python - 将 Levenshtein 比率转换为 C++

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Answer 1

我想指出，虽然这个问题是一个纯资源问题，但_levenshtein.c在 python-Levenshtein 的源代码分发中，它本身可以用作 C-only 库，前提是它是用 -DNO_PYTHON 编译的:

Levenshtein.c can be used as a pure C library, too. You only have to define NO_PYTHON preprocessor symbol (-DNO_PYTHON) when compiling it. The functionality is similar to that of the Python extension. No separate docs are provided yet, RTFS. But they are not interchangeable:

C functions exported when compiling with -DNO_PYTHON (see _levenshtein.h) are not exported when compiling as a Python extension (and vice versa) Unicode character type used with -DNO_PYTHON is wchar_t, Python extension uses Py_UNICODE, they may be the same but don't count on it

一个例子:

#define NO_PYTHON
#include <stdio.h>
#include <string.h>
#include "_levenshtein.h"

double ratio(char *s1, char *s2) {
    size_t l1 = strlen(s1);
    size_t l2 = strlen(s2);
    size_t lsum = l1 + l2;
    if (lsum == 0) {
        return 1;
    }

    size_t distance = lev_edit_distance(l1, s1, l2, s2, 1);
    return ((double)lsum - distance) / (lsum);
}

int main() {
    char *str1 = "StackOver";
    char *str2 = "Stackoverflow";

    printf("%.16f\n", ratio(str1, str2));
}

用-DNO_PYTHON编译_levenshtein.c并链接在一起，运行；输出

0.7272727272727273