Python 中的數組 TP、TN、FP 和 FN

Question

我的預測結果是這樣的

測試數組

[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],

預測數組

[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],

這是我擁有的數組的大小

TestArray.shape

Out[159]: (200, 24)

PredictionArray.shape

Out[159]: (200, 24)

我想為這些陣列獲得 TP、TN、FP 和 FN

我試過這個代碼

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)

但我得到的結果

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

我檢查了厘米的形狀

cm.shape

Out[168]: (17, 17)

125 + 5 + 0 + 1 = 131 這不等於我擁有的列數 200

我期望有 200，因為陣列中的每個單元格假設是 TF、TN、FP、TP，所以總數應該是 200

如何解決？

這是問題的一個例子

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)

輸出是

5 0 2 0 

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

數組中有 20 列和 10 行

但厘米給總共 7 ！

Answer 1

使用np.argmax ，您輸入的矩陣sklearn.metrics.confusion_matrix不再是二進制的，因為np.argmax返回第一個出現的最大值的索引。 在這種情況下，沿axis=1 。

當您的預測不是二進制時，您不會得到好的真陽性/命中、真陰性/正確拒絕等。

您應該會發現sum(sum(cm))確實等於 200。

---

如果數組的每個索引代表一個單獨的預測，即您試圖獲得 TP/TN/FP/FN 總共 200 ( 10 * 20 ) 個預測，每個預測的結果為0或1 ，那么您可以在將數組解析為confusion_matrix矩陣之前，通過展平數組來獲得 TP/TN/FP/FN。 也就是說，您可以將TestArray和PreditionArry重塑為(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)

哪個返回

74 28 73 25 = 200