Cleanly assign list values to multiple variables when there may not be enough values to unpack

Question

For example, if I want to assign a, b, c from l = [1,2,3,4,5], I can do

a, b, c = l[:3]

but what if l is only [1,2] (or [1] or []) ?

Is there a way to automatically set the rest of the variables to None or '' or 0 or some other default value?

I thought about extending the list with default values before assigning to match the number of variables, but just wondering if there's a better way.

@DeveshKumarSingh what if the list only has one value? Or none at all? — Raksha, Jun 20 '19 at 18:55
@DeveshKumarSingh the point is to do this without knowing how long the list is tho. So i might have 10 variables to assign, and the list can have 100 or 0 — Raksha, Jun 20 '19 at 18:58
I don't think the dupe applies, that is for unpacking n variables where m > n. This is for m < n. — cs95, Jun 20 '19 at 18:59
I'm reopening because the dupe doesn't apply, and I couldn't find something better. — cs95, Jun 20 '19 at 19:04

cs95 · Accepted Answer · 2019-06-20T19:24:14.123

5

In general, to unpack N elements into N separate variables from a list of size M where M <= N, then you can pad your list slice upto N and slice again:

l = [1,]
a, b, c =  (l[:3] + [None]*3)[:3]

a, b, c
# 1, None, None

If you fancy a clean generator-based approach, this will also work:

from itertools import islice, cycle, chain

def pad(seq, filler=None):
    yield from chain(seq, cycle([filler]))

a, b, c = islice(pad([1, ]), 3)
a, b, c
# 1, None, None

edited Jun 20 '19 at 19:24

answered Jun 20 '19 at 19:04

cs95

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yeah, that's what i figured ... just thought maybe there was a prettier way – Raksha Jun 20 '19 at 19:05

Cleanly assign list values to multiple variables when there may not be enough values to unpack

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