值数组在numpy数组中的位置 [英] location of array of values in numpy array

问题描述

这里有一个小代码来说明问题.

Here is a small code to illustrate the problem.

A = array([[1,2], [1,0], [5,3]])
f_of_A = f(A)   # this is precomputed and expensive


values = array([[1,2], [1,0]])


# location of values in A
# if I just had 1d values I could use numpy.in1d here
indices = array([0, 1])


# example of operation type I need (recalculating f_of_A as needed is not an option)
f_of_A[ indices ]

因此，基本上我认为我需要一些与in1d等效的东西才能获得更高的尺寸.这样的事情存在吗?还是有其他方法?

So, basically I think I need some equivalent to in1d for higher dimensions. Does such a thing exist? Or is there some other approach?

好像还有一个searchsorted()函数，但这似乎也适用于一维数组.在此示例中，我使用了2d点，但是任何解决方案也需要适用于3d点.

Looks like there is also a searchsorted() function, but that seems to work for 1d arrays also. In this example I used 2d points, but any solution would need to work for 3d points also.

推荐答案

好的，这就是我想出的.

Okay, this is what I came up with.

要找到一个多维索引的值，例如ii = np.array([1,2])，我们可以执行以下操作:

To find the value of one multi-dimensional index, let's say ii = np.array([1,2]), we can do:

n.where((A == ii).all(axis=1))[0]

让我们分解一下，我们有了A == ii，它将对A的每一行与ii进行逐元素比较.我们希望整行都是真实的，因此我们添加.all(axis=1)使其折叠.为了找到这些索引发生的位置，我们将其插入np.where并获取元组的第一个值.

Let's break this down, we have A == ii, which will give element-wise comparisons with ii for each row of A. We want an entire row to be true, so we add .all(axis=1) to collapse them. To find where these indices happen, we plug this into np.where and get the first value of the tuple.

现在，我还没有一种快速的方法来处理多个索引(尽管我有一种感觉).但是，这可以完成工作:

Now, I don't have a fast way to do this with multiple indices yet (although I have a feeling there is one). However, this will get the job done:

np.hstack([np.where((A == values[i]).all(axis=1))[0] for i in xrange(len(values))])

对于每个values值，这基本上只是调用上述方法，并将结果串联起来.

This basically just calls the above, for each value of values, and concatenates the result.

更新:

这里是多维案例(一口气，应该相当快):

Here is for the multi-dimensional case (all in one go, should be fairly fast):

np.where((np.expand_dims(A, -1) == values.T).all(axis=1).any(axis=1))[0]

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