将NumPy数组的指定元素转换为新值

I wanted to convert the specified elements of the NumPy array A: 1, 5, and 8 into 0.

我想将NumPy数组A: 1、5和8的指定元素转换为0。

So I did the following:

所以我做了如下的事:

import numpy as np

A = np.array([[1,2,3,4,5],[6,7,8,9,10]])

bad_values = (A==1)|(A==5)|(A==8)

A[bad_values] = 0

print A

Yes, I got the expected result, i.e., new array.

是的，我得到了预期的结果。新数组。

However, in my real world problem, the given array (A) is very large and is also 2-dimensional, and the number of bad_values to be converted into 0 are also too many. So, I tried the following way of doing that:

然而，在我的现实问题中，给定的数组(A)非常大，而且也是二维的，要转换为0的bad_values的数量也太多了。所以，我尝试了以下方法:

bads = [1,5,8] # Suppose they are the values to be converted into 0

bad_values = A == x for x in bads  # HERE is the problem I am facing

How can I do this?

我该怎么做呢?

Then, of course the remaining is the same as before.

那么，当然剩下的和以前一样。

A[bad_values] = 0

print A

1 个解决方案

#1

If you want to get the index of where a bad value occurs in your array A, you could use in1d to find out which values are in bads:

如果你想要得到坏值在你的数组a中出现的索引，你可以使用in1d来找出坏值在哪里:

>>> np.in1d(A, bads)
array([ True, False, False, False,  True, False, False,  True, False, False], dtype=bool)

So you can just write A[np.in1d(A, bads)] = 0 to set the bad values of A to 0.

所以你可以写一个[np]in1d(A, bads)] = 0，将A的差值设为0。

EDIT: If your array is 2D, one way would be to use the in1d method and then reshape:

编辑:如果你的数组是2D的，一种方法是使用in1d方法，然后重新塑造:

>>> B = np.arange(9).reshape(3, 3)
>>> B
array([[0, 1, 2],
       [3, 4, 5],
       [6, 7, 8]])

>>> np.in1d(B, bads).reshape(3, 3)
array([[False,  True, False],
       [False, False,  True],
       [False, False,  True]], dtype=bool)

So you could do the following:

你可以这样做:

>>> B[np.in1d(B, bads).reshape(3, 3)] = 0
>>> B
array([[0, 0, 2],
       [3, 4, 0],
       [6, 7, 0]])

#1