如何在显示匹配项的列表和未找到的元素之间进行模糊搜索？

Question

我试图对列表to_search中的值进行模糊匹配。在to_search列表中搜索choices列表中的每个值，并从result列表中显示相应的项。就像一个MS，但是带有模糊搜索。

这是我当前的代码，它几乎为不同的值打印正确的输出，但是对于在to_search中没有任何相似性的值，我想在output Not Found中显示，但是现在我得到了一些其他的输出。

我要搜索的是前缀，这个顺序是一样的。例如，值38050与此匹配的('358', 72.0, 8)一起出现，因为在38050中有3、5和8，但对我来说没有兴趣，因为3、5和8的顺序不同。如果找到的选择是380 in，至少在前缀上与38050有相似之处，这将是我的匹配。希望你能理解我的解释。

from rapidfuzz import process, fuzz

choices = [
        '237','1721','124622','334','124624','124','1246','1876','358',
        '33751','33679','599','61','230','31','65','1721','1','124623'
    ]

result = [
            'NAD','ATE','STA','SSI','GYP','RIC','EEC','AND','GIU','ANC',
            'PAI','GAR','TAL','ANI','LAN','TRI','GDO','MAR','EDE'
        ]

to_search = ['18763044','187635','23092','3162','38050','33','49185','51078','1246','1721']

for element in to_search:
    match =  process.extractOne(element, choices, scorer=fuzz.WRatio)
    print(element,result[match[2]],'         ## ',match)

电流输出

>>>
18763044    AND         ##  ('1876', 90.0, 7)
187635      AND         ##  ('1876', 90.0, 7)
23092       ANI         ##  ('230', 90.0, 13)
3162        LAN         ##  ('31', 90.0, 14)
38050       GIU         ##  ('358', 72.0, 8) // This should be marked as NOT FOUND
33          SSI         ##  ('334', 90.0, 3)
49185       MAR         ##  ('1', 90.0, 17)  // This should be marked as NOT FOUND
51078       MAR         ##  ('1', 90.0, 17)  // This should be marked as NOT FOUND
1246        EEC         ##  ('1246', 100.0, 6)
1721        ATE         ##  ('1721', 100.0, 1)

我想得到的输出：

18763044    AND
187635      AND
23092       ANI
3162        LAN
38050       NOT FOUND
33          SSI
49185       NOT FOUND
51078       NOT FOUND
1246        EEC
1721        ATE

表格格式，便于理解输入和输出。提前感谢

​

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Answer 1

你可以创造你自己的记分员：

def my_scorer(query,choice,**kwargs):
    # default score
    score=fuzz.WRatio(query,choice)

    is_prefix=False

    # if choice is a prefix of query
    if choice in query and query.index(choice)==0:
        is_prefix=True

    # if query is a prefix of choice
    if query in choice and choice.index(query)==0:
        is_prefix=True

    if not is_prefix:
        score=-1

    return score

并将其传递给process.extractOne和score_cutoff=0，以忽略分数低于0的结果

match = process.extractOne(element, choices, scorer=my_scorer, score_cutoff=0) 
if match:
  print(element, result[match[2]])
else:
  print(element, 'NOT FOUND')

Answer 2

您可以选择一个比WRatio更考虑订单的记分员。然后设置score_cutoff以排除低于给定相似性的结果。

就您的例子而言，fuzz.ratio和score_cutoff=60似乎有效。您必须在更大的数据集上进行测试，并尝试不同的评分器来确切地了解您需要的是什么：

from rapidfuzz import process, fuzz

choices = [
        '237','1721','124622','334','124624','124','1246','1876','358',
        '33751','33679','599','61','230','31','65','1721','1','124623'
    ]

result = [
            'NAD','ATE','STA','SSI','GYP','RIC','EEC','AND','GIU','ANC',
            'PAI','GAR','TAL','ANI','LAN','TRI','GDO','MAR','EDE'
        ]

to_search = ['18763044','187635','23092','3162','38050','33','49185','51078','1246','1721']

for element in to_search:
    match =  process.extractOne(element, choices, scorer=fuzz.ratio, score_cutoff=60)
    if match:
        print(element,result[match[2]],'         ## ',match)
    else:
        print(element,"NOT FOUND")

输出：

18763044 AND          ##  ('1876', 66.66666666666667, 7)
187635 AND          ##  ('1876', 80.0, 7)
23092 ANI          ##  ('230', 75.0, 13)
3162 LAN          ##  ('31', 66.66666666666667, 14)
38050 NOT FOUND
33 SSI          ##  ('334', 80.0, 3)
49185 NOT FOUND
51078 NOT FOUND
1246 EEC          ##  ('1246', 100.0, 6)
1721 ATE          ##  ('1721', 100.0, 1)