Python：将列表与字符串进行比较

Question

I want to know how to compare a string to a list. For example I have string ' abcdab ' and a list ['ab','bcd','da'] . Is there any way to compare all possible list combinations to the string, and avoid overlaping elements. so that output will be a list of tuples like [('ab','da'),('bcd'),('bcd','ab'),('ab','ab'),('ab'),('da')].

The output should avoid combinations such as ('bcd', 'da') as the character 'd' is repeated in tuple while it appears only once in the string.

As pointed out in the answer. The characters in string and list elements, must not be rearranged.

One way I tried was to split string elements in to all possible combinations and compare. Which was 2^(n-1) n being number of characters. It was very time consuming.

I am new to python programing. Thanks in advance.

Answer 1

all possible list combinations to string, and avoiding overlaping elements

Is a combination one or more complete items in its exact, current order in the list that match a pattern or subpattern of the string? I believe one of the requirements is to not rearrange the items in the list (ab doesn't get substituted for ba). I believe one of the requirements is to not rearrange the characters in the string. If the subpattern appears twice, then you want the combinations to reflect two individual copies of the subpattern by themselves as well as a list of with both items of the subpattern with other subpatterns that match too. You want multiple permutations of the matches.

Answer 2

This little recursive function should do the job:

def matches(string, words, start=-1):
    result= []
    for word in words: # for each word
        pos= start
        while True:
            pos= string.find(word, pos+1) # find the next occurence of the word
            if pos==-1: # if there are no more occurences, continue with the next word
                break

            if [word] not in result: # add the word to the result
                result.append([word])

            # recursively scan the rest of the string
            for match in matches(string, words, pos+len(word)-1):
                match= [word]+match
                if match not in result:
                    result.append(match)
    return result

output:

>>> print matches('abcdab', ['ab','bcd','da'])
[['ab'], ['ab', 'ab'], ['ab', 'da'], ['bcd'], ['bcd', 'ab'], ['da']]

Answer 3

Oops! I somehow missed Rawing's answer. Oh well. :)

Here's another recursive solution.

#! /usr/bin/env python

def find_matches(template, target, output, matches=None):
    if matches is None:
        matches = []

    for s in template:
        newmatches = matches[:]
        if s in target:
            newmatches.append(s)
            #Replace matched string with a null byte so it can't get re-matched
            # and recurse to find more matches.
            find_matches(template, target.replace(s, '\0', 1), output, newmatches)
        else:
            #No (more) matches found; save current matches
            if newmatches:
                output.append(tuple(newmatches))
            return


def main():
    target = 'abcdab'
    template = ['ab','bcd','da']

    print template
    print target

    output = []
    find_matches(template, target, output)
    print output


if __name__ == '__main__':
    main()

output

['ab', 'bcd', 'da']
abcdab
[('ab', 'ab'), ('ab',), ('bcd', 'ab'), ('bcd',), ('da', 'ab'), ('da',)]