c++ - 使用 C++ 映射来计算词频。我究竟做错了什么？

Question

所以有点背景。我应该制作一个程序，逐行检查一个段落，并在每一行中挑选出最常用的单词。我已经有效地编写了将一行内容存储在 vector 中的代码，现在我需要将频率分配给这些词。我正在使用 map ，但它没有给我想要的结果。我是 C++ 数据结构的新手，所以我将不胜感激。

map<string,int> freq;   //map for getting frequency on line

    for(int i=0;i<currLine.size();i++){   //loops through the words on the line


        string currName=currLine.at(i);  //current word
        //cout<<currName<<endl;
        if(freq.find(currName)!=freq.end()){   //if already present

                int update=freq.at(currName)+1;  //this value takes the value and adds one
                freq.insert(pair<string,int>(currName,update));  //insert back into the string with updated value
                //cout<<freq.at(currName);
        }
        else{
            freq.insert(pair<string,int>(currName,0)); //if not present, then insert with value 0
        }
    }

例如:

- 与您不同，我对 gnu 感到不满。不介意我把食物带走吗？乔治·奥威尔 (George Orwell) 对 1984 年的看法是错误的。你的猫会编码吗？如果不会，你的狗会编码吗？我喜欢玩辐射游戏。推荐你也玩。足球是美国以外最受欢迎的运动。

会返回:“不像我，乔治会踢足球”

    int main()
{
    string line;
    vector<string> result;
    while(getline(cin,line)){     //on each line
        vector<string> currLine;  //stores words on line
        string curr="";  //temp string for grasping words
        for(int i=0;i<line.length();i++){  //loops through line
            if(isalpha(line.at(i))){  //if it is a letter, add it to the temp string
                curr+=tolower(line.at(i));
            }
            else if(line.at(i)=='\''){
                    //do nothing
            }
            else{    //if not, then add the previous word to vector and reset temp string
                currLine.push_back(curr);
                curr="";
            }
        }



        vector< pair<string, int> > freq(10);
        //freq.push_back(make_pair("asd",2));
        for(int i=0;i<currLine.size();i++){
            string curr1=currLine.at(i);
            cout<<curr1<<" ";
            for(int j=0;j<freq.size();j++){
                if(freq.at(j).first==curr1){  //if present in the list
                    //cout<<"Duplicate found";
                    freq.at(j).second++;     //increment second value
                }
                else{
                    //cout<<"Pair Made";
                    freq.push_back(make_pair(curr1,1));
                    break;
                }
            }
        }

        int max=0;
        string currMost;
        for(int i =0;i<freq.size();i++){

            if(freq.at(i).second>max){
                max=freq.at(i).second;;
                currMost=freq.at(i).first;
            }
        }
        //cout<<currMost;

        cout<<endl;



       // result.push_back(currMost);

    }

    for(int i=0;i<result.size();i++){
        cout<<result.at(i);
    }

}

Answer 1

我先把句子拆分成单词，然后存入map。插入 map 的最简单和更明显的方法是使用 map[key] = value 表示法。我首先使用 count() 函数检查该值是否已经存在并插入该值，否则我会增加已经存在的键的值。

map <string, int> word_frequency;
string word="";
for (int i=0; i<=current_line.length();i++) {
    if (current_line[i]!=' ' && i< current_line.length()) {
        word = word + current_line[i];
    } else if(word_frequency.count(word) == 0) {
       word_frequency[word] = 1;
       word = "";
    } else {
        word_frequency[word]++;
        word = "";
    }
}