在每个第n个元素之后在Python列表中插入元素

问题描述

说我有这样的Python列表：

Say I have a Python list like this:

letters = ['a','b','c','d','e','f','g','h','i','j']

我想在每个第n个元素后插入一个'x'，假设该列表中有三个字符。结果应该是：

I want to insert an 'x' after every nth element, let's say three characters in that list. The result should be:

letters = ['a','b','c','x','d','e','f','x','g','h','i','x','j']

我知道我可以通过循环和插入来做到这一点。我真正想要的是一种Python方式，也许是一个单行？

I understand that I can do that with looping and inserting. What I'm actually looking for is a Pythonish-way, a one-liner maybe?

推荐答案

我有两个一个预算：

鉴于：

>>> letters = ['a','b','c','d','e','f','g','h','i','j']

1. 使用枚举获取索引，添加'x'每3个字母，例如： mod（n，3 ）== 2 ，然后连接成字符串和 list（）它。

1. Use enumerate to get index, add 'x' every 3 letter, eg: mod(n, 3) == 2, then concatenate into string and list() it.

>>> list(''.join(l + 'x' * (n % 3 == 2) for n, l in enumerate(letters)))

['a', 'b', 'c', 'x', 'd', 'e', 'f', 'x', 'g', 'h', 'i', 'x', 'j']

但是作为 如果任何元素包含多个字母，则此方法无效。

But as @sancho.s points out this doesn't work if any of the elements have more than one letter.

使用嵌套式解析来展平列表列表，以3个为一组进行切片，其中'x'在列表末尾少于3时添加。

Use nested comprehensions to flatten a list of lists, sliced in groups of 3 with 'x' added if less than 3 from end of list.

>>> [x for y in (letters[i:i+3] + ['x'] * (i < len(letters) - 2) for
     i in xrange(0, len(letters), 3)) for x in y]

['a', 'b', 'c', 'x', 'd', 'e', 'f', 'x', 'g', 'h', 'i', 'x', 'j']

（a） [子组中项目的子组项目] 展平锯齿状的列表列表。

(a) [item for subgroup in groups for item in subgroup] flattens a jagged list of lists.

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