在混乱的字符串中查找子字符串 [英] Finding a substring in a jumbled string

问题描述

我正在编写一个脚本-includes(word1，word2)-以两个字符串作为参数，并查找word2中是否包含word1.Word2是一个字母混乱.它应该返回布尔值.同样允许重复字母，我只检查两个单词中是否包含字母的顺序相同.

I am writing a script - includes(word1, word2) - that takes two strings as arguments, and finds if word1 is included in word2. Word2 is a letter jumble. It should return Boolean. Also repetition of letters are allowed, I am only checking if the letters are included in the both words in the same order.

>>>includes('queen', 'qwertyuytresdftyuiokn')
True

'queen'，' Q werty U ytr E sdftyuiok N "

'queen', 'QwertyUytrEsdftyuiokN'

我尝试将每个单词转换为列表，以便更轻松地使用每个元素.我的代码是这样的:

I tried turning each word into lists so that it is easier to work with each element. My code is this:

def includes(w1, w2):
    w1 = list(w1)
    w2 = list(w2)
    result = False
    for i in w1:
        if i in w2:
            result = True
        else:
            result = False
    return result

但是问题是我还需要检查word1的字母在word2中的顺序是否相同，而我的代码对此没有控制.我找不到用list来实现的方法.就像我不能对字符串做太多事情一样，所以我认为我需要使用其他数据结构(例如字典)，但我对它们了解不多.

But the problem is that I need to also check if the letters of word1 comes in the same order in word2, and my code doesn't controls that. I couldn't find a way to implement that with list. Just like I couldn't do this much with strings, so I think I need to use another data structure like dictionary but I don't know much about them.

推荐答案

我希望我了解您的目标是什么.
Python不是我的事，但我认为我是pythonic的:

I hope I understood what is your goal.
Python is not my thing, but I think I made it pythonic:

def is_subsequence(pattern, items_to_use):
    items_to_use = (x for x in items_to_use)
    return all(any(x == y for y in items_to_use) for x, _ in itertools.groupby(pattern))

https://ideone.com/Saz984

说明:

• itertools.groupby 传输 pattern ，以使本构重复项被丢弃
分组为 pattern 的表单中的
• 所有项必须满足条件
• any 使用生成器 items_to_use ，只要它与当前项目不匹配即可.请注意，必须在最终表达式之外定义 items_to_use ，因此每次验证 pattern 中的下一个项目时，都会保持进度.

• itertools.groupby transfers pattern in such way that constitutive duplicates are discarded
• all items form form grouped pattern must fulfill conditions
• any uses generator items_to_use as long as it doesn't matches current item. Note that items_to_use mus be defined outside of final expression so progress on it is kept every time next item from pattern is verified.

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