How to achieve python's any() with a custom predicate?

Question

>>> l = list(range(10))
>>> l
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
>>> if filter(lambda x: x > 10, l):
...     print \"foo\"
... else:                    


        

Answer 1

Use a generator expression as that one argument:

any(x > 10 for x in l)

Here the predicate is in the expression side of the generator expression, but you can use any expression there, including using functions.

Demo:

>>> l = range(10)
>>> any(x > 10 for x in l)
False
>>> l = range(20)
>>> any(x > 10 for x in l)
True

The generator expression will be iterated over until any() finds a True result, and no further:

>>> from itertools import count
>>> endless_counter = count()
>>> any(x > 10 for x in endless_counter)
True
>>> # endless_counter last yielded 11, the first value over 10:
...
>>> next(endless_counter)
12