问题
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".
回答1:
N-grams
Use N-grams, which support multiple-character transpositions across the whole text.
The general idea is that you split the two strings in question into all the possible 2-3 character substrings (n-grams) and treat the number of shared n-grams between the two strings as their similarity metric. This can be then normalized by dividing the shared number by the total number of n-grams in the longer string. This is trivial to calculate, but fairly powerful.
For the example sentences:
A. The quick brown fox
B. brown quick The fox
C. The quiet swine flu
A and B share 18 2-grams
A and C share only 8 2-grams
out of 20 total possible.
This has been discussed in more detail in the Gravano et al. paper.
tf-idf and cosine similarity
A not so trivial alternative, but grounded in information theory would be to use term term frequency–inverse document frequency (tf-idf) to weigh the tokens, construct sentence vectors and then use cosine similarity as the similarity metric.
The algorithm is:
- Calculate 2-character token frequencies (tf) per sentence.
- Calculate inverse sentence frequencies (idf), which is a logarithm of a quotient of the number of all sentences in the corpus (in this case 3) divided by the number of times a particular token appears across all sentences. In this case th is in all sentences so it has zero information content (log(3/3)=0).
- Produce the tf-idf matrix by multiplying corresponding cells in the tf and idf tables.
- Finally, calculate cosine similarity matrix for all sentence pairs, where A and B are weights from the tf-idf table for the corresponding tokens. The range is from 0 (not similar) to 1 (equal).
Levenshtein modifications and Metaphone
Regarding other answers. Damerau–Levenshtein modificication supports only the transposition of two adjacent characters. Metaphone was designed to match words that sound the same and not for similarity matching.
回答2:
Its easy. Just use the Damerau-Levenshtein distance on the words instead of letters.
回答3:
Explode on spaces, sort the array, implode, then do the Levenshtein.
回答4:
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.
回答5:
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.
回答6:
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.
回答7:
Eliminate duplicate words between the two strings and then use Levenshtein.
回答8:
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.
回答9:
If the first string is A and the second one is B:
- Split A and B into words
- For every word in A, find the best matching word in B (using levenshtein)
- Remove that word from B and put it in B* at the same index as the matching word in A.
- Now compare A and B*
Example:
A: The quick brown fox
B: Quick blue fox the
B*: the Quick blue fox
You could improve step 2 by doing it in multiple passes, finding only exact matches at first, then finding close matches for words in A that don't have a companion in B* yet, then less close matches, etc.
来源：https://stackoverflow.com/questions/828132/levenshtein-distance-how-to-better-handle-words-swapping-positions