Python中的数组TP、TN、FP和FN_Python_Arrays_Confusion Matrix

Python中的数组TP、TN、FP和FN

python arrays

Python中的数组TP、TN、FP和FN,python,arrays,confusion-matrix,Python,Arrays,Confusion Matrix,我的预测结果是这样的 TestArray [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], [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,

我的预测结果是这样的

TestArray

[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],

[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],

PredictionArray

[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],

[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)

我想得到这些阵列的TP、TN、FP和FN

我试过这个密码

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)

但是我得到的结果

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cm的形状

cm.shape

Out[168]: (17, 17)

125+5+0+1=131，这不等于我的列数，即200

我预计会有200个，因为数组中的每个单元格假设为TF、TN、FP、TP，所以总数应该是200

如何解决这个问题

下面是一个问题的例子

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

但是cm总共给了7个

使用

np.argmax

时，您输入的矩阵

sklearn.metrics.conflusion\u 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

轴=1

当你的预测不是二元预测时，你不会得到好的结果——真正的肯定/命中、真正的否定/正确的拒绝等等

你应该发现

sum（sum（cm））

确实等于200

如果数组的每个索引表示一个单独的预测，即，您试图获得总共200个（

10*20

）预测的TP/TN/FP/FN，每个预测的结果为

或

，然后，您可以通过在将数组解析为

混淆矩阵

之前展平数组来获得TP/TN/FP/FN。也就是说，您可以将

TestArray

和

PreditionArry

重塑为

（200，）

，例如：

cm=混淆矩阵（TestArray.reformate（-1），PredictionArray.reformate（-1））
TN=cm[0][0]
FN=cm[1][0]
TP=cm[1][1]
FP=cm[0][1]
打印（TN，FN，TP，FP，'='，TN+FN+TP+FP）

74 28 73 25 = 200

使用

np.argmax