假设列表如下

iterator = [None, 1, 2, 0, '', None, False, {}, (), []]

这将只返回bool(item)为True的项目

print filter(lambda item: item, iterator)
# [1, 2]

这相当于

print [item for item in iterator if item]

只过滤None:

print filter(lambda item: item is not None, iterator)
# [1, 2, 0, '', False, {}, (), []]

等价于:

print [item for item in iterator if item is not None]

获取所有值为False的项

print filter(lambda item: not item, iterator)
# Will print [None, '', 0, None, False, {}, (), []]

@jamylak的回答非常好，但是如果你不想导入几个模块来完成这个简单的任务，就在原地写你自己的lambda:

>>> L = [0, 23, 234, 89, None, 0, 35, 9]
>>> filter(lambda v: v is not None, L)
[0, 23, 234, 89, 0, 35, 9]

假设列表如下

iterator = [None, 1, 2, 0, '', None, False, {}, (), []]

这将只返回bool(item)为True的项目

print filter(lambda item: item, iterator)
# [1, 2]

这相当于

print [item for item in iterator if item]

只过滤None:

print filter(lambda item: item is not None, iterator)
# [1, 2, 0, '', False, {}, (), []]

等价于:

print [item for item in iterator if item is not None]

获取所有值为False的项

print filter(lambda item: not item, iterator)
# Will print [None, '', 0, None, False, {}, (), []]

如果列表中有NoneType和pandas._lib .missing. list。NAType对象比使用:

[i for i in lst if pd.notnull(i)]

L = [0, 23, 234, 89, None, 0, 35, 9] 
result = list(filter(lambda x: x is not None, L))

from operator import is_not
from functools import partial   

filter_null = partial(filter, partial(is_not, None))

# A test case
L = [1, None, 2, None, 3]
L = list(filter_null(L))