python 中文分词——FMM算法，python分词fmm算法,FMM算法的最简单思想是

python 中文分词——FMM算法，python分词fmm算法,FMM算法的最简单思想是

FMM算法的最简单思想是使用贪心算法向前找n个，如果这n个组成的词在词典中出现，就ok，如果没有出现，那么找n-1个...然后继续下去。假如n个词在词典中出现，那么从n+1位置继续找下去，知道句子结束。

import re   def PreProcess(sentence,edcode="utf-8"):       sentence = sentence.decode(edcode)       sentence=re.sub(u"[。，,！……!《》<>\\"':：？\\?、\\|“”‘’；]"," ",sentence)       return sentence   def FMM(sentence,diction,result = [],maxwordLength = 4,edcode="utf-8"):       i = 0      sentence = PreProcess(sentence,edcode)       length = len(sentence)       while i < length:          # find the ascii word           tempi=i           tok=sentence[i:i+1]           while re.search("[0-9A-Za-z\\-\\+#@_\\.]{1}",tok)<>None:               i= i+1              tok=sentence[i:i+1]           if i-tempi>0:               result.append(sentence[tempi:i].lower().encode(edcode))          # find chinese word           left = len(sentence[i:])           if left == 1:               """go to 4 step over the FMM"""             """should we add the last one? Yes, if not blank"""             if sentence[i:] <> " ":                   result.append(sentence[i:].encode(edcode))               return result           m = min(left,maxwordLength)           for j in xrange(m,0,-1):               leftword = sentence[i:j+i].encode(edcode)           #   print leftword.decode(edcode)               if LookUp(leftword,diction):                  # find the left word in dictionary                  # it's the right one                   i = j+i                   result.append(leftword)                   break             elif j == 1:                   """only one word, add into result, if not blank"""                 if leftword.decode(edcode) <> " ":                       result.append(leftword)                   i = i+1              else:                   continue     return result   def LookUp(word,dictionary):       if dictionary.has_key(word):           return True      return False  def ConvertGBKtoUTF(sentence):       return sentence.decode('gbk').encode('utf-8')#该片段来自于http://byrx.net