Python 使用 Sklearn 解决多类问题的精度、召回率和 F1

我有一个多班级问题，其中0是我的负面班级，1和2是正班级.判断以下代码:

import numpy as np
from sklearn.metrics import confusion_matrix
from sklearn.metrics import ConfusionMatrixDisplay
from sklearn.metrics import f1_score
from sklearn.metrics import precision_score
from sklearn.metrics import recall_score

# Outputs
y_true = np.array((1, 2, 2, 0, 1, 0))
y_pred = np.array((1, 0, 0, 0, 0, 1))
# Metrics
precision_macro = precision_score(y_true, y_pred, average='macro')
precision_weighted = precision_score(y_true, y_pred, average='weighted')
recall_macro = recall_score(y_true, y_pred, average='macro')
recall_weighted = recall_score(y_true, y_pred, average='weighted')
f1_macro = f1_score(y_true, y_pred, average='macro')
f1_weighted = f1_score(y_true, y_pred, average='weighted')
# Confusion Matrix
cm = confusion_matrix(y_true, y_pred)
disp = ConfusionMatrixDisplay(confusion_matrix=cm)
disp.plot()
plt.show()

在本例中，使用Sklearn计算的指标如下:

precision_macro = 0.25
precision_weighted = 0.25
recall_macro = 0.33333
recall_weighted = 0.33333
f1_macro = 0.27778
f1_weighted = 0.27778

这是混淆矩阵:

macro和weighted是一样的，因为我每个班级都有相同数量的样本？这是我手动完成的操作.

1-精度=TP/(TP+FP).因此，对于班级1和2，我们得到:

Precision1 = TP1/(TP1+FP1) = 1/(1+1) = 0.5
Precision2 = TP2/(TP2+FP2) = 0/(0+0) = 0 (this returns 0 according Sklearn documentation)
Precision_Macro = (Precision1 + Precision2)/2 = 0.25
Precision_Weighted = (2*Precision1 + 2*Precision2)/4 = 0.25

2-Recall=Tp/(Tp+Fn).因此，对于班级1和2，我们得到:

Recall1 = TP1/(TP1+FN1) = 1/(1+1) = 0.5
Recall2 = TP2/(TP2+FN2) = 0/(0+2) = 0
Recall_Macro = (Recall1+Recall2)/2 = (0.5+0)/2 = 0.25
Recall_Weighted = (2*Recall1+2*Recall2)/4 = (2*0.5+2*0)/4 = 0.25

3-F1=2*(精度*召回)/(精度+呼叫)

F1_Macro = 2*(Precision_Macro*Recall_Macro)/(Precision_Macro*Recall_Macro) = 0.25
F1_Weighted = 2*(Precision_Weighted*Recall_Weighted)/(Precision_Weighted*Recall_Weighted) = 0.25

因此，精度分数等于Sklearn.但Recall和F1是不同的.我在这里做错了什么？即使您使用Precision和Recall from Sklearn的值(即0.25和0.3333)，您也无法获得0.27778的F1分数.