What would be the most efficient\elegant way in Python to find the index of the first non-empty item in a list?
For example, with
list_ = [None,[],None,[1,2],'StackOverflow',[]]
the correct non-empty index should be:
3
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6 Answers
>>> lst = [None,[],None,[1,2],'StackOverflow',[]]
>>> next(i for i, j in enumerate(lst) if j)
3
if you don't want to raise a StopIteration error, just provide default value to the next function:
>>> next((i for i, j in enumerate(lst) if j == 2), 42)
42
P.S. don't use list as a variable name, it shadows built-in.
Comments
One relatively elegant way of doing it is:
map(bool, a).index(True)
(where "a" is your list... I'm avoiding the variable name "list" to avoid overriding the native "list" function)
2 Comments
SilentGhost
that only works in python-2.x and anyway it builds the whole list of boolean values.
Joe Kington
@SilentGhost - True. I didn't realize that map had changed in python3.
try:
i = next(i for i,v in enumerate(list_) if v)
except StopIteration:
# Handle...
Comments
next(i for (i, x) in enumerate(L) if x)
answered Jul 12, 2010 at 15:17
Ignacio Vazquez-Abrams
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Comments
for python 3.x @Joe Kington solution is
list(map(bool, list_)).index(True)
Also consider working with values instead of indexes
for value in filter(bool, list_):
...
1 Comment
BrokenBenchmark
This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review
next(i for i, v in enumerate(list) if v)
