在Python/numpy中环绕切片 [英] wrapping around slices in Python / numpy

问题描述

我有一个numpy数组，我想获得第i个点的邻居".通常，我使用的数组是二维的，但是下面的一维示例说明了我要查找的内容.如果

I have a numpy array, and I want to get the "neighbourhood" of the i'th point. Usually the arrays I'm using are two-dimensional, but the following 1D example illustrates what I'm looking for. If

A = numpy.array([0,10,20,30,40,50,60,70,80,90])

则元素4的(大小5)邻域为[20,30,40,50,60]，这可以通过执行A[i-2:i+3]轻松获得.

Then the (size 5) neighbourhood of element 4 is [20,30,40,50,60], and this can easily be obtained by doing A[i-2:i+3].

但是，我还需要邻域来环绕"数组的边缘，以使元素0的邻域为[80,90,0,10,20]，元素9的邻域为[70,80,90,0,10].我似乎找不到一种完美的方法来执行此操作，因此每次出现这种情况时，我最终都不得不使用一些复杂而烦人的逻辑(这对我来说是很常见的).在二维情况下，一个点的邻域将是一个矩形阵列.

However, I also need the neighbourhoods to "wrap around" the edges of the array, so that the neighbourhood of the element 0 is [80,90,0,10,20] and the neighbourhood of the element 9 is [70,80,90,0,10]. I can't seem to find an elegant way to do this, so I end up having to use some complicated, annoying logic every time this comes up (which is very often for me). In the 2D case the neighbourhood of a point would be a rectangular array.

所以我的问题是，是否有一种整洁的方法来表示这种numpy中的环绕式邻居"操作?我希望返回的是切片而不是副本，但是可读性和速度是最重要的考虑因素.

So my question is, is there a neat way to expres this "wrap-around neighbourhood" operation in numpy? I would prefer something that returns a slice rather than a copy, but readability and speed are the most important considerations.

推荐答案

numpy.take在'wrap'模式下将使用您的索引以数组的长度为模.

numpy.take in 'wrap' mode will use your indices modulo the length of the array.

indices = range(i-2,i+3)
neighbourhood = A.take(indices, mode='wrap')

有关详细信息，请参见文档 numpy.take

See documentation for details numpy.take

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