Problems

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Compute Mean Average Precision (mAP)

Metrics & Evaluation

Hard

For each retrieval query, rank items by descending score, compute Average Precision, then average the query scores to obtain mean Average Precision.

For one query:

\operatorname{AP} = \frac{1}{R}\sum_{r=1}^{n}\operatorname{Precision}(r)\operatorname{rel}(r)

Across $Q$ queries:

\operatorname{mAP} = \frac{1}{Q}\sum_{q=1}^{Q}\operatorname{AP}_q

Here, $r$ is a rank, $R$ is the total number of relevant items in that query, and $\operatorname{rel}(r)$ is one when the item at rank $r$ is relevant. If k is provided, only ranks through k contribute to the numerator, while $R$ remains the total relevant count. A query with no relevant items has AP zero. Return map_value and ap_per_query in a dictionary.

Loading visualization...

Examples

Input: y_true_list = [[1, 0, 1, 0]], y_score_list = [[0.9, 0.8, 0.7, 0.1]], k = None

Output: {"map_value": 0.833333, "ap_per_query": [0.833333]}

Explanation: Relevant items occur at ranks 1 and 3, where precision is 1 and 2/3.

Input: y_true_list = [[1, 0, 1], [1, 1, 0]], y_score_list = [[0.9, 0.8, 0.7], [0.9, 0.8, 0.7]], k = None

Output: {"map_value": 0.916667, "ap_per_query": [0.833333, 1.0]}

Hint 1

Use np.argsort(-scores, kind="stable") to obtain each ranking.

Hint 2

Use cumulative relevant counts divided by np.arange(1, n + 1) for precision at every rank.

Hint 3

Mask the precision array by the ranked relevance labels before summing.

Requirements

Sort each query by descending score
Compute precision only at relevant ranks, respecting the optional cutoff
Assign AP zero when a query has no relevant items
Return exactly map_value and ap_per_query in a dictionary

Constraints

Each relevance label is 0 or 1
Corresponding relevance and score lists have equal lengths
Use NumPy only

Try Similar Problems

Precision Recall At K Hit Rate At K Classification Metrics Iou Bounding Box Ndcg

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Compute Mean Average Precision (mAP)

Metrics & Evaluation

Hard

For each retrieval query, rank items by descending score, compute Average Precision, then average the query scores to obtain mean Average Precision.

For one query:

\operatorname{AP} = \frac{1}{R}\sum_{r=1}^{n}\operatorname{Precision}(r)\operatorname{rel}(r)

Across $Q$ queries:

\operatorname{mAP} = \frac{1}{Q}\sum_{q=1}^{Q}\operatorname{AP}_q

Loading visualization...

Examples

Input: y_true_list = [[1, 0, 1, 0]], y_score_list = [[0.9, 0.8, 0.7, 0.1]], k = None

Output: {"map_value": 0.833333, "ap_per_query": [0.833333]}

Explanation: Relevant items occur at ranks 1 and 3, where precision is 1 and 2/3.

Input: y_true_list = [[1, 0, 1], [1, 1, 0]], y_score_list = [[0.9, 0.8, 0.7], [0.9, 0.8, 0.7]], k = None

Output: {"map_value": 0.916667, "ap_per_query": [0.833333, 1.0]}

Hint 1

Use np.argsort(-scores, kind="stable") to obtain each ranking.

Hint 2

Use cumulative relevant counts divided by np.arange(1, n + 1) for precision at every rank.

Hint 3

Mask the precision array by the ranked relevance labels before summing.

Requirements

Sort each query by descending score
Compute precision only at relevant ranks, respecting the optional cutoff
Assign AP zero when a query has no relevant items
Return exactly map_value and ap_per_query in a dictionary

Constraints

Each relevance label is 0 or 1
Corresponding relevance and score lists have equal lengths
Use NumPy only

Try Similar Problems

Precision Recall At K Hit Rate At K Classification Metrics Iou Bounding Box Ndcg

Case 1

Case 2

y_true_list =

Accepts: array

y_score_list =

Accepts: array

k =

Accepts: any

For each retrieval query, rank items by descending score, compute Average Precision, then average the query scores to obtain mean Average Precision.

For one query:

\operatorname{AP} = \frac{1}{R}\sum_{r=1}^{n}\operatorname{Precision}(r)\operatorname{rel}(r)

Across $Q$ queries:

\operatorname{mAP} = \frac{1}{Q}\sum_{q=1}^{Q}\operatorname{AP}_q