Python - Finding index of first non-empty item in a list

Question

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

Answer 1

>>> 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.

Answer 2

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)

Answer 3

try:
    i = next(i for i,v in enumerate(list_) if v)
except StopIteration:
    # Handle...