Longest Common Substring without cutting a word- python

Question

Given the following, i can find the longest common substring:

s1 = "this is a foo bar sentence ."
s2 = "what the foo bar blah blah black sheep is doing ?"

def longest_common_substring(s1, s2):
  m = [[0] * (1 + len(s2)) for i in xrange(1 + len(s1))]
  longest, x_longest = 0, 0
  for x in xrange(1, 1 + len(s1)):
    for y in xrange(1, 1 + len(s2)):
      if s1[x - 1] == s2[y - 1]:
        m[x][y] = m[x - 1][y - 1] + 1
        if m[x][y] > longest:
          longest = m[x][y]
          x_longest = x
      else:
        m[x][y] = 0
  return s1[x_longest - longest: x_longest]

print longest_common_substring(s1, s2)

[out]:

foo bar

But how do i ensure that the longest common substring respect English word boundary and don't cut up a word? For example, the following sentences:

s1 = "this is a foo bar sentence ."
s2 = "what a kappa foo bar black sheep ?"
print longest_common_substring(s1, s2)

outputs the follow which is NOT desired since it breaks up the word kappa from s2:

a foo bar

The desired output is still:

foo bar

I've tried also an ngram way of getting the longest common substring respecting word boundary but is there other way that deals with strings without calculating ngrams? (see answer)

Answer 1

This is too simple to understand. I used your code to do 75% of the job. I first split the sentence into words, then pass it to your function to get the largest common substring(in this case it will be longest consecutive words), so your function gives me ['foo', 'bar'], I join the elements of that array to produce the desired result.

Here is the online working copy for you to test and verify and fiddle with it.

http://repl.it/RU0/1

def longest_common_substring(s1, s2):
  m = [[0] * (1 + len(s2)) for i in xrange(1 + len(s1))]
  longest, x_longest = 0, 0
  for x in xrange(1, 1 + len(s1)):
    for y in xrange(1, 1 + len(s2)):
      if s1[x - 1] == s2[y - 1]:
        m[x][y] = m[x - 1][y - 1] + 1
        if m[x][y] > longest:
          longest = m[x][y]
          x_longest = x
      else:
        m[x][y] = 0
  return s1[x_longest - longest: x_longest]

def longest_common_sentence(s1, s2):
    s1_words = s1.split(' ')
    s2_words = s2.split(' ')  
    return ' '.join(longest_common_substring(s1_words, s2_words))


s1 = 'this is a foo bar sentence .'
s2 = 'what a kappa foo bar black sheep ?'
common_sentence = longest_common_sentence(s1, s2)
print common_sentence
>> 'foo bar'

Edge cases

1. '.' and '?' are also treated as valid words as in your case if there is a space between last word and the punctuation mark. If you don't leave a space they will be counted as part of last word. In that case 'sheep' and 'sheep?' would not be same words anymore. Its up to you decide what to do with such characters, before calling such function. In that case

   import re
s1 = re.sub('[.?]','', s1)
s2 = re.sub('[.?]','', s2)

and then continue as usual.