如何复制向量中的字符串<pair<string, int> &gt; 到矢量<string> ? </string></pair<string,>

Question

I'm using GCC 9.2.0 and boost 1.55.

I have 2 vectors:

vector< pair< string, int > > source;
vector< string > dest;

I need to transform the source vector to the dest , such that it should contain only string elements of source vector.

Is it possible using boost::push_back and adaptors?

boost::range::push_back( dest, source | /* adaptor ??? */ );

Currently I have this workable code, but it should be changed:

transform( source.begin(), source.end(), back_inserter(dest), __gnu_cxx::select1st< std::pair< std::string, int > >() );

Answer 1

To continue the trend of posting mimimal improvements:

Live On Compiler Explorer

#include <boost/range/adaptor/map.hpp>
#include <fmt/ranges.h>

using namespace std::string_literals;
using namespace boost::adaptors;

template <typename R>
auto to_vector(R const& range) {
    return std::vector(boost::begin(range), boost::end(range));
}

int main() {
    std::vector source {std::pair {"one"s,1},{"two",2},{"three",3}};

    fmt::print("keys {}\nvalues {}\n",
            source | map_keys,
            source | map_values);
    
    // if you really need the vectors
    auto keys   = to_vector(source | map_keys);
    auto values = to_vector(source | map_values);
}

Prints

keys {"one", "two", "three"}
values {1, 2, 3}

Answer 2

You could use std::transform and std::back_inserter with a lambda instead:

#include <algorithm> // transform
#include <iterator>  // back_inserter

std::transform(source.begin(), source.end(), std::back_inserter(dest),
               [](auto& p) {
                   return p.first;
               });

Answer 3

Here's a boost solution with the transformed adaptor:

#include <boost/range/adaptor/transformed.hpp>
#include <iostream>
#include <string>
#include <utility>
#include <vector>

using boost::adaptors::transformed;

// define function object (so we can pass it around) emulating std::get
template<std::size_t N>
auto constexpr get = [](auto const& x){ return std::get<N>(x); };

int main()
{
    // prepare sample input
    std::vector<std::pair<std::string,int>> source{{"one",1},{"two",2},{"three",3}};

    // the line you look for
    auto dest = source | transformed(get<0>);

    // dest is a vector on which we can iterate
    for (auto const& i : dest) {
        std::cout << i << '\n';
    }

    // if you really need it to be a vector
    auto dest_vec = boost::copy_range<std::vector<std::string>>(dest);
}

Answer 4

Here's a solution using the range-v3 library:

auto dest = source 
          | ranges::views::keys 
          | ranges::to<std::vector<std::string>>;

Here's a demo .

---

In C++20, there's no std::ranges::to , but you can still copy over the keys to dest if it already exists:

std::ranges::copy(source | std::views::keys, 
                  std::back_inserter(dest));

Here's a demo .

Note that this doesn't work in Clang trunk, which I assume is a bug.