[英]Selecting List Elements Periodically in Python 3
Say I have a list 说我有一个清单
Q = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18]
I believe I can extract the first and every ninth value thereafter using the extended slice notation: 我相信我可以使用扩展切片表示法提取第一个和第九个值:
Q[::9]
Which should give: 哪个应该给:
[0,9,18]
But how can I similarly select all the elements apart from those? 但是,我如何才能类似地选择除这些元素之外的所有元素?
You mean this? 你是这个意思?
>>> lis =[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18]
>>> lis[1::9]
[1, 10]
Extended slice notations: 扩展切片符号:
lis[start : stop : step] #default values : start = 0, stop = len(lis), step = 1
You can pass your own value for start
(by default 0 be used) 您可以传递自己的
start
值(默认情况下,使用0)
Update: 更新:
>>> lis = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18]
>>> se = set(range(0, len(lis),9)) #use a list if the lis is not huge.
>>> [x for i,x in enumerate(lis) if i not in se]
[1, 2, 3, 4, 5, 6, 7, 8, 10, 11, 12, 13, 14, 15, 16, 17]
#for your example even this will work:
>>> [x for i,x in enumerate(lis) if i%9 != 0]
[1, 2, 3, 4, 5, 6, 7, 8, 10, 11, 12, 13, 14, 15, 16, 17]
In case you have not repeated numbers, this is a general solution for any collection of numbers (not necessarily consecutive): 如果您没有重复的数字,这是任何数字集合(不一定是连续的)的通用解决方案:
>>> Q = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18]
>>> list(set(Q).difference(Q[::9]))
[1, 2, 3, 4, 5, 6, 7, 8, 10, 11, 12, 13, 14, 15, 16, 17]
>>>
It uses set.difference
s to get the set
that is the difference between the original list and the sublist to be removed. 它使用
set.difference
来获取set
,该set
是原始列表和要删除的子列表之间的差异。
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