c++ - 具有C++ vector 的文档距离算法

Question

我试图用C++ vector 构建文档距离算法，该算法在模拟 vector 上运行良好，并给我输出0，这意味着 vector 相同为100％，但是当我用非常不同的字符串尝试时，它给我的值(value)就像一分一些小数，该值应为90或相差时在其附近，这是我的代码:

double dotProduct(vector <string> word1 , vector <string> word2) {

    double dotProductSum = 0;
    int  frequencyD1 = 0;
    int  frequencyD2 = 0;


    if (word1.size() > word2.size()) {

        for (int i = 0; i < word1.size(); i++) {

            for (int j = 0; j < word1.size(); j++) {

                if (word1[i] == word1[j]) {

                    frequencyD1++;
                }

                if (word2[i] == word2[j]) {
                    frequencyD2++;
                }

            }

            dotProductSum += frequencyD1 * frequencyD2;
            frequencyD1 , frequencyD2 = 0;
        }

    }

    else {
        
        for (int i = 0; i < word2.size(); i++) {

            for (int j = 0; j < word2.size(); j++) {

                
               if (word2[i] == word1[j]) {

                        frequencyD1++;
                }

                 if (word2[i] == word2[j]) {
                        frequencyD2++;
                    }

                }

            dotProductSum += frequencyD1 * frequencyD2;
            frequencyD1, frequencyD2 = 0;
        }
        
    }

    return dotProductSum;

}

测试案例:

    vector <string> doc1 = { "this" , "is" , "cat" };
    vector <string> doc2 = { "Here" , "test" ,"two" };
    
    double normalization = dotProduct(doc1, doc2) / sqrt(dotProduct(doc1, doc1) * dotProduct(doc2, doc2)) ;

    cout << acos(normalization)  << endl;  // 1.5708

Answer 1

frequencyD1 , frequencyD2 = 0;

这是一个 comma operator expression，评估为 (frequencyD1) , (frequencyD2 = 0);，与 frequencyD2 = 0;等效，并且 frequencyD1保持不变。
要将两个变量重新初始化为 0，请使用以下命令。

   frequencyD1 = frequencyD2 = 0;

[编辑]还要...

the value should be 90 or nearby it when it difference
...
cout << acos(normalization) << endl; // 1.5708

...请注意 acos返回 radians和 1.5708 ≈ π/2 = 90°中的 Angular 。