再次计算重叠的正则表达式匹配

2 回答

• 3

  是的，这是丑陋的，未经优化但似乎有效 . 这是对所有可能的 but unique 变体的简单尝试

  def myregex(pattern,text,dir=0):
      import re
      m = re.search(pattern, text)
      if m:
          yield m.group(0)
          if len(m.group('suffix')):
              for r in myregex(pattern, "%s%s%s" % (m.group('prefix'),m.group('suffix')[1:],m.group('end')),1):
                  yield r
              if dir<1 :
                  for r in myregex(pattern, "%s%s%s" % (m.group('prefix'),m.group('suffix')[:-1],m.group('end')),-1):
                      yield r
  
  
  def myprocess(pattern, text):    
      parts = pattern.split("*")    
      for i in range(0, len(parts)-1 ):
          res=""
          for j in range(0, len(parts) ):
              if j==0:
                  res+="(?P<prefix>"
              if j==i:
                  res+=")(?P<suffix>"
              res+=parts[j]
              if j==i+1:
                  res+=")(?P<end>"
              if j<len(parts)-1:
                  if j==i:
                      res+=".*"
                  else:
                      res+=".*?"
              else:
                  res+=")"
          for r in myregex(res,text):
              yield r
  
  def mycount(pattern, text):
      return set(myprocess(pattern, text))
  

  测试：

  >>> mycount('a*b*c','abc')
  set(['abc'])
  >>> mycount('a*k','akka')
  set(['akk', 'ak'])
  >>> mycount('b*o','bboo')
  set(['bbo', 'bboo', 'bo', 'boo'])
  >>> mycount('b*o','bb123oo')
  set(['b123o', 'bb123oo', 'bb123o', 'b123oo'])
  >>> mycount('b*o','ffbfbfffofoff')
  set(['bfbfffofo', 'bfbfffo', 'bfffofo', 'bfffo'])
  

  回复于 2024-05-06T03:45:26+08:00
• 2

  我认为你正在寻找的东西最好用一个像lepl这样的解析库来完成：

  >>> from lepl import *
  >>> parser = Literal('a') + Any()[:] + Literal('k')
  >>> parser.config.no_full_first_match()
  >>> list(parser.parse_all('akka'))
  [['akk'], ['ak']]
  >>> parser = Literal('a') + Any()[:] + Literal('k') + Any()[:] + Literal('a')
  >>> list(parser.parse_all('akka'))
  [['akka'], ['akka']]
  

  我相信 parser.parse_all 的输出长度是你正在寻找的 .

  请注意，如果模式与整个字符串不匹配，则需要使用 parser.config.no_full_first_match() 来避免错误 .

  编辑：根据@ Shamanu4的评论，我看到你想从任何位置开始匹配结果，你可以这样做：

  >>> text = 'bboo'
  >>> parser = Literal('b') + Any()[:] + Literal('o')
  >>> parser.config.no_full_first_match()
  >>> substrings = [text[i:] for i in range(len(text))]
  >>> matches = [list(parser.parse_all(substring)) for substring in substrings]
  >>> matches = filter(None, matches) # Remove empty matches
  >>> matches = list(itertools.chain.from_iterable(matches)) # Flatten results
  >>> matches = list(itertools.chain.from_iterable(matches)) # Flatten results (again)
  >>> matches
  ['bboo', 'bbo', 'boo', 'bo']
  

  回复于 2024-05-06T03:45:26+08:00