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Home/ Questions/Q 668001
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Editorial Team
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Editorial Team
Asked: May 13, 20262026-05-13T23:59:30+00:00 2026-05-13T23:59:30+00:00

I got a function like def f(): … … return [list1, list2] this returns

  • 0

I got a function like

def f():
    ...
    ...
    return [list1, list2]

this returns a list of lists

[[list1.item1,list1.item2,...],[list2.item1,list2.item2,...]]

now when I do the following:

for i in range(0,2):print f()[i][0:10]

it works and print the lists sliced

but if i do

print f()[0:2][0:10]

then it prints the lists ignoring the [0:10] slicing.

Is there any way to make the second form work or do I have to loop every time to get the desired result?

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  1. Editorial Team
    Editorial Team
    2026-05-13T23:59:30+00:00Added an answer on May 13, 2026 at 11:59 pm

    The reason why these two behave differently is because f()[0:2][0:10] works like this:

    1. f() gives you a list of lists.
    2. [0:2] gives you a list containing the first two elements in the list of lists. Since the elements in the list of lists are lists, this is also a list of lists.
    3. [0:10] gives you a list containing the first ten elements in the list of lists that was produced in step 2.

    In other words, f()[0:2][0:10] starts with a list of lists, then takes a sublist of that list of lists (which is also a list of lists), and then takes a sublist of the second list of lists (which is also a list of lists).

    In contrast, f()[i] actually extracts the i-th element out of your list of lists, which is just a simple list (not a list of lists). Then, when you apply [0:10], you are applying it to the simple list that you got from f()[i] and not to a list of lists.

    The bottom line is that any solution that gives the desired behavior will have to access a single array element like [i] at some point, rather than working only with slices like [i:j].

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