Levenshtein distance: how to better handle words swapping positions?

Question

I've had some success comparing strings using the PHP levenshtein function.

However, for two strings which contain substrings that have swapped positions, the algorithm counts those as whole new substrings.

For example:

levenshtein("The quick brown fox", "brown quick The fox"); // 10 differences

are treated as having less in common than:

levenshtein("The quick brown fox", "The quiet swine flu"); // 9 differences

I'd prefer an algorithm which saw that the first two were more similar.

How could I go about coming up with a comparison function that can identify substrings which have switched position as being distinct to edits?

One possible approach I've thought of is to put all the words in the string into alphabetical order, before the comparison. That takes the original order of the words completely out of the comparison. A downside to this, however, is that changing just the first letter of a word can create a much bigger disruption than a changing a single letter should cause.

What I'm trying to achieve is to compare two facts about people which are free text strings, and decide how likely these facts are to indicate the same fact. The facts might be the school someone attended, the name of their employer or publisher, for example. Two records may have the same school spelled differently, words in a different order, extra words, etc, so the matching has to be somewhat fuzzy if we are to make a good guess that they refer to the same school. So-far it is working very well for spelling errors (I am using a phoenetic algorithm similar to metaphone on top of this all) but very poorly if you switch the order of words around which seem common in a school: "xxx college" vs "college of xxx".

Answer 1

Its easy. Just use the Damerau-Levenshtein distance on the words instead of letters.

Answer 2

You can also try this. (just an extra suggestion)

$one = metaphone("The quick brown fox"); // 0KKBRNFKS
$two = metaphone("brown quick The fox"); // BRNKK0FKS
$three = metaphone("The quiet swine flu"); // 0KTSWNFL

similar_text($one, $two, $percent1); // 66.666666666667
similar_text($one, $three, $percent2); // 47.058823529412
similar_text($two, $three, $percent3); // 23.529411764706

This will show that the 1st and 2nd are more similar than one and three and two and three.

Answer 3

Explode on spaces, sort the array, implode, then do the Levenshtein.

Answer 4

Take this answer and make the following change:

void match(trie t, char* w, string s, int budget){
  if (budget < 0) return;
  if (*w=='\0') print s;
  foreach (char c, subtrie t1 in t){
    /* try matching or replacing c */
    match(t1, w+1, s+c, (*w==c ? budget : budget-1));
    /* try deleting c */
    match(t1, w, s, budget-1);
  }
  /* try inserting *w */
  match(t, w+1, s + *w, budget-1);
  /* TRY SWAPPING FIRST TWO CHARACTERS */
  if (w[1]){
    swap(w[0], w[1]);
    match(t, w, s, budget-1);
    swap(w[0], w[1]);
  }
}

This is for dictionary search in a trie, but for matching to a single word, it's the same idea. You're doing branch-and-bound, and at any point, you can make any change you like, as long as you give it a cost.

Answer 5

Eliminate duplicate words between the two strings and then use Levenshtein.

Answer 6

I've been implementing levenshtein in a spell checker.

What you're asking for is counting transpositions as 1 edit.

This is easy if you only wish to count transpositions of one word away. However for transposition of words 2 or more away, the addition to the algorithm is worst case scenario !(max(wordorder1.length(), wordorder2.length())). Adding a non-linear subalgorithm to an already quadratic algorithm is not a good idea.

This is how it would work.

if (wordorder1[n] == wordorder2[n-1])
{
  min(workarray[x-1, y] + 1, workarray[x, y-1] + 1, workarray[x-2, y-2]);
}
  else
{
  min(workarray[x-1, y] + 1, workarray[x, y-1] + 1);
}

JUST for touching transpositions. If you want all transpositions, you'd have to for every position work backwards from that point comparing

1[n] == 2[n-2].... 1[n] == 2[0]....

So you see why they don't include this in the standard method.

Answer 7

i believe this is a prime example for using a vector-space search engine.

in this technique, each document essentially becomes a vector with as many dimensions as there are different words in the entire corpus; similar documents then occupy neighboring areas in that vector space. one nice property of this model is that queries are also just documents: to answer a query, you simply calculate their position in vector space, and your results are the closest documents you can find. i am sure there are get-and-go solutions for PHP out there.

to fuzzify results from vector space, you could consider to do stemming / similar natural language processing technique, and use levenshtein to construct secondary queries for similar words that occur in your overall vocabulary.