Question

所以有点背景。我应该制作一个程序，通过一个段落，逐行，并挑选出每行最常用的单词。我已经有效地编写了一个代码，用于在一个向量中存储一行的内容，现在我需要将frequncey分配给这些单词。我正在使用地图，但它并没有给我预期的结果。我是C ++数据结构的新手，所以一些建议将不胜感激。

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：

- 和你不一样，我对gnu感到不快。别介意我和我一起吃饭？乔治奥威尔在1984年错了。你的猫代码可以吗？如果没有，你的狗代码可以吗？我喜欢玩后果游戏。建议你也玩它。足球是美国以外最受欢迎的运动。

会回归：“不像我，乔治可以踢足球”

    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

我首先将句子分成单词，然后将它们存储到地图中。地图插入最简单，最明显的方法是使用map[key] = value表示法。我首先使用count()函数检查该值是否已存在并插入该值，否则我将增加已存在的键的值。

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] = 0;
       word = "";
   }
else{
    word_frequency[word]++;
    word = "";
}

}

使用C ++映射计算字频率。我做错了什么？

1 个答案: