Python 3:从元组列表中删除空元组

Question

我有一个元组列表,如下所示:

>>>myList
[(), (), ('',), ('c', 'e'), ('ca', 'ea'), ('d',), ('do',), ('dog', 'ear', 'eat', 'cat', 'car'), ('dogs', 'cars', 'done', 'eats', 'cats', 'ears'), ('don',)]

我想这样读:

>>>myList
[('',), ('c', 'e'), ('ca', 'ea'), ('d',), ('do',), ('dog', 'ear', 'eat', 'cat', 'car'), ('dogs', 'cars', 'done', 'eats', 'cats', 'ears'), ('don',)]

即我想()从列表中删除空元组.在这样做时我想保留元组('',).我似乎无法找到从列表中删除这些空元组的方法.

我已经尝试myList.remove(())并使用for循环来执行此操作,但要么不起作用,要么我的语法错误.任何帮助,将不胜感激.

Answer 1

您可以过滤"空"值:

filter(None, myList)

或者你可以使用列表理解.在Python 3上,filter()返回一个生成器; list comprehension在Python 2或3上返回一个列表:

[t for t in myList if t]

如果您的列表包含的不仅仅是元组,您可以显式测试空元组:

[t for t in myList if t != ()]

Python 2演示:

>>> myList = [(), (), ('',), ('c', 'e'), ('ca', 'ea'), ('d',), ('do',), ('dog', 'ear', 'eat', 'cat', 'car'), ('dogs', 'cars', 'done', 'eats', 'cats', 'ears'), ('don',)]
>>> filter(None, myList)
[('',), ('c', 'e'), ('ca', 'ea'), ('d',), ('do',), ('dog', 'ear', 'eat', 'cat', 'car'), ('dogs', 'cars', 'done', 'eats', 'cats', 'ears'), ('don',)]
>>> [t for t in myList if t]
[('',), ('c', 'e'), ('ca', 'ea'), ('d',), ('do',), ('dog', 'ear', 'eat', 'cat', 'car'), ('dogs', 'cars', 'done', 'eats', 'cats', 'ears'), ('don',)]
>>> [t for t in myList if t != ()]
[('',), ('c', 'e'), ('ca', 'ea'), ('d',), ('do',), ('dog', 'ear', 'eat', 'cat', 'car'), ('dogs', 'cars', 'done', 'eats', 'cats', 'ears'), ('don',)]

在这些选项中,filter()功能最快:

>>> timeit.timeit('filter(None, myList)', 'from __main__ import myList')
0.637274980545044
>>> timeit.timeit('[t for t in myList if t]', 'from __main__ import myList')
1.243359088897705
>>> timeit.timeit('[t for t in myList if t != ()]', 'from __main__ import myList')
1.4746298789978027

在Python 3上,坚持使用列表理解:

>>> timeit.timeit('list(filter(None, myList))', 'from __main__ import myList')
1.5365421772003174
>>> timeit.timeit('[t for t in myList if t]', 'from __main__ import myList')
1.29734206199646

Answer 2

myList = [x for x in myList if x != ()]