ndcg_score#

sklearn.metrics.ndcg_score(y_true, y_score, *, k=None, sample_weight=None, ignore_ties=False)[source]#

计算归一化折现累积增益。

计算真实分数根据预测分数排序后的对数折扣累加和。然后除以最佳可能分数（理想DCG，通过完美排序获得）以获得0到1之间的分数。

如果真实标签被 y_score 排在高位，此排名指标返回高值。

参数:

y_true形状为 (n_samples, n_labels) 的类数组对象: 多标签分类的真实目标，或待排序实体的真实分数。 y_true 中的负值可能导致输出不在0到1之间。
y_score形状为 (n_samples, n_labels) 的类数组对象: 目标分数，可以是概率估计、置信值，或非阈值决策度量（如某些分类器返回的“decision_function”）。
kint, default=None: 仅考虑排名中最高的k个分数。如果为 None，则使用所有输出。
sample_weightshape 为 (n_samples,) 的 array-like, default=None: 样本权重。如果为 None，则所有样本具有相同的权重。
ignore_tiesbool, default=False: 假设 y_score 中没有平局（如果 y_score 是连续的，这种情况很可能发生），以提高效率。

返回:

normalized_discounted_cumulative_gainfloat in [0., 1.]: 所有样本的平均NDCG分数。

另请参阅

dcg_score: 折扣累积增益（未归一化）。

References

维基百科关于折损累计增益的条目

Jarvelin, K., & Kekalainen, J. (2002). Cumulated gain-based evaluation of IR techniques. ACM Transactions on Information Systems (TOIS), 20(4), 422-446.

Wang, Y., Wang, L., Li, Y., He, D., Chen, W., & Liu, T. Y. (2013, May). A theoretical analysis of NDCG ranking measures. In Proceedings of the 26th Annual Conference on Learning Theory (COLT 2013)

McSherry, F., & Najork, M. (2008, March). Computing information retrieval performance measures efficiently in the presence of tied scores. In European conference on information retrieval (pp. 414-421). Springer, Berlin, Heidelberg.

示例

>>> import numpy as np
>>> from sklearn.metrics import ndcg_score
>>> # we have ground-truth relevance of some answers to a query:
>>> true_relevance = np.asarray([[10, 0, 0, 1, 5]])
>>> # we predict some scores (relevance) for the answers
>>> scores = np.asarray([[.1, .2, .3, 4, 70]])
>>> ndcg_score(true_relevance, scores)
0.69
>>> scores = np.asarray([[.05, 1.1, 1., .5, .0]])
>>> ndcg_score(true_relevance, scores)
0.49
>>> # we can set k to truncate the sum; only top k answers contribute.
>>> ndcg_score(true_relevance, scores, k=4)
0.35
>>> # the normalization takes k into account so a perfect answer
>>> # would still get 1.0
>>> ndcg_score(true_relevance, true_relevance, k=4)
1.0...
>>> # now we have some ties in our prediction
>>> scores = np.asarray([[1, 0, 0, 0, 1]])
>>> # by default ties are averaged, so here we get the average (normalized)
>>> # true relevance of our top predictions: (10 / 10 + 5 / 10) / 2 = .75
>>> ndcg_score(true_relevance, scores, k=1)
0.75
>>> # we can choose to ignore ties for faster results, but only
>>> # if we know there aren't ties in our scores, otherwise we get
>>> # wrong results:
>>> ndcg_score(true_relevance,
...           scores, k=1, ignore_ties=True)
0.5...

ndcg_score#

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