如何在Python中获得一个字符串与另一个字符串相似的概率?

我想要得到一个十进制值,比如0.9(意思是90%)等等。最好是标准的Python和库。

e.g.

similar("Apple","Appel") #would have a high prob.

similar("Apple","Mango") #would have a lower prob.

当前回答

还添加了Spacy NLP库;

@profile
def main():
    str1= "Mar 31 09:08:41  The world is beautiful"
    str2= "Mar 31 19:08:42  Beautiful is the world"
    print("NLP Similarity=",nlp(str1).similarity(nlp(str2)))
    print("Diff lib similarity",SequenceMatcher(None, str1, str2).ratio()) 
    print("Jellyfish lib similarity",jellyfish.jaro_distance(str1, str2))

if __name__ == '__main__':

    #python3 -m spacy download en_core_web_sm
    #nlp = spacy.load("en_core_web_sm")
    nlp = spacy.load("en_core_web_md")
    main()

使用Robert Kern的line_profiler运行

kernprof -l -v ./python/loganalysis/testspacy.py

NLP Similarity= 0.9999999821467294
Diff lib similarity 0.5897435897435898
Jellyfish lib similarity 0.8561253561253562

然而,时间的启示

Function: main at line 32

Line #      Hits         Time  Per Hit   % Time  Line Contents
==============================================================
    32                                           @profile
    33                                           def main():
    34         1          1.0      1.0      0.0      str1= "Mar 31 09:08:41  The world is beautiful"
    35         1          0.0      0.0      0.0      str2= "Mar 31 19:08:42  Beautiful is the world"
    36         1      43248.0  43248.0     99.1      print("NLP Similarity=",nlp(str1).similarity(nlp(str2)))
    37         1        375.0    375.0      0.9      print("Diff lib similarity",SequenceMatcher(None, str1, str2).ratio()) 
    38         1         30.0     30.0      0.1      print("Jellyfish lib similarity",jellyfish.jaro_distance(str1, str2))

其他回答

你可以创建这样一个函数:

def similar(w1, w2):
    w1 = w1 + ' ' * (len(w2) - len(w1))
    w2 = w2 + ' ' * (len(w1) - len(w2))
    return sum(1 if i == j else 0 for i, j in zip(w1, w2)) / float(len(w1))

这是内置的。

from difflib import SequenceMatcher

def similar(a, b):
    return SequenceMatcher(None, a, b).ratio()

使用它:

>>> similar("Apple","Appel")
0.8
>>> similar("Apple","Mango")
0.0

这是我想到的:

import string

def match(a,b):
    a,b = a.lower(), b.lower()
    error = 0
    for i in string.ascii_lowercase:
            error += abs(a.count(i) - b.count(i))
    total = len(a) + len(b)
    return (total-error)/total

if __name__ == "__main__":
    print(match("pple inc", "Apple Inc."))

Python3.6 + = 没有导入图书馆 在大多数情况下工作良好

在堆栈溢出,当你试图添加一个标签或发布一个问题,它会带来所有相关的东西。这是如此方便,正是我正在寻找的算法。因此,我编写了一个查询集相似度过滤器。

def compare(qs, ip):
    al = 2
    v = 0
    for ii, letter in enumerate(ip):
        if letter == qs[ii]:
            v += al
        else:
            ac = 0
            for jj in range(al):
                if ii - jj < 0 or ii + jj > len(qs) - 1: 
                    break
                elif letter == qs[ii - jj] or letter == qs[ii + jj]:
                    ac += jj
                    break
            v += ac
    return v


def getSimilarQuerySet(queryset, inp, length):
    return [k for tt, (k, v) in enumerate(reversed(sorted({it: compare(it, inp) for it in queryset}.items(), key=lambda item: item[1])))][:length]
        


if __name__ == "__main__":
    print(compare('apple', 'mongo'))
    # 0
    print(compare('apple', 'apple'))
    # 10
    print(compare('apple', 'appel'))
    # 7
    print(compare('dude', 'ud'))
    # 1
    print(compare('dude', 'du'))
    # 4
    print(compare('dude', 'dud'))
    # 6

    print(compare('apple', 'mongo'))
    # 2
    print(compare('apple', 'appel'))
    # 8

    print(getSimilarQuerySet(
        [
            "java",
            "jquery",
            "javascript",
            "jude",
            "aja",
        ], 
        "ja",
        2,
    ))
    # ['javascript', 'java']

解释

compare takes two string and returns a positive integer. you can edit the al allowed variable in compare, it indicates how large the range we need to search through. It works like this: two strings are iterated, if same character is find at same index, then accumulator will be added to a largest value. Then, we search in the index range of allowed, if matched, add to the accumulator based on how far the letter is. (the further, the smaller) length indicate how many items you want as result, that is most similar to input string.

如上所述,有许多指标可以定义字符串之间的相似性和距离。我将给出我的5美分,通过展示一个Jaccard与Q-Grams相似的例子和一个编辑距离的例子。

from nltk.metrics.distance import jaccard_distance
from nltk.util import ngrams
from nltk.metrics.distance  import edit_distance

Jaccard相似

1-jaccard_distance(set(ngrams('Apple', 2)), set(ngrams('Appel', 2)))

我们得到:

0.33333333333333337

还有苹果和芒果

1-jaccard_distance(set(ngrams('Apple', 2)), set(ngrams('Mango', 2)))

我们得到:

0.0

编辑距离

edit_distance('Apple', 'Appel')

我们得到:

2

最后,

edit_distance('Apple', 'Mango')

我们得到:

5

q - grams上的余弦相似度(q=2)

另一个解决方案是使用textdistance库。我将提供一个余弦相似度的例子

import textdistance
1-textdistance.Cosine(qval=2).distance('Apple', 'Appel')

我们得到:

0.5