I have found some Python
implementations of the Levenshtein distance .
I am wondering though how these algorithms can be efficiently modified so that they break if the Levenshtein distance is greater than n
(eg 3) instead of running until the end?
So essentially I do not want to let the algorithm run for too long to calculate the final distance if I simply want to know if the distance is greater than a threshold or not.
I have found some relevant posts here:
but still, I do not see any Python code which does what I describe above (which is more or less what these posts describe too).
PS: The solution provided by @amirouche below is based on the fastest implementation that I have tested with some benchmarking (from here: https://en.wikibooks.org/wiki/Algorithm_Implementation/Strings/Levenshtein_distance#Python , https://stackoverflow.com/a/32558749/9024698 ) and its bounded version is the fastest one of its kind from my tests (without excluding that there may be even faster ones).
As described in Levenstein distance limit , you can add a test over the row that is computed to return early:
def levenshtein(s1, s2, maximum):
if len(s1) > len(s2):
s1, s2 = s2, s1
distances = range(len(s1) + 1)
for i2, c2 in enumerate(s2):
distances_ = [i2+1]
for i1, c1 in enumerate(s1):
if c1 == c2:
distances_.append(distances[i1])
else:
distances_.append(1 + min((distances[i1], distances[i1 + 1], distances_[-1])))
if all((x >= maximum for x in distances_)):
return False
distances = distances_
return distances[-1]
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