How to achieve python's any() with a custom predicate?
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)TrueThe 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
Use a generator expression inside of any():
pred = lambda x: x > 10if any(pred(i) for i in l): print "foo"else: print "bar"This assumes you already have some predicate function you want to use, of course if it is something simple like this you can just use the Boolean expression directly: any(i > 10 for i in l).