Question

我在Java中有与此类似的代码，但是让它在C ++中运行却是另一回事。目前，此功能确实可以正确找到密钥，并将该密钥推到向量的背面。我想同时获取键及其值对，将其全部插入向量的相同索引中。我已经研究过将一对用于向量，所以它会是（键，一对值），但我不确定我能做得到。

vector<string> getIfPresentPartOfSpeech(multimap<string, pair<string, string>> dictionary, string name) {
    vector<string> response;
    auto it = dictionary.begin();
    while (it != dictionary.end()) {
        // Check if value of this entry matches with given value
        if (it->second.first == name) {
            // Yes found
            // Push the key in given map
            response.push_back(it->first);
        }
        // Go to next entry in map
        it++;
    }
    return response;
}

Answer 1

为您的response定义一个结构，并返回该类型的std::vector。

struct Response {
  std::string key;
  std::pair<std::string, std::string> resp_pair;
};

vector<Response> getIfPresentPartOfSpeech(multimap<string, pair<string, string>> dictionary, string name) {
    vector<Response> response;
    auto it = dictionary.begin();
    while (it != dictionary.end()) {
        // Check if value of this entry matches with given value
        if (it->second.first == name) {
            // Yes found
            // Create a Response object
            Response resp;
            resp.key = it->first;
            resp.resp_pair = it->second;
            // Push it in the vector
            response.push_back(resp);
        }
        // Go to next entry in map
        it++;
    }
    return response;
}

还可以通过使用emplace_back而不是push_back来进行优化，以避免容器元素的多个副本通过。

Answer 2

您可以在std::copy_if中使用<algorithm>，而不必手动编写显式循环。还要通过（常量）引用传递dictionary以避免不必要的复制：

#include <algorithm>
#include <map>
#include <utility>
#include <vector>

// You'll need C++14 to use `auto const& kv` in your lambda expression
std::vector<std::pair<const std::string, std::pair<std::string, std::string>>>
getIfPresentPartOfSpeech(
    const std::multimap<std::string, std::pair<std::string, std::string>>&
        dictionary,
    const std::string& name)
{
    std::vector<
        std::pair<const std::string, std::pair<std::string, std::string>>>
        res;
    std::copy_if(begin(dictionary), end(dictionary), std::back_inserter(res),
                 [&](auto const& kv) { return kv.second.first == name; });
    return res;
}

//test
int main()
{
    std::multimap<std::string, std::pair<std::string, std::string>> dictionary{
        {"first", {"abc", "def"}},
        {"second", {"123", "456"}},
        {"first", {"asdf", "fdas"}},
        {"thrid", {"abc", "123"}},
    };
    auto res = getIfPresentPartOfSpeech(dictionary, "abc");
    for (auto& x : res)
        std::cout << x.first << " " << x.second.first << " " << x.second.second
                  << "\n";
}

使用类似@aep的结构建议会更清晰。您需要一个额外的转换构造函数才能与std::copy_if一起使用：

struct Response {
    std::string key;
    std::pair<std::string, std::string> value;
    // Provide a conversion constructor that convert your `dictionary` multimap
    // key-value pair into a Response object
    Response(const std::pair<const std::string,
                             std::pair<std::string, std::string>>& kv)
    : key{kv.first}, value{kv.second}
    {}
};

std::vector<Response> getIfPresentPartOfSpeech(
    const std::multimap<std::string, std::pair<std::string, std::string>>&
        dictionary,
    const std::string& name)
{
    std::vector<Response> res;
    std::copy_if(begin(dictionary), end(dictionary), std::back_inserter(res),
                 [&](auto const& kv) { return kv.second.first == name; });
    return res;
}

寻找一种将映射的键和值从地图推入向量的方法

2 个答案: