Levenstein
Publicado por Carlos (20 intervenciones) el 19/01/2018 10:18:27
Alguna idea para optimizar el algoritmo de levenstein?
def levenshtein_distance(first, second):
if len(first) > len(second):
first, second = second, first
if len(second) == 0:
return len(first)
first_length = len(first) + 1
second_length = len(second) + 1
distance_matrix = [[0] * second_length for x in
range(first_length)]
for i in range(first_length): distance_matrix[i][0] = i
for j in range(second_length): distance_matrix[0][j] = j
for i in xrange(1, first_length):
for j in range(1, second_length):
deletion = distance_matrix[i-1][j] + 1
insertion = distance_matrix[i][j-1] + 1
substitution = distance_matrix[i-1][j-1]
if first[i-1] != second[j-1]:
substitution += 1
distance_matrix[i][j] = min(insertion,
deletion, substitution)
return distance_matrix[first_length-1][second_length-1]
def levenshtein_distance(first, second):
if len(first) > len(second):
first, second = second, first
if len(second) == 0:
return len(first)
first_length = len(first) + 1
second_length = len(second) + 1
distance_matrix = [[0] * second_length for x in
range(first_length)]
for i in range(first_length): distance_matrix[i][0] = i
for j in range(second_length): distance_matrix[0][j] = j
for i in xrange(1, first_length):
for j in range(1, second_length):
deletion = distance_matrix[i-1][j] + 1
insertion = distance_matrix[i][j-1] + 1
substitution = distance_matrix[i-1][j-1]
if first[i-1] != second[j-1]:
substitution += 1
distance_matrix[i][j] = min(insertion,
deletion, substitution)
return distance_matrix[first_length-1][second_length-1]
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