在 Python 中使用一个回调函数 any()

Python 标准库定义了一个 any()函数

如果迭代的任何元素为 True，则返回 True。如果迭代的元素为空，则返回 False。

它只检查元素的值是否为 True。我希望它能够指定一个回调函数来判断一个元素是否符合条件，比如:

any([1, 2, 'joe'], lambda e: isinstance(e, int) and e > 0)
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最佳答案

How about:

>>> any(isinstance(e, int) and e > 0 for e in [1,2,'joe'])
True

It also works with all() of course:

>>> all(isinstance(e, int) and e > 0 for e in [1,2,'joe'])
False
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You should use a "generator expression" - that is, a language construct that can consume iterators and apply filter and expressions on then on a single line:

For example (i ** 2 for i in xrange(10)) is a generator for the square of the first 10 natural numbers (0 to 9)

They also allow an "if" clause to filter the itens on the "for" clause, so for your example you can use:

any (e for e in [1, 2, 'joe'] if isinstance(e, int) and e > 0)
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any function returns True when any condition is True.

>>> any(isinstance(e, int) and e > 0 for e in [0 ,0, 1])
True # Returns True because 1 is greater than 0.




>>> any(isinstance(e, int) and e > 0 for e in [0 ,0, 0])
False # Returns False because not a single condition is True.

Actually,the concept of any function is brought from Lisp or you can say from the function programming approach. There is another function which is just opposite to it is all

>>> all(isinstance(e, int) and e > 0 for e in [1, 33, 22])
True # Returns True when all the condition satisfies.


>>> all(isinstance(e, int) and e > 0 for e in [1, 0, 1])
False # Returns False when a single condition fails.

These two functions are really cool when used properly.

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filter can work, plus it returns you the matching elements

>>> filter(lambda e: isinstance(e, int) and e > 0, [1,2,'joe'])
[1, 2]
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While the others gave good Pythonic answers (I'd just use the accepted answer in most cases), I just wanted to point out how easy it is to make your own utility function to do this yourself if you really prefer it:

def any_lambda(iterable, function):
return any(function(i) for i in iterable)


In [1]: any_lambda([1, 2, 'joe'], lambda e: isinstance(e, int) and e > 0
Out[1]: True
In [2]: any_lambda([-1, '2', 'joe'], lambda e: isinstance(e, int) and e > 0)
Out[2]: False

I think I'd at least define it with the function parameter first though, since that'd more closely match existing built-in functions like map() and filter():

def any_lambda(function, iterable):
return any(function(i) for i in iterable)
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Slight improvement to Antoine P's answer

>>> any(type(e) is int for e in [1,2,'joe'])
True

For all()

>>> all(type(e) is int for e in [1,2,'joe'])
False
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You can use a combination of any and map if you really want to keep your lambda notation like so :

any(map(lambda e: isinstance(e, int) and e > 0, [1, 2, 'joe']))

But it is better to use a generator expression because it will not build the whole list twice.

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If you really want to inline a lambda in any() you can do this: 

>>> any((lambda: isinstance(e, int))() for e in [1,2,'joe'])
True
>>> any((lambda: isinstance(e, int))() for e in ['joe'])
False

You just have to wrap up the unnamed lambda and ensure it is invoked on each pass by appending the ()

The advantage here is that you still get to take advantage of short circuiting the evaluation of any when you hit the first int


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