FuzzyWuzzy

• 簡(jiǎn)單比較

>>> from fuzzywuzzy import fuzz
>>> fuzz.ratio("this is pass","this is a poce!")
67

• 部分比

>>> fuzz.partial_ratio("this is a text", "this is a test!")
    93

• 單詞排序比

>>> fuzz.ratio("fuzzy wuzzy was ","wuzzy,fuzzy as dfd")
53
>>> fuzz.token_sort_ratio("fuzzy wuzzy was ","wuzzy,fuzzy as dfd")
67

• 單詞集合比

>>> 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

• Process

>>> 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

• Levenshtein.hamming(str1,str2)

計(jì)算漢明距離，要求str1很str2的長(zhǎng)度必須一致。是描述兩個(gè)等長(zhǎng)字串之間對(duì)應(yīng)位置上不同字符的個(gè)數(shù)

• Levenshtein.distance(str1,str2)

計(jì)算編輯距離(也稱為L(zhǎng)evenshtein距離)。是描述由一個(gè)字符轉(zhuǎn)化為另一個(gè)字符串最少的操作次數(shù)，在其中包括插入、刪除、替換

  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)
> 計(jì)算萊文斯坦比。計(jì)算公式
```math
(sum - idist)/sum

其中sum是指str1和str2的字符串長(zhǎng)度總和。idist是類編輯距離
注：這里的類編輯距離不是2中所講的編輯距離，2中三種操作中的每個(gè)操作+1，而此處，刪除、插入依然加+1，但是替換加2
這樣做的目的是：ratio('a','c'),sum=2 按2中計(jì)算為(2-1)/2=0.5,'a'/'c'沒(méi)有重合，顯然不合算，但是替換操作+2，就會(huì)解決這個(gè)問(wèn)題

• Levenshtein.jaro(str1,str2)

計(jì)算jaro距離

其中m為s1,s2的匹配長(zhǎng)度，當(dāng)某位置的認(rèn)為匹配當(dāng)該位置字符相同，或者不超過(guò)

t是調(diào)換次數(shù)的一般

• Levenshtein.jaro_winkler(str1,str2)

計(jì)算jaro_Winkler距離