如何计算多标签分类的 F1-Score? [英] How To Calculate F1-Score For Multilabel Classification?

问题描述

我尝试计算 f1_score，但是当我使用 sklearn f1_score 方法时，我会收到一些警告.

I try to calculate the f1_score but I get some warnings for some cases when I use the sklearn f1_score method.

我有一个用于预测的多标签 5 类问题.

I have a multilabel 5 classes problem for a prediction.

import numpy as np
from sklearn.metrics import f1_score

y_true = np.zeros((1,5))
y_true[0,0] = 1 # => label = [[1, 0, 0, 0, 0]]

y_pred = np.zeros((1,5))
y_pred[:] = 1 # => prediction = [[1, 1, 1, 1, 1]]

result_1 = f1_score(y_true=y_true, y_pred=y_pred, labels=None, average="weighted")

print(result_1) # prints 1.0

result_2 = f1_score(y_true=y_ture, y_pred=y_pred, labels=None, average="weighted")

print(result_2) # prints: (1.0, 1.0, 1.0, None) for precision/recall/fbeta_score/support

当我使用 average="samples" 而不是 "weighted" 时，我得到 (0.1, 1.0, 0.1818..., None)."weighted" 选项是否对多标签问题没有用，或者我如何正确使用 f1_score 方法?

When I use average="samples" instead of "weighted" I get (0.1, 1.0, 0.1818..., None). Is the "weighted" option not useful for a multilabel problem or how do I use the f1_score method correctly?

我在使用 average="weighted" 时也收到警告:

I also get a warning when using average="weighted":

UndefinedMetricWarning:召回率和 F 分数定义不明确，在没有真实样本的标签中被设置为 0.0."

"UndefinedMetricWarning: Recall and F-score are ill-defined and being set to 0.0 in labels with no true samples."

推荐答案

如果你把数据稍微加起来就可以了:

It works if you slightly add up data:

y_true = np.array([[1,0,0,0], [1,1,0,0], [1,1,1,1]])
y_pred = np.array([[1,0,0,0], [1,1,1,0], [1,1,1,1]])

recall_score(y_true=y_true, y_pred=y_pred, average='weighted')
>>> 1.0
precision_score(y_true=y_true, y_pred=y_pred, average='weighted')
>>> 0.9285714285714286

f1_score(y_true=y_true, y_pred=y_pred, average='weighted')
>>> 0.95238095238095244

数据表明我们没有遗漏任何真阳性，也没有预测任何假阴性(recall_score 等于 1).然而，我们在第二次观察中预测了一个误报，导致 precision_score 等于 ~0.93.

The data suggests we have not missed any true positives and have not predicted any false negatives (recall_score equals 1). However, we have predicted one false positive in the second observation that lead to precision_score equal ~0.93.

由于precision_score 和recall_score 在weighted 参数下都不为零，因此存在f1_score.由于示例中缺少信息，我认为您的案例无效.

As both precision_score and recall_score are not zero with weighted parameter, f1_score, thus, exists. I believe your case is invalid due to lack of information in the example.

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