在每个第 n 个元素之后插入 Python 列表中的元素

问题描述

假设我有一个这样的 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 个元素之后插入一个x"，比如说该列表中的三个字符.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.我真正在寻找的是一种 Python 方式，也许是单线?

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

我有两个一体机.

给定:

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

1. 使用enumerate获取索引，每3^rd个字母添加'x'，eg: mod(n, 3) == 2，然后拼接成字符串并list().

1. Use enumerate to get index, add 'x' every 3^rd 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 @sancho.s 指出如果任何元素有多个字母，这将不起作用.

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

使用嵌套推导来展平列表列表^(a)，以 3 个为一组进行切片，如果距离末尾小于 3，则添加 'x'列表.

Use nested comprehensions to flatten a list of lists^(a), 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) [item for subgroup in groups for item in subgroup] 展平一个锯齿状的列表列表.

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

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