问写一个以正类为输入的混淆矩阵函数
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Stack Overflow用户

提问于 2021-11-14 11:50:05

回答 2查看 58关注 0票数 0

我想创建一个混淆矩阵，而不依赖于任何包。我有两个列表(预测值和实际值)，我想把它们和一个正类的指示器一起输入到一个函数中。

例如，当1为正类时：

predicted_lst = [1, 0, 1, 0, 0]
actual_lst = [1, 0, 0, 1, 1]

我的函数目前看起来是这样的，但它的效率很低：

def confusion_matrix(predicted, actual, pos_class):
    
    TP = 0
    TN = 0 
    FP = 0
    FN = 0
    
    for i in range(len(actual)):
        if actual[i] == pos_class and predicted[i] == pos_class:
            TP +=1
        elif actual[i] == pos_class and predicted[i] != pos_class:
            FN +=1
        elif actual[i] != pos_class and predicted[i] == pos_class:
            FP +=1
        else:
            TN +=1
    return TP, FP, TN, FN

我的问题是，有没有更有效的方法来编写这些代码？我看到了these posts，但它们并不像我希望的那样将正类作为函数输入。我也根本不想使用任何包(包括numpy)

python

machine-learning

confusion-matrix

回答 2

Stack Overflow用户

回答已采纳

发布于 2021-11-14 22:56:43

请看下面的几个替代解决方案。

选项1：

def confmat_1(actual, predicted, positive, negative):

    tn = len([x for x in zip(predicted, actual) if x[0] == negative and x[1] == negative])
    fp = len([x for x in zip(predicted, actual) if x[0] == positive and x[1] == negative])
    fn = len([x for x in zip(predicted, actual) if x[0] == negative and x[1] == positive])
    tp = len([x for x in zip(predicted, actual) if x[0] == positive and x[1] == positive])

    return tn, fp, fn, tp

选项2：

def confmat_2(actual, predicted, positive, negative):

    tn = 0
    fp = 0
    fn = 0
    tp = 0

    for x in zip(predicted, actual):

        if x[0] == negative and x[1] == negative:
            tn += 1

        elif x[0] == positive and x[1] == negative:
            fp += 1

        elif x[0] == negative and x[1] == positive:
            fn += 1

        else:
            tp += 1

    return tn, fp, fn, tp

示例：

from sklearn.metrics import confusion_matrix

actual = [1, 0, 0, 1, 1]
predicted = [1, 0, 1, 0, 0]

# Option 1
tn, fp, fn, tp = confmat_1(actual, predicted, positive=1, negative=0)
print(tn, fp, fn, tp)
# 1 1 2 1

# Option 2
tn, fp, fn, tp = confmat_2(actual, predicted, positive=1, negative=0)
print(tn, fp, fn, tp)
# 1 1 2 1

# Scikit-learn
tn, fp, fn, tp = confusion_matrix(actual, predicted).ravel()
print(tn, fp, fn, tp)
# 1 1 2 1

票数 0

Stack Overflow用户

发布于 2021-11-14 13:27:38

您可以假设1是正数类来编写函数，如果0是正数类，则只需相应地更改返回值的顺序就可以适应pos_class参数。

在循环内部，您可以删除FP和FN的增量计算，因为这些值可以从循环外部的真值派生：

def confusion_matrix(predicted, actual, pos_class):
    
    TP = 0
    TN = 0 
    
    for pred, act in zip(predicted, actual):
        if pred == act:
            if act == 0:
                TN += 1
            else:
                TP += 1
    
    positive = sum(predicted)
    negative = len(predicted) - positive        
    FP = positive - TP 
    FN = negative - TN
    
    if pos_class == 1:
        return TP, FP, TN, FN
    else:
        return TN, FN, TP, FP