Problems
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Implement Cross-Entropy Loss

Loss Functions
Medium

Compute the average cross-entropy loss for multi-class classification.

y_true contains the correct class labels (like [0, 2, 1])

y_pred contains the predicted probabilities for each class (each row sums to ~1)

For each sample, the loss is:

lossi=−log⁡(pi,yi)\text{loss}_i = -\log(p_{i, y_i})

Where:

yi=correct class label for sample iy_i = \text{correct class label for sample } i

pi,yi=predicted probability for that classp_{i, y_i} = \text{predicted probability for that class}

The final cross-entropy loss is the average:

CrossEntropy=−1N∑i=1Nlog⁡(pi,yi)\text{CrossEntropy} = -\frac{1}{N} \sum_{i=1}^{N} \log(p_{i, y_i})

Note: You can assume all probabilities are valid (greater than 0), so no need to worry about log(0) .log(0) .

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Examples

Input: y_true = [0, 1], y_pred = [[0.9, 0.1], [0.3, 0.7]]

Output: 0.231018

Input: y_true = [2], y_pred = [[0.1, 0.1, 0.8]]

Output: 0.223144

Input: y_true = [1, 0, 1], y_pred = [[0.2, 0.8], [0.6, 0.4], [0.49, 0.51]]

Output: 0.469105

Hint 1

Use np.arrange() with advanced indexing to extract probabilities for the correct classes.

Hint 2

Use np.log() and np.mean() to compute the negative average of logarithms.

Requirements

  • y_true and y_pred must have matching first dimension N
  • y_true contains valid class indices for the second dimension of y_pred
  • Vectorized (no Python loops over samples)
  • Input is probabilities (not logits)
  • All probabilities are guaranteed to be > 0

Constraints

  • 1 ≤ N ≤ 105, 2 ≤ K ≤ 100
  • ∑k pi,k ≈ 1 for each row
  • NumPy only; time limit: 200 ms
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