Is it possible to unzip previously zipped vectors using the C++ Range-v3 library? I would expect it to behave similarly to Haskell's unzip function or Python's zip(*list).
It would be convenient, for instance, when sorting a vector by values of another vector:
using namespace ranges;
std::vector<std::string> names {"john", "bob", "alice"};
std::vector<int> ages {32, 19, 35};
// zip names and ages
auto zipped = view::zip(names, ages);
// sort the zip by age
sort(zipped, [](auto &&a, auto &&b) {
return std::get<1>(a) < std::get<1>(b);
});
// put the sorted names back into the original vector
std::tie(names, std::ignore) = unzip(zipped);
When passed container arguments, view::zip in range-v3 creates a view consisting of tuples of references to the original elements. Passing the zipped view to sort sorts the elements in place. I.e., this program:
#include <vector>
#include <string>
#include <iostream>
#include <range/v3/algorithm.hpp>
#include <range/v3/view.hpp>
using namespace ranges;
template <std::size_t N>
struct get_n {
template <typename T>
auto operator()(T&& t) const ->
decltype(std::get<N>(std::forward<T>(t))) {
return std::get<N>(std::forward<T>(t));
}
};
namespace ranges {
template <class T, class U>
std::ostream& operator << (std::ostream& os, common_pair<T, U> const& p) {
return os << '(' << p.first << ", " << p.second << ')';
}
}
int main() {
std::vector<std::string> names {"john", "bob", "alice"};
std::vector<int> ages {32, 19, 35};
auto zipped = view::zip(names, ages);
std::cout << "Before: Names: " << view::all(names) << '\n'
<< " Ages: " << view::all(ages) << '\n'
<< " Zipped: " << zipped << '\n';
sort(zipped, less{}, get_n<1>{});
std::cout << " After: Names: " << view::all(names) << '\n'
<< " Ages: " << view::all(ages) << '\n'
<< " Zipped: " << zipped << '\n';
}
Outputs:
Before: Names: [john,bob,alice]
Ages: [32,19,35]
Zipped: [(john, 32),(bob, 19),(alice, 35)]
After: Names: [bob,john,alice]
Ages: [19,32,35]
Zipped: [(bob, 19),(john, 32),(alice, 35)]