John Misoskian

def levenshtein(a,b):
    n, m = len(a), len(b)
    if n > m:
        a,b = b,a
        n,m = m,n  

    current = range(n+1)
    for i in range(1,m+1):
        previous, current = current, [i]+[0]*n
        for j in range(1,n+1):
            add, delete = previous[j]+1, current[j-1]+1
            change = previous[j-1]
            if a[j-1] != b[i-1]:
                change = change + 1
            current[j] = min(add, delete, change)
    perc =   (m - current[n]) / (m/100.0)
    return perc

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