>>> from operator import itemgetter
>>> a = range(100)
>>> itemgetter(5,13,25)(a)
(5, 13, 25)

All you need is list slicing starting from index 1 and then you need to jump by 3 indexes:

>>> my_list = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]
>>> my_list[1::3]
[2, 5, 8, 11]

General syntax of list slicing is your_list[start:end:jump] where:

• start: is the starting index for slicing the list. Empty value means start of the list i.e. index 0

• end: is the index till which you want to slice the list. Empty value means end of the list

• jump: is used to jump the elements from start such that you'll get values in the order start, start+jump, start+2*jump, so on till your list reaches end. Empty value means 1

As mentioned by Mr. cᴏʟᴅsᴘᴇᴇᴅ in comment, for large lists one should prefer itertools.islice which returns iterator object and could be used in this case as:

from itertools import islice  # import islice

islice(my_list, 1, None, 3)

Very rarely using slice objects is easier to read than employing a list comprehension, and this is not one of those cases.

>>> A = [[1,2,3,4,5],[1,2,3,4,5],[1,2,3,4,5]]
>>> [sublist[:3] for sublist in A]
[[1, 2, 3], [1, 2, 3], [1, 2, 3]]

This is very clear. For every sublist in A, give me the list of the first three elements.

that's what slice() is for:

a = range(10)
s = slice(2,4)
print a[s]

That's the same as using a[2:4].