如何按元素比較3個numpy數組，並獲得具有最大值的數組的結果？

Question

numpy數組包含如下所示的預測概率：

predict_prob1 =([[0.95602106, 0.04397894],
                 [0.93332366, 0.06667634],
                 [0.97311459, 0.02688541],
                 [0.97323962, 0.02676038]])

predict_prob2 =([[0.70425144, 0.29574856],
                 [0.69751251, 0.30248749],
                 [0.7072872 , 0.2927128 ],
                 [0.68683139, 0.31316861]])

predict_prob3 =([[0.56551921, 0.43448079],
                 [0.93321106, 0.06678894],
                 [0.92345399, 0.07654601],
                 [0.88396842, 0.11603158]])

我想將這三個numpy.ndarray逐元素進行比較，並找出哪個數組具有最大的概率。 三個數組的長度相同。 我試圖實現這種不正確的方法。

for i in range(len(predict_prob1)):
    if(predict_prob1[i] > predict_prob2[i])
        c = predict_prob1[i]
    else
        c = predict_prob2[i]
    if(c > predict_prob3[i])
        result = c
    else
        result = array[i]

請幫忙！！

Answer 1

您可以使用np.maximum.reduce ：

np.maximum.reduce([A, B, C])

其中A ， B ， C是numpy.ndarray

對於您的示例，結果為：

[[0.95602106 0.43448079]
 [0.93332366 0.30248749]
 [0.97311459 0.2927128 ]
 [0.97323962 0.31316861]]

Answer 2

對我來說，還不清楚您要問的是什么— 如果您想要的結果是一個4x2數組，該數組索引三個數組中哪個數組在i,j位置具有最大值i,j那么您想使用np.argmax

>>> import numpy as np
>>> predict_prob1 =([[0.95602106, 0.04397894],
    [0.93332366, 0.06667634],
    [0.97311459, 0.02688541],
    [0.97323962, 0.02676038]])
>>> predict_prob2 =([[0.70425144, 0.29574856],
    [0.69751251, 0.30248749],
    [0.7072872 , 0.2927128 ],
    [0.68683139, 0.31316861]])
>>> predict_prob3 =([[0.56551921, 0.43448079],
    [0.93321106, 0.06678894],
    [0.92345399, 0.07654601],
    [0.88396842, 0.11603158]])
>>> np.argmax((predict_prob1,predict_prob2,predict_prob3), 0)
array([[0, 2],
       [0, 1],
       [0, 1],
       [0, 1]])
>>>

---

附錄

閱讀了OP的評論后，我在回答中添加了以下內容

>>> names = np.array(['predict_prob%d'%(i+1) for i in range(3)])
>>> names[np.argmax((predict_prob1,predict_prob2,predict_prob3),0)]
array([['predict_prob1', 'predict_prob3'],
       ['predict_prob1', 'predict_prob2'],
       ['predict_prob1', 'predict_prob2'],
       ['predict_prob1', 'predict_prob2']], dtype='<U13')
>>> 

Answer 3

假設您要為每一行分配類別0概率最高的數組索引：

which = 0

np.stack([predict_prob1, predict_prob2, predict_prob3], axis=2)[:, which, :].argmax(axis=1)

輸出：

array([0, 0, 0, 0])

對於第1類：

array([2, 1, 1, 1])

Answer 4

您可以使用操作數>和<產生數組的布爾掩碼的事實。

import numpy as np

predict_prob1 =np.array([[0.95602106, 0.04397894],
   [0.93332366, 0.06667634],
   [0.97311459, 0.02688541],
   [0.97323962, 0.02676038]])

predict_prob2 =np.array([[0.70425144, 0.29574856],
   [0.69751251, 0.30248749],
   [0.7072872 , 0.2927128 ],
   [0.68683139, 0.31316861]])

predict_prob3 =np.array([[0.56551921, 0.43448079],
   [0.93321106, 0.06678894],
   [0.92345399, 0.07654601],
   [0.88396842, 0.11603158]])

predict_prob = (predict_prob1>predict_prob2)*predict_prob1 + (predict_prob1<predict_prob2)*predict_prob2
predict_prob = (predict_prob>predict_prob3)*predict_prob + (predict_prob<predict_prob3)*predict_prob3

print(predict_prob)

結果是：

[[0.95602106 0.43448079]
 [0.93332366 0.30248749]
 [0.97311459 0.2927128 ]
 [0.97323962 0.31316861]]