I haven\'t found anything relevant on Google, so I\'m hoping to find some help here :)
I\'ve got a Python list as follows:
[[\'hoose\', 200], [\"Ba
In common with the other comments, I'm not sure that doing this makes much sense, but here's a solution that does what you want, I think. It's very inefficient - O(n2) where n is the number of words in your list - but I'm not sure there's a better way of doing it:
data = [['hoose', 200],
["Bananphone", 10],
['House', 200],
["Bonerphone", 10],
['UniqueValue', 777]]
already_merged = []
for word, score in data:
added_to_existing = False
for merged in already_merged:
for potentially_similar in merged[0]:
if levenshtein(word, potentially_similar) < 5:
merged[0].add(word)
merged[1] += score
added_to_existing = True
break
if added_to_existing:
break
if not added_to_existing:
already_merged.append([set([word]),score])
print already_merged
The output is:
[[set(['House', 'hoose']), 400], [set(['Bonerphone', 'Bananphone']), 20], [set(['UniqueValue']), 777]]
One of the obvious problems with this approach is that the word that you're considering might be close enough to many of the different sets of words that you've already considered, but this code will just lump it into the first one it finds. I've voted +1 for Space_C0wb0y's answer ;)