在numpy数组中将min值替换为另一个

Question

假设我们有这个数组,我想用50号替换最小值

import numpy as np
numbers = np.arange(20)
numbers[numbers.min()] = 50

所以输出是 [50,1,2,3,....20]

但是现在我遇到了这个问题:

numbers = np.arange(20).reshape(5,4)
numbers[numbers.min(axis=1)]=50

要得到 [[50,1,2,3],[50,5,6,7],....]

但是我收到此错误:

IndexError:索引8超出轴0的范围,大小为5 ....

有任何帮助的想法吗？

Answer 1

您需要使用numpy.argmin而不是numpy.min:

In [89]: numbers = np.arange(20).reshape(5,4)

In [90]: numbers[np.arange(len(numbers)), numbers.argmin(axis=1)] = 50
In [91]: numbers
Out[91]: 
array([[50,  1,  2,  3],
       [50,  5,  6,  7],
       [50,  9, 10, 11],
       [50, 13, 14, 15],
       [50, 17, 18, 19]])

In [92]: numbers = np.arange(20).reshape(5,4)

In [93]: numbers[1,3] = -5 # Let's make sure that mins are not on same column

In [94]: numbers[np.arange(len(numbers)), numbers.argmin(axis=1)] = 50

In [95]: numbers
Out[95]: 
array([[50,  1,  2,  3],
       [ 4,  5,  6, 50],
       [50,  9, 10, 11],
       [50, 13, 14, 15],
       [50, 17, 18, 19]])

(我相信我的原始答案是不正确的,我混淆了行和列,这是对的)