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FuzzyWuzzy and Levenshtein

2017-07-17  本文已影响278人  羽恒

FuzzyWuzzy

>>> from fuzzywuzzy import fuzz
>>> fuzz.ratio("this is pass","this is a poce!")
67
>>> fuzz.partial_ratio("this is a text", "this is a test!")
    93
>>> fuzz.ratio("fuzzy wuzzy was ","wuzzy,fuzzy as dfd")
53
>>> fuzz.token_sort_ratio("fuzzy wuzzy was ","wuzzy,fuzzy as dfd")
67
>>> fuzz.token_sort_ratio("fuzzy was a ced", "fuzzy fuzzy wer a bear")
    59
>>> fuzz.token_set_ratio("fuzzy was a ced", "fuzzy fuzzy wer a bear")
    71
>>> from suzzywuzzy import process
>>> choices = ["Atlanta hello", "New York Jets", "New York Giants", "Dallas bob_dd"]
>>> process.extract("new york jets", choices, limit=2)
    [('New York Jets', 100), ('New York Giants', 79)]
>>> process.extractOne("cowboys", choices)
    ("Dallas Cowboys", 90)

Levenshtein

计算汉明距离,要求str1很str2的长度必须一致。是描述两个等长字串之间对应位置上不同字符的个数

计算编辑距离(也称为Levenshtein距离)。是描述由一个字符转化为另一个字符串最少的操作次数,在其中包括插入、删除、替换

  def levenshtein(first,second):
      if len(first)>len(second):
          first,second = second,first
  
      if len(first) == 0:
          return len(second)
      if len(second) == 0:
          return len(first)
  
      first_length = len(first)+1
      second_length = len(second)+1
  
      distance_matrix = [range(second_length) for x in range(first_length)]
      print distance_matrix[1][1],distance_matrix[1][2],distance_matrix[1][3],distance_matrix[1][4]
  
      for i in range(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)
  
      print distan_matrix
      return distance_matrix[first_length-1][second_length-1]
  ```
- Levenshtein.ratio(str1,str2)
> 计算莱文斯坦比。计算公式
```math
(sum - idist)/sum

其中sum是指str1和str2的字符串长度总和。idist是类编辑距离
注:这里的类编辑距离不是2中所讲的编辑距离,2中三种操作中的每个操作+1,而此处,删除、插入依然加+1,但是替换加2
这样做的目的是:ratio('a','c'),sum=2 按2中计算为(2-1)/2=0.5,'a'/'c'没有重合,显然不合算,但是替换操作+2,就会解决这个问题

计算jaro距离


其中m为s1,s2的匹配长度,当某位置的认为匹配当该位置字符相同,或者不超过



t是调换次数的一般

计算jaro_Winkler距离
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