How can I get a Python generator to return None rather than StopIteration?
I am using generators to perform searches in lists like this simple example:
>>> a = [1,2,3,4]
>>> (i for i, v in enumerate(a) if v == 4).next()
3
(Just to frame the example a bit, I am using very much longer lists compared to the one above, and the entries are a little bit more complicated than int
. I do it this way so the entire lists won't be traversed each time I search them)
Now if I would instead change that to i == 666
, it would return a StopIteration
because it can't find any 666
entry in a
.
How can I make it return None
instead? I could of course wrap it in a try ... except
clause, but is there a more pythonic way to do it?
Solution 1:
If you are using Python 2.6+ you should use the next
built-in function, not the next
method (which was replaced with __next__
in 3.x). The next
built-in takes an optional default argument to return if the iterator is exhausted, instead of raising StopIteration
:
next((i for i, v in enumerate(a) if i == 666), None)
Solution 2:
You can chain the generator with (None,):
from itertools import chain
a = [1,2,3,4]
print chain((i for i, v in enumerate(a) if v == 6), (None,)).next()
but I think a.index(2) will not traverse the full list, when 2 is found, the search is finished. you can test this:
>>> timeit.timeit("a.index(0)", "a=range(10)")
0.19335955439601094
>>> timeit.timeit("a.index(99)", "a=range(100)")
2.1938486138533335