machine-learning - 为什么不能从宏精度和召回率计算宏 F1 度量？

Question

我对通过宏精度和手动召回计算宏 f1-score 感兴趣。但结果并不相等。代码中 f1 和 f1_new 的最终公式有什么区别？

from sklearn.metrics import precision_score, recall_score, f1_score

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

p = precision_score(y_true, y_pred, average='macro')
r = recall_score(y_true, y_pred, average='macro')
f1_new = (2 * p * r) / (p + r) # 0.6291390728476821

f1 = f1_score(y_true, y_pred, average='macro') #  0.6190476190476191

print(f1_new == f1) 
# False

Answer 1

f1_score在 scikit-learn 中计算如下:

all_positives = 4
all_negatives = 4
true_positives = 2
true_negatives = 3

true_positive_rate = true_positives/all_positives = 2/4
true_negative_rate = true_negatives/all_negatives = 3/4

pred_positives = 3
pred_negatives = 5

positive_predicted_value = true_positives/pred_positives = 2/3
negative_predicted_value = true_negatives/pred_negatives = 3/5

f1_score_pos = 2 * true_positive_rate * positive_predicted_value / (true_positive_rate + positive_predicted_value)
             = 2 * 2/4 * 2/3 / (2/4 + 2/3)

f1_score_neg = 2 * true_negative_rate * negative_predicted_value / (true_negative_rate + negative_predicted_value)
             = 2 * 3/4 * 3/5 / (3/4 + 3/5)

f1 = average(f1_score_pos, f1_score_neg)
   = 2/4 * 2/3 / (2/4 + 2/3) + 3/4 * 3/5 / (3/4 + 3/5)
   = 0.6190476190476191

这与 documentation 中给出的定义相匹配为 'macro' Skicit-learn 的参数 f1_score :计算每个标签的指标，并找到它们的未加权平均值。此定义也适用于 precision_score和 recall_score .
您手动计算的 F1 分数如下:

precision = average(positive_predicted_value, negative_predicted_value)
          = average(2/3, 3/5)
          = 19/30

recall = average(true_positive_rate, true_negative_rate)
       = average(2/4, 3/4)
       = 5/8

f1_new = 2 * precision * recall / (precision + recall)
       = 2 * 19/30 * 5/8 / (19/30 + 5/8)
       = 0.6291390728476821

其实通式 F1 = 2 * (precision * recall) / (precision + recall)如 docs 中所述仅对 average='binary' 有效和 average='micro' ，但不适用于 average='macro'和 average='weighted' .从这个意义上说，正如目前在 scikit-learn 中提出的那样，该公式具有误导性，因为它表明它与所选参数无关，但事实并非如此。