Filter a Python list by predicate

Question

I would want to do something like:

>>> lst = [1, 2, 3, 4, 5] >>> lst.find(lambda x: x % 2 == 0) 2 >>> lst.findall(lambda x: x % 2 == 0) [2, 4] 

Is there anything nearing such behavior in Python's standard libraries?

I know it's very easy to roll-your-own here, but I'm looking for a more standard way.

Answer 1

You can use the filter method:

>>> lst = [1, 2, 3, 4, 5] >>> filter(lambda x: x % 2 == 0, lst) [2, 4] 

or a list comprehension:

>>> lst = [1, 2, 3, 4, 5] >>> [x for x in lst if x %2 == 0] [2, 4] 

to find a single element, you could try:

>>> next(x for x in lst if x % 2 == 0) 2 

Though that would throw an exception if nothing matches, so you'd probably want to wrap it in a try/catch. The () brackets make this a generator expression rather than a list comprehension.

Personally though I'd just use the regular filter/comprehension and take the first element (if there is one).

These raise an exception if nothing is found

filter(lambda x: x % 2 == 0, lst)[0] [x for x in lst if x %2 == 0][0] 

These return empty lists

filter(lambda x: x % 2 == 0, lst)[:1] [x for x in lst if x %2 == 0][:1]