如何加入字符串列表并删除重复的字母(将它们链接在一起) [英] How can I join a list of strings and remove duplicated letters (keep them chained)

问题描述

l = ["volcano", "noway", "lease", "sequence", "erupt"]

所需的输出:

'volcanowayleasequencerupt'

我尝试过:

使用itertools.groupby，但当行中有2个重复的字母(即leasesequence-> sese停留)时，它似乎无法正常工作:

I have tried:

using itertools.groupby but it seems like it doesn't work well when there is 2 repeated letters in row (i.e. leasesequence -> sese stays):

>>> from itertools import groupby
>>> "".join([i[0] for i in groupby("".join(l))])
'volcanonowayleasesequencerupt'

如您所见，它仅在最后一个'e'处被删除，因此这不是理想的，因为如果一个字母包含双字符，它们将缩小为1，即'suddenly'变为'sudenly'.

As you can see it got rid only for the last 'e', and this is not ideal because if a letter has double characters they will be shrunk to 1. i.e 'suddenly' becomes 'sudenly'.

我正在寻找最适合Python的方法.

I'm looking for the most Pythonic approach for this.

谢谢.

编辑

我的列表中没有重复的项目.

My list does not have any duplicated items in it.

推荐答案

我认为是更具可读性的版本:

A more readable version, in my opinion:

from functools import reduce


def max_overlap(s1, s2):

    return next(
        i
        for i in reversed(range(len(s2) + 1))
        if s1.endswith(s2[:i])
    )


def overlap(strs):

    return reduce(
        lambda s1, s2:
            s1 + s2[max_overlap(s1, s2):],
        strs, '',
    )


overlap(l)
#> 'volcanowayleasequencerupt'

但是，它还会考虑之前重叠的单词中的累积"字符:

However, it also considers "accumulated" characters from previous words that overlapped:

overlap(['split', 'it', 'lit'])
#> 'split'

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