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

前端 未结 2 1021
长发绾君心
长发绾君心 2021-02-07 01:12
>>> 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:                    


        
相关标签:
2条回答
  • 2021-02-07 02:04

    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
    
    0 讨论(0)
  • 2021-02-07 02:04

    Use a generator expression inside of any():

    pred = lambda x: x > 10
    if 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).

    0 讨论(0)
提交回复
热议问题