Python implementation of the Levenshtein distance with edits

levenshtein.py

# Calculates the levenshtein distance and the edits between two strings
def levenshtein(s1, s2, key=hash):
  rows = costmatrix(s1, s2, key)
  edits = backtrace(s1, s2, rows, key)
 
  return rows[-1][-1], edits
 
# Generate the cost matrix for the two strings
# Based on http://en.wikibooks.org/wiki/Algorithm_Implementation/Strings/Levenshtein_distance#Python
def costmatrix(s1, s2, key=hash):
  rows = []
 
  previous_row = xrange(len(s2) + 1)
  rows.append(list(previous_row))
 
  for i, c1 in enumerate(s1):
    current_row = [i + 1]
    for j, c2 in enumerate(s2):
      insertions = previous_row[j + 1] + 1
      deletions = current_row[j] + 1
      substitutions = previous_row[j] + (key(c1) != key(c2))
      current_row.append(min(insertions, deletions, substitutions))
    previous_row = current_row
 
    rows.append(previous_row)
 
  return rows
 
# Trace back through the cost matrix to generate the list of edits
def backtrace(s1, s2, rows, key=hash):
  i, j = len(s1), len(s2)
 
  edits = []
 
  while(not (i == 0  and j == 0)):
    prev_cost = rows[i][j]
 
    neighbors = []
 
    if(i!=0 and j!=0):
      neighbors.append(rows[i-1][j-1])
    if(i!=0):
      neighbors.append(rows[i-1][j])
    if(j!=0):
      neighbors.append(rows[i][j-1])
 
    min_cost = min(neighbors)
 
    if(min_cost == prev_cost):
      i, j = i-1, j-1
      edits.append({'type':'match', 'i':i, 'j':j})
    elif(i!=0 and j!=0 and min_cost == rows[i-1][j-1]):
      i, j = i-1, j-1
      edits.append({'type':'substitution', 'i':i, 'j':j})
    elif(i!=0 and min_cost == rows[i-1][j]):
      i, j = i-1, j
      edits.append({'type':'deletion', 'i':i, 'j':j})
    elif(j!=0 and min_cost == rows[i][j-1]):
      i, j = i, j-1
      edits.append({'type':'insertion', 'i':i, 'j':j})
 
  edits.reverse()
 
  return edits