如何按第一个字符串对成对的字符串向量进行分组? [英] How to group a vector of pairs of strings by first string?

问题描述

我有一个包含字符串对的向量:

I have a vector containing pairs of strings:

vector<pair<string, string>> list;

我想对具有相同 list[n].first

const size_t nbElements = list.size();
for (size_t n = 0; n < nbElements ; n++)
{
    const string& name = list[n].first;
    const string& type = list[n].second;
}

<小时>

考虑这个例子:

---

Consider this example:

(big; table) (normal; chair) (small; computer) (big; door) (small; mouse)

会导致:

(big; table, door) (normal; chair) (small; computer, mouse)

<小时>

你知道怎么做吗?

---

Do you have any idea how to do that?

推荐答案

您可以使用 std::map

示例:

#include <boost/algorithm/string/join.hpp>
#include <boost/format.hpp>

#include <iostream>
#include <map>
#include <vector>

int main() {
    // define original data
    std::vector<std::pair<std::string, std::string> > v = 
            {{"a", "b"}, {"a", "c"}, {"b", "a"}, {"b", "d"}, {"c", "e"}};

    // populate map
    std::map<std::string, std::vector<std::string> > grouped;
    for (auto it = v.begin(); it != v.end(); ++it) {
        grouped[(*it).first].push_back((*it).second);
    }

    // output        
    for (auto it = grouped.begin(); it != grouped.end(); ++it) {
        std::cout << boost::format("(%s: %s)\n")
                % (*it).first 
                % boost::algorithm::join((*it).second, ", ");
    }
}

输出为:

(a: b, c)
(b: a, d)
(c: e)

<小时>

注意，这段代码使用了 C++11 特性(初始化列表、auto 关键字).查看上面的链接示例以了解成功编译.

---

Note, this code makes use of C++11 features (initializer lists, auto keyword). Have a look at the linked example above for the successful compilation.

为了自己编译，请确保您使用的编译器支持这些功能或将它们替换为适当的 C++03 等效项.

In order to compile this yourself, make sure that the compiler you use supports these features or replace them with the appropriate C++03 equivalents.

例如，这里是迭代器类型(在上面的代码中使用 auto 关键字进行了美化):

For example, here are the iterator types (that are beautified using the auto keyword in above code):

// the iterator on the vector `v`
std::vector<std::pair<std::string, std::string> >::iterator it_v;

// the iterator on the map `grouped`
std::map<std::string, std::vector<std::string> >::iterator it_grouped;

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