood for feature j given class c is: -0.5 * log(2 * pi * var) - (x_j - mean)^2 / (2 * var). Remember to add epsilon to var before using it.

Requirements

Compute the prior P(c) = n_c / n for each class
Compute the mean and population variance (divide by n_c, not n_c - 1) for each feature per class
Add epsilon = 1e-9 to all variances to handle zero-variance features
Compute log posteriors and predict the class with the highest value
Return a list of predicted class labels for the test set

Constraints

X_train has at least 2 rows with at least 2 distinct classes in y_train
X_test has at least 1 row
All feature values are numeric
Return a list of integer class labels with the same length as X_test
Time limit: 300 ms

Try Similar Problems

Naive Bayes Bernoulli Linear Regression Closed Form Ridge Regression Logistic Regression Training Bernoulli Pmf

Prev Next

Gaussian Naive Bayes

Classic ML

Hard

Given labeled training data and unlabeled test data, predict the class for each test sample by computing the posterior probability for each class.

Algorithm

For each class c, compute the prior probability:

P(c) = \frac{n_c}{n}

For each class c and feature j, compute the mean and population variance:

\mu_{cj} = \frac{1}{n_c} \sum_{i \in c} x_{ij}, \quad \sigma^2_{cj} = \frac{1}{n_c} \sum_{i \in c} (x_{ij} - \mu_{cj})^2

For each test sample, compute the log posterior for each class:

\log P(c | x) \propto \log P(c) + \sum_j \left[ -\frac{1}{2} \log(2\pi\sigma^2_{cj}) - \frac{(x_j - \mu_{cj})^2}{2\sigma^2_{cj}} \right]

Predict the class with the highest log posterior. Add a small epsilon (1e-9) to variances to avoid division by zero.

Loading visualization...

Examples

Input:

X_train = [[1], [2], [3], [10], [11], [12]], y_train = [0, 0, 0, 1, 1, 1], X_test = [[2], [11], [6]]

Output:

[0, 1, 0]

Class 0 has mean 2 and class 1 has mean 11. Test points near class 0's mean are classified as 0, near class 1's mean as 1. The midpoint sample x=6 is closer to class 0 in this Gaussian model.

Input:

X_train = [[0, 0], [1, 0], [0, 1], [10, 10], [11, 10], [10, 11]], y_train = [0, 0, 0, 1, 1, 1], X_test = [[0.5, 0.5], [10.5, 10.5]]

Output:

[0, 1]

With 2D features, each class has a cluster of points. The test points fall clearly within each cluster's distribution.

Hint 1

Hint 2

The Gaussian log-likelihood for feature j given class c is: -0.5 * log(2 * pi * var) - (x_j - mean)^2 / (2 * var). Remember to add epsilon to var before using it.

Requirements

Compute the prior P(c) = n_c / n for each class
Compute the mean and population variance (divide by n_c, not n_c - 1) for each feature per class
Add epsilon = 1e-9 to all variances to handle zero-variance features
Compute log posteriors and predict the class with the highest value
Return a list of predicted class labels for the test set

Constraints

X_train has at least 2 rows with at least 2 distinct classes in y_train
X_test has at least 1 row
All feature values are numeric
Return a list of integer class labels with the same length as X_test
Time limit: 300 ms

Try Similar Problems

Naive Bayes Bernoulli Linear Regression Closed Form Ridge Regression Logistic Regression Training Bernoulli Pmf

Case 1

Case 2

X_train =

Accepts: array

y_train =

Accepts: array

X_test =

Accepts: array

Given labeled training data and unlabeled test data, predict the class for each test sample by computing the posterior probability for each class.

Algorithm

For each class c, compute the prior probability:

P(c) = \frac{n_c}{n}

For each class c and feature j, compute the mean and population variance:

\mu_{cj} = \frac{1}{n_c} \sum_{i \in c} x_{ij}, \quad \sigma^2_{cj} = \frac{1}{n_c} \sum_{i \in c} (x_{ij} - \mu_{cj})^2

For each test sample, compute the log posterior for each class:

\log P(c | x) \propto \log P(c) + \sum_j \left[ -\frac{1}{2} \log(2\pi\sigma^2_{cj}) - \frac{(x_j - \mu_{cj})^2}{2\sigma^2_{cj}} \right]

Predict the class with the highest log posterior. Add a small epsilon (1e-9) to variances to avoid division by zero.