In numpy, for an array of arrays, how do I use an array or start indices and an array of end indices to slice out a unique slice of each array [duplicate]

Question

Say that I have an array of arrays

array = np.random.randint(0, 6, (4, 6))
array

array([[3, 5, 2, 5, 1, 3],
       [5, 3, 0, 1, 4, 3],
       [2, 1, 0, 2, 2, 4],
       [2, 1, 0, 4, 2, 2]])

And I also have arrays for desired start and end indices for slicing out this array for each row

starts = np.random.randint(0, 3, (4,))
ends = starts + 3
ends

array([6, 3, 4, 3])

How do I slice out the array of arrays using these indices?

For the example, the desired result will be

array([[5, 1, 3],
       [5, 3, 0],
       [1, 0, 2],
       [2, 1, 0]])

something like array[starts:ends] does not work

Either iterate row by row, or index once with [5,6,7, 3,4,5, 4,5,6, 3,4,5] — hpaulj
– hpaulj, Commented Jul 24, 2022 at 15:51
I think you made a mistake in the example: it is either array([6, 3, 4, 3]) or array([5, 2, 3, 2]), but the former is impossible with np.random.randint(0, 3, (4,)) and the later impossible with ends = starts + 3. So I expect the desired result not to be possible with the provided information: the first line should be 2, 5, 1. — Jérôme Richard
– Jérôme Richard, Commented Jul 24, 2022 at 15:59
@JérômeRichard sorry, I think it was an issue with double clicking a jupyter cell and not tracking the changes, I updated the post to reflect your statement. — SantoshGupta7
– SantoshGupta7, Commented Jul 24, 2022 at 16:06
It is still possible due to the second part of the previous comment. The last list is the one that matters the most ;) . — Jérôme Richard
– Jérôme Richard, Commented Jul 24, 2022 at 16:07

4 revs · Accepted Answer · 2022-07-24 16:47:03Z

0

I am not sure but if this may help you:

a = []
i = 0
for s, e in zip(starts, ends):
    a.append(array[i][s:e])
    i += 1

Or use enumerate()

a = []
for i,(s, e) in enumerate(zip(starts, ends)):
    a.append(array[i][s:e])

then convert the list into an array to get the desired output

print(np.array(a))

It looks like this:

array([[2, 4, 2],
       [4, 3, 0],
       [2, 2, 0],
       [4, 0, 4]])

edited Jul 24, 2022 at 16:47

community wiki

4 revs
Dejene T.

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5 Comments

Nin17 Over a year ago

it's better to use enumerate than to do += on a counting variable in a for loop

Dejene T. Over a year ago

actually they have the same number of rows that's why I didn't use enumerate

Nin17 Over a year ago

what do you mean? I'm saying you should do for i, (s, e) in enumerate(zip(starts, ends)): and then get rid of i=0 and i+=1

Dejene T. Over a year ago

yeah, that's a good option. I was thinking differently

Jérôme Richard Over a year ago

Comprehension lists should be faster: a = [array[i][s:e] for i,(s, e) in enumerate(zip(starts, ends))]. np.fromiter without the list may be also even faster.

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