slice an array without knowing its dimensions

Question

I want to slice an array without knowing its dimensions. Indexes(start and end are given in list format. How to do this? Thank you.

Slice 1d array

import numpy as np
a = np.array([1, 2, 3, 4, 5, 6])
# idx: [2, 5)
print(a[2:5])
# [3, 4, 5]

Slice 2d array

import numpy as np
a = np.array([
[1, 2, 3],
[2, 3, 2],
[4, 5, 2]
])

start = [0, 1]
end = [1, 2]

print(a[start[0]:end[0], start[1]:end[1]])

Slice N-d array?

a = np.array([
    [[1, 0, 2],
    [2, 1, 0],
    [5, 6, 3]],
    
    [[2, 1, 3],
    [3, 2, 1],
    [1, 4, 6]]
])

start = [0, 0, 1]
end = [1, 1, 2]

result = a[start[0]:end[0], start[1]:end[1], start[2]:end[2]] # key point: how to adapt this?
print(result)

You need to get len(a.shape) to find out the number of dimensions and then you can loop through all of the dimensions to find out the number of values in the dimension. — NotAName
– NotAName, Commented Dec 17, 2020 at 5:20

Kapocsi · Accepted Answer · 2020-12-17 06:12:59Z

1

If you have a list of start and end indexes, you can construct a tuple of slice objects and do the following to slice an array without knowing its dimensions:

a[tuple(slice(*indexes) for indexes in zip(start, end))]

Or:

a[tuple(slice(st, en) for st, en in zip(start, end))]

Or:

from itertools import starmap

a[tuple(starmap(slice, zip(start, end)))]

In action:

a = np.array([
    [[1, 0, 2],
    [2, 1, 0],
    [5, 6, 3]],
    
    [[2, 1, 3],
    [3, 2, 1],
    [1, 4, 6]]
])

start = [0, 0, 1]
end = [1, 1, 2]

result1 = a[start[0]:end[0], start[1]:end[1], start[2]:end[2]] # key point: how to adapt this?

result2 = a[tuple(slice(*indexes) for indexes in zip(start, end))] 

assert np.all(result2 == result2)

edited Dec 17, 2020 at 6:12

answered Dec 17, 2020 at 6:04

Kapocsi

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