3 回答
TA贡献1831条经验 获得超10个赞
filter
:
str_list = filter(None, str_list) # fastest
str_list = filter(bool, str_list) # fastest
str_list = filter(len, str_list) # a bit slower
str_list = filter(lambda item: item, str_list) # slower than list comprehension
Python 3返回一个迭代器。filter,所以应该在调用list()
str_list = list(filter(None, str_list)) # fastest
(等。)
测试:
>>> timeit('filter(None, str_list)', 'str_list=["a"]*1000', number=100000)
2.4797441959381104
>>> timeit('filter(bool, str_list)', 'str_list=["a"]*1000', number=100000)
2.4788150787353516
>>> timeit('filter(len, str_list)', 'str_list=["a"]*1000', number=100000)
5.2126238346099854
>>> timeit('[x for x in str_list if x]', 'str_list=["a"]*1000', number=100000)
13.354584932327271
>>> timeit('filter(lambda item: item, str_list)', 'str_list=["a"]*1000', number=100000)
17.427681922912598
添加回答
举报