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.

内置的SequenceMatcher在大输入时非常慢，下面是如何用diff-match-patch完成的:

from diff_match_patch import diff_match_patch

def compute_similarity_and_diff(text1, text2):
    dmp = diff_match_patch()
    dmp.Diff_Timeout = 0.0
    diff = dmp.diff_main(text1, text2, False)

    # similarity
    common_text = sum([len(txt) for op, txt in diff if op == 0])
    text_length = max(len(text1), len(text2))
    sim = common_text / text_length

    return sim, diff

TheFuzz是一个用python实现Levenshtein距离的包，在某些情况下，当你希望两个不同的字符串被认为是相同的时，它带有一些帮助函数来提供帮助。例如:

>>> fuzz.ratio("fuzzy wuzzy was a bear", "wuzzy fuzzy was a bear")
    91
>>> fuzz.token_sort_ratio("fuzzy wuzzy was a bear", "wuzzy fuzzy was a bear")
    100

我想你们可能在寻找一种描述字符串之间距离的算法。这里有一些你可以参考的:

汉明距离 Levenshtein距离 Damerau-Levenshtein距离 Jaro-Winkler距离

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.

出于我的目的，我有自己的quick_ratio()，它比difflib SequenceMatcher的quick_ratio()快2倍，同时提供类似的结果。A和b是字符串:

    score = 0
    for letters in enumerate(a):
        score = score + b.count(letters[1])