如何在python中用另一个numpy数组索引一个numpy数组 [英] How to index a numpy array with another numpy array in python

问题描述

我试图用另一个数组索引一个 np.array 以便在某个索引之后我可以在任何地方都有零，但它给了我错误

I am trying to index an np.array with another array so that I can have zeros everywhere after a certain index but it gives me the error

TypeError: 只有整数标量数组可以转换为标量索引

TypeError: only integer scalar arrays can be converted to a scalar index

基本上我希望我的代码做的是，如果我有:

Basically what I would like my code to do is that if I have:

a = np.array([[1, 2, 3], [4, 5, 6], [7, 8, 9], [10, 11, 12]])
d = np.array([2, 1, 3])

我可以做类似的事情

a[d:] = 0

给出输出

a = [[ 1  2  3]
     [ 4  0  6]
     [ 0  0  9]
     [ 0  0  0]]

推荐答案

对于此类问题，一种广泛使用的方法是构造一个布尔掩码，将索引数组与适当的arange 进行比较:

A widely used approach for this kind of problem is to construct a boolean mask, comparing the index array with the appropriate arange:

In [619]: mask = np.arange(4)[:,None]>=d
In [620]: mask
Out[620]: 
array([[False, False, False],
       [False,  True, False],
       [ True,  True, False],
       [ True,  True,  True]])
In [621]: a[mask]
Out[621]: array([ 5,  7,  8, 10, 11, 12])
In [622]: a[mask] = 0
In [623]: a
Out[623]: 
array([[1, 2, 3],
       [4, 0, 6],
       [0, 0, 9],
       [0, 0, 0]])

这不一定比行(或在本例中为列)迭代快.由于切片是基本的索引编制，因此即使进行多次也可能会更快.

That's not necessarily faster than a row (or in this case column) iteration. Since slicing is basic indexing, it may be faster, even if done several times.

In [624]: for i,v in enumerate(d):
     ...:     print(a[v:,i])
     ...: 
[0 0]
[0 0 0]
[0]

通常，如果结果涉及多个不同长度的数组或列表，则不会出现整洁"的结果.多维解决方案.要么迭代这些列表，要么退后一步跳出框框思考".

Generally if a result involves multiple arrays or lists with different lengths, there isn't a "neat" multidimensional solution. Either iterate over those lists, or step back and "think outside the box".

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