Python:通过删除第n个元素从现有列表构建新列表 [英] Python: building new list from existing by dropping every n-th element

问题描述

我想建立一个新列表，其中初始列表的第n个元素都被忽略了，例如:

I want to build up a new list in which every n-th element of an initial list is left out, e.g.:

之所以使用['second', 'third', 'fifth', 'sixth'是因为n = 3

该怎么做? 首先，通过建立新列表而不是尝试删除列表来完成此操作是否正确?对于后者，我尝试使用deque和rotate，但最终陷入混乱.
为了建立一个新列表，我尝试使用range(1,len(list),n)进行操作，但这是要删除的元素位置，而不是要为新列表保留的元素位置.

How to do that? Is it - first of all - correct to accomplish this by building up a new list, instead of trying to delete? For the latter I tried with deque and rotate but that ended up in confusion.
To build up a new list I was trying something with range(1,len(list),n) but that are the element positions to be deleted and not the ones which are to be kept for the new list.

如何获取所需列表?

推荐答案

>>> [s for (i,s) in enumerate(['first', 'second', 'third', 'fourth', 'fifth', 'sixth', 'seventh']) if i%3]
['second', 'third', 'fifth', 'sixth']

答案分几个步骤:

enumerate函数提供一个元组列表，其索引后面是项目:

The enumerate function gives a list of tuples with the index followed by the item:

>>> list(enumerate(['first', 'second', 'third', 'fourth', 'fifth', 'sixth', 'seventh']))
[(0, 'first'), (1, 'second'), (2, 'third'), (3, 'fourth'), (4, 'fifth'), (5, 'sixth'), (6, 'seventh')]

然后检查索引是否不被三除，如果是，则包括该项.

then you check if the index does not divide by three, and if so you include the item.

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