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Does the STL have a way to apply a function before calling less than?



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10















A lot of algorithms accept a comparison object. Often, I end up with something like



std::sort(begin, end, [&](auto const& lhs, auto const& rhs) 
return Function(lhs) < Function(rhs);
);


Is there anything in the STL to apply a Function before calling less than? So I could write:



std::sort(begin, end, std::DoesThisExist(Function));


I know I could write my own, but I wonder if this already exists. I glanced through cpprefence but didn't see it. Could easily have missed it.










share|improve this question







New contributor




Fomar putes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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  • I don't think this exists, however, it should be very easy to write.

    – JVApen
    Apr 14 at 20:03











  • std::bind, see details below

    – Helmut Zeisel
    Apr 16 at 7:24


















10















A lot of algorithms accept a comparison object. Often, I end up with something like



std::sort(begin, end, [&](auto const& lhs, auto const& rhs) 
return Function(lhs) < Function(rhs);
);


Is there anything in the STL to apply a Function before calling less than? So I could write:



std::sort(begin, end, std::DoesThisExist(Function));


I know I could write my own, but I wonder if this already exists. I glanced through cpprefence but didn't see it. Could easily have missed it.










share|improve this question







New contributor




Fomar putes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




















  • I don't think this exists, however, it should be very easy to write.

    – JVApen
    Apr 14 at 20:03











  • std::bind, see details below

    – Helmut Zeisel
    Apr 16 at 7:24














10












10








10








A lot of algorithms accept a comparison object. Often, I end up with something like



std::sort(begin, end, [&](auto const& lhs, auto const& rhs) 
return Function(lhs) < Function(rhs);
);


Is there anything in the STL to apply a Function before calling less than? So I could write:



std::sort(begin, end, std::DoesThisExist(Function));


I know I could write my own, but I wonder if this already exists. I glanced through cpprefence but didn't see it. Could easily have missed it.










share|improve this question







New contributor




Fomar putes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












A lot of algorithms accept a comparison object. Often, I end up with something like



std::sort(begin, end, [&](auto const& lhs, auto const& rhs) 
return Function(lhs) < Function(rhs);
);


Is there anything in the STL to apply a Function before calling less than? So I could write:



std::sort(begin, end, std::DoesThisExist(Function));


I know I could write my own, but I wonder if this already exists. I glanced through cpprefence but didn't see it. Could easily have missed it.







c++






share|improve this question







New contributor




Fomar putes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|improve this question







New contributor




Fomar putes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|improve this question




share|improve this question






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Fomar putes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked Apr 14 at 17:59









Fomar putesFomar putes

534




534




New contributor




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New contributor





Fomar putes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Fomar putes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • I don't think this exists, however, it should be very easy to write.

    – JVApen
    Apr 14 at 20:03











  • std::bind, see details below

    – Helmut Zeisel
    Apr 16 at 7:24


















  • I don't think this exists, however, it should be very easy to write.

    – JVApen
    Apr 14 at 20:03











  • std::bind, see details below

    – Helmut Zeisel
    Apr 16 at 7:24

















I don't think this exists, however, it should be very easy to write.

– JVApen
Apr 14 at 20:03





I don't think this exists, however, it should be very easy to write.

– JVApen
Apr 14 at 20:03













std::bind, see details below

– Helmut Zeisel
Apr 16 at 7:24






std::bind, see details below

– Helmut Zeisel
Apr 16 at 7:24













3 Answers
3






active

oldest

votes


















4














The STL should have a sort that works on a transform of the elements rather than the elements themselves. The reason for this being that Function could actually be costly. By simply incorporating it into the comparison as you did you invoke Function nlog(n) times rather than the optimal n.




To sort arrays in parallel using STL algorithm :




std::sort(std::execution::par, container.begin(), container.end(), comparison_object);


Anyway, I think if you try to just sort with ranges::view::transform it will probably still call your function ~n log n many times. But you could just do something like:



auto values = /* some container */;
auto keys = values | ranges::view::transform(f) | ranges::to_vector;
ranges::sort(ranges::view::zip(keys, values),
[](auto const& x, auto const& y) return std::get<0>(x) < std::get<0>(y); );





share|improve this answer























  • Calling a transform n log n times matters only if its expense is significant compared to the comparison and permutation. In common cases where the “transform” is just selecting a member or performing simple arithmetic, allocating the space for the key values would be much worse than the repeated work.

    – Davis Herring
    Apr 14 at 22:10


















2














The Ranges TS (which has been merged for C++20) defines variations of many of the standard algorithms that include projections with exactly this behavior.






share|improve this answer






























    0














    In principle, you could use std::bind, but it is very verbose:



     typedef std::remove_reference<decltype(*begin)>::type T;
    std::sort(begin, end, std::bind(std::less<T>(),
    std::bind(Function,std::placeholders::_1),
    std::bind(Function,std::placeholders::_2)));





    share|improve this answer























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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      4














      The STL should have a sort that works on a transform of the elements rather than the elements themselves. The reason for this being that Function could actually be costly. By simply incorporating it into the comparison as you did you invoke Function nlog(n) times rather than the optimal n.




      To sort arrays in parallel using STL algorithm :




      std::sort(std::execution::par, container.begin(), container.end(), comparison_object);


      Anyway, I think if you try to just sort with ranges::view::transform it will probably still call your function ~n log n many times. But you could just do something like:



      auto values = /* some container */;
      auto keys = values | ranges::view::transform(f) | ranges::to_vector;
      ranges::sort(ranges::view::zip(keys, values),
      [](auto const& x, auto const& y) return std::get<0>(x) < std::get<0>(y); );





      share|improve this answer























      • Calling a transform n log n times matters only if its expense is significant compared to the comparison and permutation. In common cases where the “transform” is just selecting a member or performing simple arithmetic, allocating the space for the key values would be much worse than the repeated work.

        – Davis Herring
        Apr 14 at 22:10















      4














      The STL should have a sort that works on a transform of the elements rather than the elements themselves. The reason for this being that Function could actually be costly. By simply incorporating it into the comparison as you did you invoke Function nlog(n) times rather than the optimal n.




      To sort arrays in parallel using STL algorithm :




      std::sort(std::execution::par, container.begin(), container.end(), comparison_object);


      Anyway, I think if you try to just sort with ranges::view::transform it will probably still call your function ~n log n many times. But you could just do something like:



      auto values = /* some container */;
      auto keys = values | ranges::view::transform(f) | ranges::to_vector;
      ranges::sort(ranges::view::zip(keys, values),
      [](auto const& x, auto const& y) return std::get<0>(x) < std::get<0>(y); );





      share|improve this answer























      • Calling a transform n log n times matters only if its expense is significant compared to the comparison and permutation. In common cases where the “transform” is just selecting a member or performing simple arithmetic, allocating the space for the key values would be much worse than the repeated work.

        – Davis Herring
        Apr 14 at 22:10













      4












      4








      4







      The STL should have a sort that works on a transform of the elements rather than the elements themselves. The reason for this being that Function could actually be costly. By simply incorporating it into the comparison as you did you invoke Function nlog(n) times rather than the optimal n.




      To sort arrays in parallel using STL algorithm :




      std::sort(std::execution::par, container.begin(), container.end(), comparison_object);


      Anyway, I think if you try to just sort with ranges::view::transform it will probably still call your function ~n log n many times. But you could just do something like:



      auto values = /* some container */;
      auto keys = values | ranges::view::transform(f) | ranges::to_vector;
      ranges::sort(ranges::view::zip(keys, values),
      [](auto const& x, auto const& y) return std::get<0>(x) < std::get<0>(y); );





      share|improve this answer













      The STL should have a sort that works on a transform of the elements rather than the elements themselves. The reason for this being that Function could actually be costly. By simply incorporating it into the comparison as you did you invoke Function nlog(n) times rather than the optimal n.




      To sort arrays in parallel using STL algorithm :




      std::sort(std::execution::par, container.begin(), container.end(), comparison_object);


      Anyway, I think if you try to just sort with ranges::view::transform it will probably still call your function ~n log n many times. But you could just do something like:



      auto values = /* some container */;
      auto keys = values | ranges::view::transform(f) | ranges::to_vector;
      ranges::sort(ranges::view::zip(keys, values),
      [](auto const& x, auto const& y) return std::get<0>(x) < std::get<0>(y); );






      share|improve this answer












      share|improve this answer



      share|improve this answer










      answered Apr 14 at 19:59









      Ben Chaliah AyoubBen Chaliah Ayoub

      2,757319




      2,757319












      • Calling a transform n log n times matters only if its expense is significant compared to the comparison and permutation. In common cases where the “transform” is just selecting a member or performing simple arithmetic, allocating the space for the key values would be much worse than the repeated work.

        – Davis Herring
        Apr 14 at 22:10

















      • Calling a transform n log n times matters only if its expense is significant compared to the comparison and permutation. In common cases where the “transform” is just selecting a member or performing simple arithmetic, allocating the space for the key values would be much worse than the repeated work.

        – Davis Herring
        Apr 14 at 22:10
















      Calling a transform n log n times matters only if its expense is significant compared to the comparison and permutation. In common cases where the “transform” is just selecting a member or performing simple arithmetic, allocating the space for the key values would be much worse than the repeated work.

      – Davis Herring
      Apr 14 at 22:10





      Calling a transform n log n times matters only if its expense is significant compared to the comparison and permutation. In common cases where the “transform” is just selecting a member or performing simple arithmetic, allocating the space for the key values would be much worse than the repeated work.

      – Davis Herring
      Apr 14 at 22:10













      2














      The Ranges TS (which has been merged for C++20) defines variations of many of the standard algorithms that include projections with exactly this behavior.






      share|improve this answer



























        2














        The Ranges TS (which has been merged for C++20) defines variations of many of the standard algorithms that include projections with exactly this behavior.






        share|improve this answer

























          2












          2








          2







          The Ranges TS (which has been merged for C++20) defines variations of many of the standard algorithms that include projections with exactly this behavior.






          share|improve this answer













          The Ranges TS (which has been merged for C++20) defines variations of many of the standard algorithms that include projections with exactly this behavior.







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Apr 14 at 18:11









          Davis HerringDavis Herring

          9,1501736




          9,1501736





















              0














              In principle, you could use std::bind, but it is very verbose:



               typedef std::remove_reference<decltype(*begin)>::type T;
              std::sort(begin, end, std::bind(std::less<T>(),
              std::bind(Function,std::placeholders::_1),
              std::bind(Function,std::placeholders::_2)));





              share|improve this answer



























                0














                In principle, you could use std::bind, but it is very verbose:



                 typedef std::remove_reference<decltype(*begin)>::type T;
                std::sort(begin, end, std::bind(std::less<T>(),
                std::bind(Function,std::placeholders::_1),
                std::bind(Function,std::placeholders::_2)));





                share|improve this answer

























                  0












                  0








                  0







                  In principle, you could use std::bind, but it is very verbose:



                   typedef std::remove_reference<decltype(*begin)>::type T;
                  std::sort(begin, end, std::bind(std::less<T>(),
                  std::bind(Function,std::placeholders::_1),
                  std::bind(Function,std::placeholders::_2)));





                  share|improve this answer













                  In principle, you could use std::bind, but it is very verbose:



                   typedef std::remove_reference<decltype(*begin)>::type T;
                  std::sort(begin, end, std::bind(std::less<T>(),
                  std::bind(Function,std::placeholders::_1),
                  std::bind(Function,std::placeholders::_2)));






                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered Apr 15 at 8:32









                  Helmut ZeiselHelmut Zeisel

                  1387




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