creating a new list with subset of list using index in python

Question

A list:

a = [\'a\', \'b\', \'c\', 3, 4, \'d\', 6, 7, 8]

I want a list using a subset of a using a[0:2],a[4], a[6:],

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Answer 1

Try new_list = a[0:2] + [a[4]] + a[6:].

Or more generally, something like this:

from itertools import chain
new_list = list(chain(a[0:2], [a[4]], a[6:]))

This works with other sequences as well, and is likely to be faster.

Or you could do this:

def chain_elements_or_slices(*elements_or_slices):
    new_list = []
    for i in elements_or_slices:
        if isinstance(i, list):
            new_list.extend(i)
        else:
            new_list.append(i)
    return new_list

new_list = chain_elements_or_slices(a[0:2], a[4], a[6:])

But beware, this would lead to problems if some of the elements in your list were themselves lists. To solve this, either use one of the previous solutions, or replace a[4] with a[4:5] (or more generally a[n] with a[n:n+1]).