将最小值替换为 numpy 数组中的另一个

Question

假设我们有这个数组，我想用数字 50 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]

但现在我遇到了问题：

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

获得[[50,1,2,3],[50,5,6,7],....]

但是我得到这个错误：

IndexError: index 8 is out of bounds for axis 0 with size 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]])

（我认为我原来的答案是错误的，我混淆了行和列，这是对的）

将最小值替换为 numpy 数组中的另一个

replace min value to another in numpy array

python

arrays

numpy

min