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Precision, Recall, and F1 Score

Compute precision, recall, and F1 score for binary classification.

Given lists of true and predicted labels:

$\text{Precision} = \frac{TP}{TP + FP}$ $\text{Recall} = \frac{TP}{TP + FN}$ $\text{F1} = 2 \cdot \frac{\text{Precision} \cdot \text{Recall}}{\text{Precision} + \text{Recall}}$

Return a tuple (precision, recall, f1), each rounded to 4 decimal places. If a denominator is 0, return 0.0 for that metric.

Example:

Input:

y_true = [1, 0, 1, 1, 0, 1]
y_pred = [1, 0, 0, 1, 1, 1]

Output:

(0.75, 0.75, 0.75)

Reasoning:

First, we identify the true positives ( $TP$ ), false positives ( $FP$ ), and false negatives ( $FN$ ) by comparing y_true and y_pred: $TP = 3$ , $FP = 1$ , $FN = 1$
Then, we calculate precision and recall using the given formulas: $\text{Precision} = \frac{3}{3 + 1} = \frac{3}{4} = 0.75$ , $\text{Recall} = \frac{3}{3 + 1} = \frac{3}{4} = 0.75$
Next, we calculate the F1 score using the precision and recall values: $\text{F1} = 2 \cdot \frac{0.75 \cdot 0.75}{0.75 + 0.75} = 2 \cdot \frac{0.5625}{1.5} = 0.75$
The final output is a tuple of the calculated metrics, each rounded to 4 decimal places: $(0.75, 0.75, 0.75)$

Constraints:

y_true and y_pred are lists of 0s and 1s
Return tuple (precision, recall, f1) rounded to 4 decimal places
Handle edge case where denominator is 0 (return 0.0)

Editor

Python 3.13.1

Test Results

0/0

Run code to see test results.