Python 中的数组 TP、TN、FP 和 FN

Question

我的预测结果是这样的

测试数组

[1,0,0,0,1,0,1,...,1,0,1,1],
[1,0,1,0,0,1,0,...,0,1,1,1],
[0,1,1,1,1,1,0,...,0,1,1,1],
.
.
.
[1,1,0,1,1,0,1,...,0,1,1,1],

预测数组

[1,0,0,0,0,1,1,...,1,0,1,1],
[1,0,1,1,1,1,0,...,1,0,0,1],
[0,1,0,1,0,0,0,...,1,1,1,1],
.
.
.
[1,1,0,1,1,0,1,...,0,1,1,1],

这是我拥有的数组的大小

TestArray.shape

Out[159]: (200, 24)

PredictionArray.shape

Out[159]: (200, 24)

我想为这些阵列获得 TP、TN、FP 和 FN

我试过这个代码

cm=confusion_matrix(TestArray.argmax(axis=1), PredictionArray.argmax(axis=1))
TN = cm[0][0]
FN = cm[1][0]
TP = cm[1][1]
FP = cm[0][1]
print(TN,FN,TP,FP)

但我得到的结果

TN = cm[0][0]
FN = cm[1][0]
TP = cm[1][1]
FP = cm[0][1]
print(TN,FN,TP,FP)

125 5 0 1

我检查了厘米的形状

cm.shape

Out[168]: (17, 17)

125 + 5 + 0 + 1 = 131 这不等于我拥有的列数 200

我期望有 200，因为阵列中的每个单元格假设是 TF、TN、FP、TP，所以总数应该是 200

如何解决？

这是问题的一个例子

import numpy as np
from sklearn.metrics import confusion_matrix


TestArray = np.array(
[
[1,0,0,1,0,1,1,0,1,0,1,1,0,0,1,1,1,0,0,1],
[0,1,1,0,1,0,0,1,0,0,0,1,0,1,0,1,1,0,1,1],
[1,0,1,1,1,1,0,0,1,1,1,1,0,0,1,0,0,0,0,0],
[0,1,1,1,0,0,0,0,0,1,0,0,1,0,0,1,0,1,1,1],
[0,0,0,0,1,1,0,1,1,0,0,1,0,1,1,0,1,1,1,1],
[1,0,0,1,1,1,0,1,1,0,1,0,0,1,1,0,0,1,0,0],
[1,1,1,0,0,1,0,0,1,1,0,1,0,1,1,1,1,1,0,1],
[0,0,0,1,0,0,1,0,1,0,1,0,0,0,0,1,0,0,1,1],
[1,0,1,0,0,0,0,1,0,1,0,1,0,0,0,0,1,0,1,0],
[1,1,0,1,1,1,1,0,1,0,1,0,1,1,1,1,0,1,0,0]
])

TestArray.shape



PredictionArray = np.array(
[
[0,0,0,1,1,1,1,0,0,0,1,0,0,0,1,0,1,0,1,1],
[0,1,0,0,1,0,1,1,0,0,0,1,1,0,0,1,1,0,0,1],
[1,1,0,1,1,1,0,0,0,0,0,1,0,0,1,0,0,1,0,0],
[0,1,0,1,0,0,1,0,0,1,0,1,1,0,0,1,0,0,1,1],
[0,0,1,0,0,1,0,1,1,1,0,1,1,1,0,0,1,1,0,1],
[1,0,0,1,0,1,1,1,1,0,0,1,0,1,1,1,0,1,1,0],
[1,1,0,0,1,1,0,0,0,1,0,1,0,0,1,1,0,1,0,1],
[0,0,0,0,0,0,0,1,1,0,1,0,0,1,0,1,1,0,1,1],
[1,0,1,1,0,0,0,1,0,1,0,1,1,1,1,0,0,0,1,0],
[1,1,0,1,1,1,1,1,1,0,1,0,0,0,0,1,1,1,0,0]
])

PredictionArray.shape

cm=confusion_matrix(TestArray.argmax(axis=1), PredictionArray.argmax(axis=1))
TN = cm[0][0]
FN = cm[1][0]
TP = cm[1][1]
FP = cm[0][1]

print(TN,FN,TP,FP)

输出是

5 0 2 0 

= 5+0+2+0 = 7 !!

数组中有 20 列和 10 行

但厘米给总共 7 ！

Answer 1

使用np.argmax ，您输入的矩阵sklearn.metrics.confusion_matrix不再是二进制的，因为np.argmax返回第一个出现的最大值的索引。 在这种情况下，沿axis=1 。

当您的预测不是二进制时，您不会得到好的真阳性/命中、真阴性/正确拒绝等。

您应该会发现sum(sum(cm))确实等于 200。

---

如果数组的每个索引代表一个单独的预测，即您试图获得 TP/TN/FP/FN 总共 200 ( 10 * 20 ) 个预测，每个预测的结果为0或1 ，那么您可以在将数组解析为confusion_matrix矩阵之前，通过展平数组来获得 TP/TN/FP/FN。 也就是说，您可以将TestArray和PreditionArry重塑为(200,) ，例如：

cm = confusion_matrix(TestArray.reshape(-1), PredictionArray.reshape(-1))

TN = cm[0][0]
FN = cm[1][0]
TP = cm[1][1]
FP = cm[0][1]

print(TN, FN, TP, FP, '=', TN + FN + TP + FP)

哪个返回

74 28 73 25 = 200