String search by coincidence?

Question

I just wanted to know if there's a simple way to search a string by coincidence with another one in Python. Or if anyone knows how it could be done.

To make myself clear I'll do an example.

text_sample = "baguette is a french word"
words_to_match = ("baguete","wrd")

letters_to_match = ('b','a','g','u','t','e','w','r','d')   #   With just one 'e'
coincidences = sum(text_sample.count(x) for x in letters_to_match)

#    coincidences = 14 Current output
#    coincidences = 10 Expected output

My current method breaks the words_to_match into single characters as in letters_to_match but then it is matched as follows: "baguette is a french word" (coincidences = 14).

But I want to obtain (coincidences = 10) where "baguette is a french word" were counted as coincidences. By checking the similarity between words_to_match and the words in text_sample.

How do I get my expected output?

Answer 1

It looks like you need the length of the longest common subsequence (LCS). See the algorithm in the Wikipedia article for computing it. You may also be able to find a C extension which computes it quickly. For example, this search has many results, including pylcs. After installation (pip install pylcs):

import pylcs
text_sample = "baguette is a french word"
words_to_match = ("baguete","wrd")
print(pylcs.lcs2(text_sample, ' '.join(words_to_match.join)))  #: 14

Answer 2

first, split words_to_match with

    words = ''
    for item in words_to_match:
        words += item
    letters = [] # create a list
    for letter in words:
        letters.append(letter)
    letters = tuple(letters)

then, see if its in it

    x = 0
    for i in sample_text:
        if letters[x] == i:
            x += 1
            coincidence += 1

also if it's not in sequence just do:

    for i in sample_text:
        if i in letters: coincidence += 1

(note that some versions of python you'l need a newline)