如何按元素比较 3 个 numpy 数组并将结果作为具有最大值的数组？

Question

numpy 数组包含如下所示的预测概率：

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]])

我想按元素比较这三个 numpy.ndarray 并找出哪个数组的概率最大。其中三个数组的长度相同。我试图实现这样的东西，但这是不正确的。

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]

请帮忙！！

Answer 1

假设您想要，对于每一行，class 0:

的概率最高的数组索引

which = 0

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

输出：

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

对于class 1:

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

Answer 2

你可以用 np.maximum.reduce:

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

其中 A、B、C 是 numpy.ndarray

对于您的示例，结果为：

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

Answer 3

~~对我来说，我不太清楚你在问什么 —~~ 如果你想要的结果是一个 4x2 数组，索引三个数组中哪一个在位置 i,j 那么你想使用 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]])
>>>

附录

阅读我将以下内容添加到我的回答中

>>> 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 4

您可以利用操作数 > 和 < 产生数组的布尔掩码这一事实。

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)

结果是：

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

如何按元素比较 3 个 numpy 数组并将结果作为具有最大值的数组？

How to compare 3 numpy arrays elementwise and get the results as the array with maximum values?

python

arrays

numpy

numpy-ndarray