【问题标题】:How to create table of grouped multimap key and associated values in c++如何在 C++ 中创建分组多映射键和关联值的表
【发布时间】:2018-01-23 12:31:53
【问题描述】:

在下面的代码中,在基于 Group1 和 Group2 将元素插入映射后,我使用 equal_range 将多映射键与关联值分组。 根据我的要求,我需要根据以下格式的键值关联在 API 接口中设置这些值。

     {"group1","pointcloud_min_z", "int_t","24"};
     {"group1","pointcloud_max_z", "bool_t","true"};
     {"group2","pointcloud_max_z", "double_t","13"};
     {"group2","pointcloud_min_z", "double_t","20.0"};

现在的问题是如何根据多个分组的关键数据创建表。在这种情况下,列将始终保持不变,但基于多图行中的数据插入/删除将被添加/删除。这里 group1 和 group2 是键。请建议选择容器和将组转换为表格的方法。

代码 sn-p

#include <iostream>
#include <string>
#include <map>
using namespace std;

typedef multimap<string, string> StringToIntMap;
typedef StringToIntMap::iterator mapIter;

int main ()
{
    StringToIntMap mymap;

    mymap.insert(make_pair("Group2", "pointcloud_max_z"));
    mymap.insert(make_pair("Group1",  "pointcloud_min_z"));
    mymap.insert(make_pair("Group2", "double_t"));
    mymap.insert(make_pair("Group1",  "int_t"));
    mymap.insert(make_pair("Group2", "13"));
    mymap.insert(make_pair("Group1",  "pointcloud_min_z"));
    mymap.insert(make_pair("Group2",  "pointcloud_min_z"));
    mymap.insert(make_pair("Group1",  "bool_t"));
    mymap.insert(make_pair("Group1",  "true"));
    mymap.insert(make_pair("Group1",  "pointcloud_min_max_z"));

    cout << "mymap contains:" << endl;

    mapIter m_it, s_it;

    for (m_it = mymap.begin();  m_it != mymap.end();  m_it = s_it)
    {
        string theKey = (*m_it).first;

        cout << endl;
        cout << "  key = '" << theKey << "'" << endl;

        pair<mapIter, mapIter> keyRange = mymap.equal_range(theKey);

        // Iterate over all map elements with key == theKey

        for (s_it = keyRange.first;  s_it != keyRange.second;  ++s_it)
        {
           cout << "    value = " << (*s_it).second << endl;
        }
    }

    return 0;

}   //  end main

enter code here

【问题讨论】:

    标签: c++ boost stl multimap


    【解决方案1】:

    从您的问题看来,您希望以(键、值)格式存储记录。为此,您可以将非键列封装在对象中并将其存储在多映射中,例如 (key, object*)。

    我已将您的代码修改为与下面给出的相同-

    class DATA
    {
    public:
       //control access to data members as needed
       string name;
       string type;
       string value;
    
       DATA(string name, string type, string value): name(name), type(type), value(value) {}   
     };
    
    typedef multimap<string, DATA*> StringToIntMap;
    typedef StringToIntMap::iterator mapIter;
    
    int main ()
    {
        StringToIntMap mymap;
    
        //insert record {"group1","pointcloud_min_z", "int_t","24"}
        DATA *elem = new DATA("pointcloud_min_z", "int_t","24");
        mymap.insert(make_pair("Group1", elem));
        //insert record {"group1","pointcloud_max_z", "bool_t","true"}
        elem = new DATA("pointcloud_max_z", "bool_t","true");
        mymap.insert(make_pair("Group1", elem));
        //insert record {"group2","pointcloud_max_z", "double_t","13"}
        elem = new DATA("pointcloud_max_z", "double_t","13");
        mymap.insert(make_pair("Group2", elem));
        //insert record {"group2","pointcloud_min_z", "double_t","20.0"}
        elem = new DATA("pointcloud_min_z", "double_t","20.0");
        mymap.insert(make_pair("Group2", elem));
    
        cout << "mymap contains:" << endl;
    
        mapIter m_it, s_it;
    
        for (m_it = mymap.begin();  m_it != mymap.end();  m_it = s_it)
        {
            string theKey = (*m_it).first;
    
            cout << endl;
            cout << "  key = '" << theKey << "'" << endl;
    
            pair<mapIter, mapIter> keyRange = mymap.equal_range(theKey);
    
            // Iterate over all map elements with key == theKey
    
            for (s_it = keyRange.first;  s_it != keyRange.second;  ++s_it)
            {
               cout << "    value = " << (*s_it).second->name << ", " 
                    << (*s_it).second->type << ", " 
                    << (*s_it).second->value << endl;
            }
        }
    
        //when done delete the record pointers from the multimap
        return 0;
    
    }   //  end main 
    

    【讨论】:

    • 感谢 .+1 的回答。
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