使用C ++映射对单词频率进行计数.我究竟做错了什么? [英] Using a C++ map to count word frequency. What am i doing wrong?

问题描述

那么一点背景.我应该编写一个程序，逐段浏览段落，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行的内容存储在向量中的代码，现在我需要将频率分配给这些单词.我正在使用地图，但这并没有给我想要的结果.我是C ++数据结构的新手，所以请教一些建议.

So a little background. I am supposed to make a program that goes through a paragraph, line by line and picks out the most frequently used word on each line. I have effectively written code that stores the contents of a line in a vector, now i need to assign the frequncey to these words. I am using a map, but it isnt giving me the desired outcome. I am new to C++ data structures, so some advice would be appreciated.

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:

-与您不同的是，我对gnu表示不满.不介意我随身带食物吗?乔治·奥威尔(George Orwell)大约在1984年错了.您的猫码可以吗?如果没有，可以输入狗码吗?我喜欢玩辐射游戏.建议您也玩.足球是美国以外最受欢迎的运动.

-Unlike you I take umbrage with gnu. Don’t mind me taking the food with me? George Orwell was wrong about 1984. Can your cat code? If not, can your dog code? I like to play the fallout games. Recommend you play it as well. Soccer is the most popular sport outside America.

会回来的:与我不同，乔治可以踢足球"

Would return: "unlike me george can play soccer"

    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);
    }

}

推荐答案

我首先将句子拆分为单词，然后将其存储到map中.插入地图最简单，最明显的方法是使用 map [key] = value 表示法.我先使用 count()函数检查该值是否已经存在并插入该值，否则我增加了已经存在的键的值.

I first split the sentence into words and then store them into map. The easiest and more obvious way for map insertion is to use map[key] = value notation. I checked first whether the value is already present by using count() function and insert the value, otherwise I incremented the already present key's value.

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 = "";
    }
}

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