Python - 从列表列表中删除列表（类似于 .pop() 的功能）
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Python - Remove list(s) from list of lists (Similar functionality to .pop() )
提问by Phoenix
a=[[1,2,3],[4,5,6],[7,8,9]]
.pop() has the capacity to not only remove an element of a list but also return that element.
.pop() 不仅可以删除列表中的元素，还可以返回该元素。
I am looking for a similar function that can remove and return a whole list that could exist in the middle of another list.
我正在寻找一个类似的函数，它可以删除并返回可能存在于另一个列表中间的整个列表。
E.g is there a function that will remove [4,5,6]from the above list a, and return it.
例如，是否有一个函数可以[4,5,6]从上面的列表中删除a并返回它。
The reason for the question is that I'm sorting a list through itemgetterand there's a collision between the headings row (string) and the rest of the data (datetime). As such, I'm looking to effectively pop the list which represents the headings, do a sort, then insert it back in.
问题的原因是我正在对列表进行排序，itemgetter并且标题行（字符串）和其余数据（datetime）之间存在冲突。因此，我希望有效地弹出代表标题的列表，进行排序，然后将其重新插入。
采纳答案by Martijn Pieters
The nested lists are just values in the outer list. Just use .pop()on that outer list:
嵌套列表只是外部列表中的值。只需.pop()在该外部列表上使用：
inner_list = a.pop(1)
Demo:
演示：
>>> a = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
>>> a.pop(1)
[4, 5, 6]
>>> a
[[1, 2, 3], [7, 8, 9]]
You could just use a slice to remove the first row from consideration if a header row is in the way:
如果标题行挡住了，您可以使用切片从考虑中删除第一行：
result = rows[:1] + sorted(rows[1:], key=itemgetter(1))
