lcs.py 最长公共子串算法

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感觉用来匹配相似文件比最短编辑距离更靠谱,最短编辑应该是用来纠错的


http://www.unixuser.org/~euske/python/
这个网站还有不少好脚本

http://www.unixuser.org/~euske/python/lcs.py

zuroc@frodo ~/dev/douban $ cat lcs.py
#!/usr/bin/env python
# find an LCS (Longest Common Subsequence).
# *public domain*

def find_lcs_len(s1, s2):
  m = [ [ 0 for x in s2 ] for y in s1 ]
  for p1 in range(len(s1)):
    for p2 in range(len(s2)):
      if s1[p1] == s2[p2]:
        if p1 == 0 or p2 == 0:
          m[p1][p2] = 1
        else:
          m[p1][p2] = m[p1-1][p2-1]+1
      elif m[p1-1][p2] < m[p1][p2-1]:
        m[p1][p2] = m[p1][p2-1]
      else:                             # m[p1][p2-1] < m[p1-1][p2]
        m[p1][p2] = m[p1-1][p2]
  return m[-1][-1]

def find_lcs(s1, s2):
  # length table: every element is set to zero.
  m = [ [ 0 for x in s2 ] for y in s1 ]
  # direction table: 1st bit for p1, 2nd bit for p2.
  d = [ [ None for x in s2 ] for y in s1 ]
  # we don't have to care about the boundery check.
  # a negative index always gives an intact zero.
  for p1 in range(len(s1)):
    for p2 in range(len(s2)):
      if s1[p1] == s2[p2]:
        if p1 == 0 or p2 == 0:
          m[p1][p2] = 1
        else:
          m[p1][p2] = m[p1-1][p2-1]+1
        d[p1][p2] = 3                   # 11: decr. p1 and p2
      elif m[p1-1][p2] < m[p1][p2-1]:
        m[p1][p2] = m[p1][p2-1]
        d[p1][p2] = 2                   # 10: decr. p2 only
      else:                             # m[p1][p2-1] < m[p1-1][p2]
        m[p1][p2] = m[p1-1][p2]
        d[p1][p2] = 1                   # 01: decr. p1 only
  (p1, p2) = (len(s1)-1, len(s2)-1)
  # now we traverse the table in reverse order.
  s = []
  while 1:
    print p1,p2
    c = d[p1][p2]
    if c == 3: s.append(s1[p1])
    if not ((p1 or p2) and m[p1][p2]): break
    if c & 2: p2 -= 1
    if c & 1: p1 -= 1
  s.reverse()
  return ''.join(s)

if __name__ == '__main__':
  print find_lcs('abcoisjf','axbaoeijf')
  print find_lcs_len('abcoisjf','axbaoeijf')