使用python中的索引创建一个包含列表子集的新列表
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creating a new list with subset of list using index in python
提问by user2783615
A list:
一个列表：
a = ['a', 'b', 'c', 3, 4, 'd', 6, 7, 8]
I want a list using a subset of a using a[0:2],a[4], a[6:],
我想要一个使用 using 的子集的列表a[0:2],a[4], a[6:]，
that is I want a list ['a', 'b', 4, 6, 7, 8]
那是我想要一个清单 ['a', 'b', 4, 6, 7, 8]
采纳答案by rlms
Try new_list = a[0:2] + [a[4]] + a[6:].
试试new_list = a[0:2] + [a[4]] + a[6:]。
Or more generally, something like this:
或者更一般地说，是这样的：
from itertools import chain
new_list = list(chain(a[0:2], [a[4]], a[6:]))
This works with other sequences as well, and is likely to be faster.
这也适用于其他序列，并且可能会更快。
Or you could do this:
或者你可以这样做：
def chain_elements_or_slices(*elements_or_slices):
new_list = []
for i in elements_or_slices:
if isinstance(i, list):
new_list.extend(i)
else:
new_list.append(i)
return new_list
new_list = chain_elements_or_slices(a[0:2], a[4], a[6:])
But beware, this would lead to problems if some of the elements in your list were themselves lists.
To solve this, either use one of the previous solutions, or replace a[4]with a[4:5](or more generally a[n]with a[n:n+1]).
但请注意，如果列表中的某些元素本身就是列表，则会导致问题。要解决此问题，请使用以前的解决方案之一，或者替换a[4]为a[4:5]（或更一般地替换a[n]为a[n:n+1]）。
回答by Aurélien Ooms
The following definition might be more efficient than the first solution proposed
以下定义可能比提出的第一个解决方案更有效
def new_list_from_intervals(original_list, *intervals):
n = sum(j - i for i, j in intervals)
new_list = [None] * n
index = 0
for i, j in intervals :
for k in range(i, j) :
new_list[index] = original_list[k]
index += 1
return new_list
then you can use it like below
然后你可以像下面一样使用它
new_list = new_list_from_intervals(original_list, (0,2), (4,5), (6, len(original_list)))
回答by G. Cohen
Suppose
认为
a = ['a', 'b', 'c', 3, 4, 'd', 6, 7, 8]
and the list of indexes is stored in
索引列表存储在
b= [0, 1, 2, 4, 6, 7, 8]
then a simple one-line solution will be
那么一个简单的单行解决方案将是
c = [a[i] for i in b]
