NumPy - Updating all values in array, except for slice

Question

I have the following array:

arr = np.array([[1, 1, 1, 1, 1, 1],
            [2, 0, 0, 0, 0, 2],
            [3, 0, 0, 0, 0, 3],
            [4, 4, 4, 4, 4, 4]])

inverse_slice = arr[1:3, 1:5]

I want to update the values of all values in the array, execpt for the values in the slice. For example, this could be multiplying all values with 2, exepct for values in the slice (the rectangle of 0's in this case).

How can I achieve this in an efficient manner?

NOTE: Performance is critical, as the actual array I'm processing is very large, so iteration using Python for-loops is not sufficient.

are your inverse_slice indices always found in a rectangle shape? — lemon
– lemon, Commented Apr 1, 2022 at 14:11

Salvatore Daniele Bianco · Accepted Answer · 2022-04-01 13:25:30Z

3

Why not this?

arr2 = arr+1000
arr2[1:3, 1:5] = arr[1:3, 1:5]

arr2:

array([[1001, 1001, 1001, 1001, 1001, 1001],
       [1002,    0,    0,    0,    0, 1002],
       [1003,    0,    0,    0,    0, 1003],
       [1004, 1004, 1004, 1004, 1004, 1004]])

answered Apr 1, 2022 at 13:25

Salvatore Daniele Bianco

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2 Comments

Rudy Over a year ago

That seems like a good solution, however there is a bit of memory overhead since you have to copy the entire array, but that is perhaps unavoidable?

Salvatore Daniele Bianco Over a year ago

You right. I'm sure that there is another way. Let me think.

lemon · Accepted Answer · 2022-04-01 17:34:01Z

1

You could do this working by chunks.

Probably not the most elegant way to do it, though you can cycle over the exact cells you need to affect, while keeping a less than O(n^2) complexity depending on the size of the untouched rectangle:

x_min, x_max = (1, 3)
y_min, y_max = (1, 5)
f = <your function>

arr[:x_min, :] = f(arr[:x_min, :])
arr[x_min:x_max, :y_min] = f(arr[x_min:x_max, :y_min])
arr[x_min:x_max, y_max:] = f(arr[x_min:x_max, y_max:])
arr[x_max:, :] = f(arr[x_max:, :])

answered Apr 1, 2022 at 17:34

lemon

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1 Comment

Rudy Over a year ago

This is probably a viable solution, given the low memory complexity. It's not elegant, but if it works it works!

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