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Is there a function or more concise notation to get b or bb in the code snippet below?

import numpy as np

a = np.arange(12).reshape(3, 4)
c = (1, 2)
r = 1
b = a[c[0]-r:c[0]+r, c[1]-r:c[1]+r]
bb = a[c[0]-r:c[0]+r+1, c[1]-r:c[1]+r+1]

The objective is to get a subarray defined by a center index c and radius r (L1-norm).

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  • What's wrong with your current approach? Commented Oct 11, 2022 at 20:00
  • What exactly is wrong with bb?. It seems like a pythonic solution and how numpy is intended to work. Commented Oct 11, 2022 at 20:02
  • @mozway @Timo Imagine a higher dimensional array, e.g. 8, and all the cruft hiding intended meaning, which is c-r:c+r. Commented Oct 11, 2022 at 20:05
  • 1
    Not much better in my opinion, but you might consider a[tuple(slice(i-r, i+r) for i in c)] for higher dimensions. Commented Oct 11, 2022 at 20:10
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