需要帮助修复代码以从文件输出正确的唯一字数

Question

我试图从文本文件中找到唯一的单词计数。但由于某种原因，我的号码总是关闭。我的常规字数很好。

我的字符串数组wordArr包含文件中的所有单词。

我发现每个单词在被发现不唯一之后尝试将其分配给另一个数组，然后我查看我已经通过的单词列表，看它是否与当前显示的单词匹配。如果单词匹配我将oldWord设置为true，并且该单词不计入我的unique计数。

//New portion
int main(int argc, char *argv[]) {
    //File Paths
    ifstream fp;
    fp.open(argv[1]);
    if (fp.fail()) {
        cout << "Error No file" << endl;
        return 0;
    }
    string wordArr[10000];
    string words;
    string temp;
    int wordCount = 0;


    while (fp >> words) {
        int newWord = 0;
        for (int i; i < words.length(); i++) {
            if (isalpha(words[i])) {

            } else {
                wordArr[wordCount++] = words.substr(0, i);
                //wordCount++;
                newWord = 1;
                if(words[i] + 1 != '\0') {
                    for (int j = i + 1; j <  words.length(); j++) {
                        temp = temp +words[j];
                    }
                    wordArr[wordCount++] = temp;
                    //wordCount++;
                }

            }

        }
        if (newWord == 0) {
            wordArr[wordCount] = words;
            wordCount++;
        }
    }
    cout << "Number of words found was: " << wordCount << endl;
    //New portion


    // makes all lower
    for(int k=0; k<wordCount;k++){ //need to find size of array
        for(int l=0; l<wordArr[k].length(); l++){
            tolower(wordArr[k].at(l));
        }

    }



    //unique count
    string tempArr[10000];
    int unique=0;
    int oldWord=0;
    for(int m=0; m<wordCount;m++ ) {
        for (int n = 0; n < wordCount; n++) {
            if (wordArr[m] == tempArr[n]) {


                oldWord = 1;
            }
        }
        if(oldWord==0){
            wordArr[m] = tempArr[n];
            unique++;
        }
    }
    cout << "Unique word count is: " << unique << endl;
}

我希望从我的测试用例中获得52个独特的单词，但最终只得到37个。

测试用例：

密码学是用于私下和安全地通信和/或存储信息或数据的技术的实践和研究，而不是被第三方拦截。这可以包括加密，散列和隐写等过程。直到现代，密码学几乎全部称为加密，但现在密码学是一个广泛的领域，应用于我们生活的许多关键领域。

Answer 1

您的解析代码具有错误的逻辑（事实上，它甚至不编译）。如何分解非字母字符上的单词，查找和跟踪重复单词的方式，甚至是如何使用较低级别的单词，都存在逻辑错误。

简而言之，这整个代码充满了需要修复的错误，例如：

#include <iostream>
#include <fstream>
#include <string>
#include <ctype.h>
using namespace std;

int main(int argc, char *argv[]) {
    //File Paths
    ifstream fp;
    fp.open(argv[1]);
    if (!fp.is_open()) {
        cout << "Error No file" << endl;
        return 0;
    }

    string wordArr[10000];
    string words;
    int wordCount = 0;
    while ((fp >> words) && (wordCount < 10000)) {
        for (int i = 0; i < words.length(); ++i) {
            if (!isalpha(words[i])) {
                wordArr[wordCount++] = words.substr(0, i);
                if (wordCount == 10000) break;
                ++i;
                while ((i < words.length()) && (!isalpha(words[i]))) {
                    ++i;
                }
                words.erase(0, i);
                i = -1;
            }
        }
        if (words.length() > 0) {
            wordArr[wordCount++] = words;
        }
    }
    cout << "Number of words found was: " << wordCount << endl;

    // makes all lower
    for(int k=0; k<wordCount;k++){ //need to find size of array
        for(int l=0; l<wordArr[k].length(); l++){
            wordArr[k][l] = tolower(wordArr[k][l]);
        }
    }

    //unique count
    string tempArr[10000];
    int unique=0;
    for(int m=0; m<wordCount;m++ ) {
        int oldWord=0;
        for (int n = 0; n < unique; n++) {
            if (wordArr[m] == tempArr[n]) {
                oldWord = 1;
                break;
            }
        }
        if(oldWord==0){
            tempArr[unique++] = wordArr[m];
        }
    }
    cout << "Unique word count is: " << unique << endl;
}

Now the code works as expected：

Number of words found was: 64
Unique word count is: 52

Answer 2

您需要在每次迭代中重置oldWord：

 //unique count
string tempArr[10000];
int unique=0;
int oldWord=0;
for(int m=0; m<wordCount;m++ ) {
    for (int n = 0; n < wordCount; n++) {
        if (wordArr[m] == tempArr[n]) {


            oldWord = 1;
        }
    }
    if(oldWord==0){
        wordArr[m] = tempArr[n];
        unique++;
    }
    // reset the oldWord variable here
    oldWord=0;

}
cout << "Unique word count is: " << unique << endl;

}