In Python 2.6+ and Python 3:
If you want StopIteration
to be raised if no matching element is found:
next(x for x in the_iterable if x > 3)
If you want default_value
(e.g. None
) to be returned instead:
next((x for x in the_iterable if x > 3), default_value)
Note that you need an extra pair of parentheses around the generator expression in this case ? they are needed whenever the generator expression isn't the only argument.
I see most answers resolutely ignore the next
built-in and so I assume that for some mysterious reason they're 100% focused on versions 2.5 and older -- without mentioning the Python-version issue (but then I don't see that mention in the answers that do mention the next
built-in, which is why I thought it necessary to provide an answer myself -- at least the "correct version" issue gets on record this way;-).
In 2.5, the .next()
method of iterators immediately raises StopIteration
if the iterator immediately finishes -- i.e., for your use case, if no item in the iterable satisfies the condition. If you don't care (i.e., you know there must be at least one satisfactory item) then just use .next()
(best on a genexp, line for the next
built-in in Python 2.6 and better).
If you do care, wrapping things in a function as you had first indicated in your Q seems best, and while the function implementation you proposed is just fine, you could alternatively use itertools
, a for...: break
loop, or a genexp, or a try/except StopIteration
as the function's body, as various answers suggested. There's not much added value in any of these alternatives so I'd go for the starkly-simple version you first proposed.
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