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Say you have scanned a document with names on it. Due to mistakes in the scanning process, you want to look up the names in a dictionary. Therefore, you need a function that takes in a possible name and outputs a list with every possible string variation of the input within a Levenshtein-Distance of 1.

I modified an implementation (https://rosettacode.org/wiki/Levenshtein_distance#Python) but didn't get the right result, yet. Since Levenshtein implementations usually take in two strings and compare them to give out an int for the L-Distance, I am wondering how to change that to get the variations of one string?

def levenshteinVariation(n_possible):


m = n_possible
n = n_correct

d = []           

for i in range(len(m)+1):
    d.append([i])        
del d[0][0]    

for j in range(len(n)+1):
    d[0].append(j)       

for j in range(1,len(n)+1):

    for i in range(1,len(m)+1):

        if m[i-1] == n[j-1]:
            d[i].insert(j,d[i-1][j-1])           

        else:
            minimum = min(d[i-1][j]+1, d[i][j-1]+1, d[i-1][j-1]+2)         
            d[i].insert(j, minimum)

return d

The expected result would be a match in the dictionary to all variations within a L-Distance of 1.

for n_correct, n_possible in [('Marcus','Maacus'), ('David','Davide'), ('Steve', 'Steven')]:
print(f"{n_correct} found: {n_correct in levenshteinVariation(n_possible)}")

But I got:

Marcus found: False
David found: False
Steve found: False
J. Jones
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1 Answers1

0

Thanks to Dan D. I was able to solve it myself: 

def Variations1(name):

letters    = 'abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ'

splits     = [(name[:i], name[i:])    for i in range(len(name) + 1)]
deletes    = [L + R[1:]               for L, R in splits if R]
transposes = [L + R[1] + R[0] + R[2:] for L, R in splits if len(R) > 1]
replaces   = [L + c + R[1:]           for L, R in splits if R for c in letters]
inserts    = [L + c + R               for L, R in splits for c in letters]

return set(deletes + transposes + replaces + inserts)
J. Jones
  • 19
  • 4