How to return nothing for slicing a Python list

Question

I want to have variables to control the number of items at the beginning and end of the list to skip for example.

skip_begin = 1
skip_end = 2
my_list = [9,8,7,6,5]
print(my_list[skip_begin:-skip_end])

returns [8, 7].

How Why does slice [: -0] return an empty list in Python shows that skip_end = 0 gives an empty list.

In this case, I really want to just my_list [skip_begin:].

my_list[skip_begin:-skip_end if skip_end != 0 else '']

does not work. How can I return empty value in ternary operation?

I would rather use the ternary operation inside "[]" rather than

my_list[skip_begin:-skip_end] if skip_end != 0 else my_list[skip_begin:]

or an if-else block.

+3

Gnubie June 30. 17 at 14:33

3 answers

my_list[skip_begin:-skip_end if skip_end else None]

Your line:

my_list[skip_begin:-skip_end if skip_end != 0 else '']

could get you on the right track because it gives:

TypeError: slice indexes must be integer or None or have an __index__ Method

I am inspired by the previous answer, but the if skip_end != 0

same asskip_end

+6

Owl Max June 30. 17 at 14:46

 my_list[skip_begin:-skip_end if skip_end != 0 else skip_begin:]

-1

KR29 June 30. 17 at 14:48

yinnonsanders · Accepted Answer · 2017-06-30T14:41:36+0000

my_list[skip_begin:-skip_end if skip_end != 0 else None]

The problem with your first solution is that the index slicing expects a number (or null), not a string.

As an alternative:

my_list[skip_begin:-skip_end or None]