list(filter(None,[1,0,False,True]))
返回[1, True]
为什么不是0，False？？不应该是这两个符合条件的吗？

这种事情就找python官方文档，在3.9.6文档的内置函数中有如下描述

从 filter()底层代码来看

class filter(object):
    """
    filter(function or None, iterable) --> filter object
    
    Return an iterator yielding those items of iterable for which function(item)
    is true. If function is None, return the items that are true.
    """
    def __getattribute__(self, *args, **kwargs): # real signature unknown
        """ Return getattr(self, name). """
        pass

    def __init__(self, function_or_None, iterable): # real signature unknown; restored from __doc__
        pass

    def __iter__(self, *args, **kwargs): # real signature unknown
        """ Implement iter(self). """
        pass

    @staticmethod # known case of __new__
    def __new__(*args, **kwargs): # real signature unknown
        """ Create and return a new object.  See help(type) for accurate signature. """
        pass

    def __next__(self, *args, **kwargs): # real signature unknown
        """ Implement next(self). """
        pass

    def __reduce__(self, *args, **kwargs): # real signature unknown
        """ Return state information for pickling. """
        pass

注释译文:

filter(function or None, iterable) 会返回一个迭代器，该迭代器产生iterable函数中的(item)为真的那些项。
如果function为None，则返回为true的项,也就是为真的项。
相当于-->示例中的结果：

def func(item):
    if 1 == item:
        return item


a = list(filter(func, [1, 0, False, True]))
print(a)

#结果为：
[1, True]

也就是当你传入None 时代表的就是返回真， 且也只返回真

这是过滤掉类型为false的，可以记得采纳哦