您可以使用 scikit-learn 计算每个类的 recall 和 precision。

示例：

from sklearn.metrics import classification_report

y_true = [0,1,2,2]
y_pred = [0,1]
target_names = ['class 0','class 1','class 2']
print(classification_report(y_true,y_pred,target_names=target_names))

              precision    recall  f1-score   support

     class 0       0.50      1.00      0.67         1
     class 1       0.00      0.00      0.00         1
     class 2       1.00      0.67      0.80         3

    accuracy                           0.60         5
   macro avg       0.50      0.56      0.49         5
weighted avg       0.70      0.60      0.61         5

参考here

正如其他人已经推荐的那样，您可以直接计算 y_actual 和 y_predictedusingsklearn.metricswithprecision_scoreandrecall_score` 来计算您需要的值。在此处阅读有关 precision 和 recall 分数的更多信息。

但是，IIUC，您希望直接使用混淆矩阵做同样的事情。以下是直接使用混淆矩阵计算准确率和召回率的方法。

1. 首先，我将使用一个虚拟示例进行演示，显示来自 SKLEARN API 的结果，然后直接计算它们。

注意：通常计算的精度和召回率有 2 种类型 -

• 微精度：所有类别的所有 TP 相加并除以 TP+FP
• 宏观精度：对每个类分别计算TP/TP+FP，然后取平均值（忽略nans）
• 您可以在 types of precision (and recall) here 上找到更多详细信息。

我在下面展示了两种方法供您理解 -

import numpy as np
from sklearn.metrics import confusion_matrix,precision_score,recall_score

####################################################
#####Using SKLEARN API on TRUE & PRED Labels########
####################################################

y_true = [0,1]
y_pred = [0,2]
confusion_matrix(y_true,y_pred)

precision_micro = precision_score(y_true,average="micro")
precision_macro = precision_score(y_true,average="macro")
recall_micro = recall_score(y_true,average='micro')
recall_macro = recall_score(y_true,average="macro")

print("Sklearn API")
print("precision_micro:",precision_micro)
print("precision_macro:",precision_macro)
print("recall_micro:",recall_micro)
print("recall_macro:",recall_macro)

####################################################
####Calculating directly from confusion matrix######
####################################################

cf = confusion_matrix(y_true,y_pred)
TP = cf.diagonal()

precision_micro = TP.sum()/cf.sum()
recall_micro = TP.sum()/cf.sum()

#NOTE: The sum of row-wise sums of a matrix = sum of column-wise sums of a matrix = sum of all elements of a matrix
#Therefore,the micro-precision and micro-recall is mathematically the same for a multi-class problem.


precision_macro = np.nanmean(TP/cf.sum(0))
recall_macro = np.nanmean(TP/cf.sum(1))

print("")
print("Calculated:")
print("precision_micro:",recall_macro)

Sklearn API
precision_micro: 0.6666666666666666
precision_macro: 0.8333333333333334
recall_micro: 0.6666666666666666
recall_macro: 0.7777777777777777

Calculated:
precision_micro: 0.6666666666666666
precision_macro: 0.8333333333333334
recall_micro: 0.6666666666666666
recall_macro: 0.7777777777777777

2. 既然我已经证明 API 背后的定义如所描述的那样工作，让我们为您的案例计算准确率和召回率。

cf = [[748,4,5,16,9,8,0],[  0,869,6,12,3],[  6,19,642,33,13,7,15,31,6],[  5,3,30,679,44,23,12],[  4,704,10,43],39,11,566,18,10],17,737,14,752,42],[  7,34,28,29,600,18],21,50,680]]

cf = np.array(cf)
TP = cf.diagonal()

precision_micro = TP.sum()/cf.sum()
recall_micro = TP.sum()/cf.sum()


precision_macro = np.nanmean(TP/cf.sum(0))
recall_macro = np.nanmean(TP/cf.sum(1))

print("Calculated:")
print("precision_micro:",recall_macro)

Calculated:
precision_micro: 0.872125
precision_macro: 0.8702549015235986
recall_micro: 0.872125
recall_macro: 0.8696681555022805