Delete with multiple indices is extremely slow--workaround?











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Delete is unbelievably slow when deleting multiple elements from a non-packed array.



Is there a robust workaround that will work on any non-packed array?



inds = List /@ RandomSample[Range[100000], 50000];
Delete[Developer`FromPackedArray@Range[100000], inds]; // AbsoluteTiming
(* {17.8957, Null} *)


On packed arrays it performs as expected, but my array cannot be packed. It does not necessarily contain numbers.



inds = List /@ RandomSample[Range[100000], 50000];
Delete[Range[100000], inds]; // AbsoluteTiming
(* {0.005767, Null} *)




I did not try to test for this, but one possible explanation is that even when given multiple indices, Delete will delete elements one-by-one, re-allocating the array after each step. If someone feels like testing it, you can try to see if the timing is quadratic in the number of elements deleted.










share|improve this question




























    up vote
    13
    down vote

    favorite
    2












    Delete is unbelievably slow when deleting multiple elements from a non-packed array.



    Is there a robust workaround that will work on any non-packed array?



    inds = List /@ RandomSample[Range[100000], 50000];
    Delete[Developer`FromPackedArray@Range[100000], inds]; // AbsoluteTiming
    (* {17.8957, Null} *)


    On packed arrays it performs as expected, but my array cannot be packed. It does not necessarily contain numbers.



    inds = List /@ RandomSample[Range[100000], 50000];
    Delete[Range[100000], inds]; // AbsoluteTiming
    (* {0.005767, Null} *)




    I did not try to test for this, but one possible explanation is that even when given multiple indices, Delete will delete elements one-by-one, re-allocating the array after each step. If someone feels like testing it, you can try to see if the timing is quadratic in the number of elements deleted.










    share|improve this question


























      up vote
      13
      down vote

      favorite
      2









      up vote
      13
      down vote

      favorite
      2






      2





      Delete is unbelievably slow when deleting multiple elements from a non-packed array.



      Is there a robust workaround that will work on any non-packed array?



      inds = List /@ RandomSample[Range[100000], 50000];
      Delete[Developer`FromPackedArray@Range[100000], inds]; // AbsoluteTiming
      (* {17.8957, Null} *)


      On packed arrays it performs as expected, but my array cannot be packed. It does not necessarily contain numbers.



      inds = List /@ RandomSample[Range[100000], 50000];
      Delete[Range[100000], inds]; // AbsoluteTiming
      (* {0.005767, Null} *)




      I did not try to test for this, but one possible explanation is that even when given multiple indices, Delete will delete elements one-by-one, re-allocating the array after each step. If someone feels like testing it, you can try to see if the timing is quadratic in the number of elements deleted.










      share|improve this question















      Delete is unbelievably slow when deleting multiple elements from a non-packed array.



      Is there a robust workaround that will work on any non-packed array?



      inds = List /@ RandomSample[Range[100000], 50000];
      Delete[Developer`FromPackedArray@Range[100000], inds]; // AbsoluteTiming
      (* {17.8957, Null} *)


      On packed arrays it performs as expected, but my array cannot be packed. It does not necessarily contain numbers.



      inds = List /@ RandomSample[Range[100000], 50000];
      Delete[Range[100000], inds]; // AbsoluteTiming
      (* {0.005767, Null} *)




      I did not try to test for this, but one possible explanation is that even when given multiple indices, Delete will delete elements one-by-one, re-allocating the array after each step. If someone feels like testing it, you can try to see if the timing is quadratic in the number of elements deleted.







      list-manipulation performance-tuning






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      edited Dec 4 at 6:56









      xzczd

      25.7k469245




      25.7k469245










      asked Dec 3 at 9:40









      Szabolcs

      158k13432926




      158k13432926






















          2 Answers
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          up vote
          14
          down vote













          Exploiting the fact that Delete works fine on packed arrays, we can first construct an index vector, delete the unneeded indices, then finally use the remaining ones to index into the main array.



          arr = Developer`FromPackedArray@Range[100000];
          inds = List /@ RandomSample[Range[100000], 50000];

          Part[arr, Delete[Range@Length[arr], inds]]; // AbsoluteTiming
          (* {0.006371, Null} *)





          share|improve this answer

















          • 1




            Embarrassing for Delete that this method which ought to have terrible time complexity is so much faster... Have you tested some of the other things like delete on unpacked arrays?
            – b3m2a1
            Dec 3 at 9:52








          • 3




            @b3m2a1 I don't think that this has larger complexity... Still it is pretty bad that Delete is not clever enough to do that automatically. I'd suggest to inform Wolfram Support.
            – Henrik Schumacher
            Dec 3 at 10:00




















          up vote
          7
          down vote













          Why not use Part assignment (to Sequence) instead?



          arr = Developer`FromPackedArray@Range[100000];
          inds = List /@ RandomSample[Range[100000],50000];

          r1 = Part[arr, Delete[Range@Length[arr], inds]]; //RepeatedTiming
          (r2 = arr; r2[[Flatten @ inds]] = Sequence;) //RepeatedTiming

          r1 === r2



          {0.0059, Null}



          {0.0019, Null}



          True







          share|improve this answer





















          • Nothing is slightly faster than Sequence on my laptop.
            – Sjoerd C. de Vries
            Dec 3 at 20:11












          • @SjoerdC.deVries and Carl, to fully reduce the array the procedure needs to be followed by r2;. Both Sequence and Nothing will now produce same timings as Part+Delete. p.s. to see what I mean try this: r = {1, 2}; r[[1]] = Nothing; Information@r
            – Kuba
            Dec 4 at 7:28













          Your Answer





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          2 Answers
          2






          active

          oldest

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          2 Answers
          2






          active

          oldest

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          active

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          up vote
          14
          down vote













          Exploiting the fact that Delete works fine on packed arrays, we can first construct an index vector, delete the unneeded indices, then finally use the remaining ones to index into the main array.



          arr = Developer`FromPackedArray@Range[100000];
          inds = List /@ RandomSample[Range[100000], 50000];

          Part[arr, Delete[Range@Length[arr], inds]]; // AbsoluteTiming
          (* {0.006371, Null} *)





          share|improve this answer

















          • 1




            Embarrassing for Delete that this method which ought to have terrible time complexity is so much faster... Have you tested some of the other things like delete on unpacked arrays?
            – b3m2a1
            Dec 3 at 9:52








          • 3




            @b3m2a1 I don't think that this has larger complexity... Still it is pretty bad that Delete is not clever enough to do that automatically. I'd suggest to inform Wolfram Support.
            – Henrik Schumacher
            Dec 3 at 10:00

















          up vote
          14
          down vote













          Exploiting the fact that Delete works fine on packed arrays, we can first construct an index vector, delete the unneeded indices, then finally use the remaining ones to index into the main array.



          arr = Developer`FromPackedArray@Range[100000];
          inds = List /@ RandomSample[Range[100000], 50000];

          Part[arr, Delete[Range@Length[arr], inds]]; // AbsoluteTiming
          (* {0.006371, Null} *)





          share|improve this answer

















          • 1




            Embarrassing for Delete that this method which ought to have terrible time complexity is so much faster... Have you tested some of the other things like delete on unpacked arrays?
            – b3m2a1
            Dec 3 at 9:52








          • 3




            @b3m2a1 I don't think that this has larger complexity... Still it is pretty bad that Delete is not clever enough to do that automatically. I'd suggest to inform Wolfram Support.
            – Henrik Schumacher
            Dec 3 at 10:00















          up vote
          14
          down vote










          up vote
          14
          down vote









          Exploiting the fact that Delete works fine on packed arrays, we can first construct an index vector, delete the unneeded indices, then finally use the remaining ones to index into the main array.



          arr = Developer`FromPackedArray@Range[100000];
          inds = List /@ RandomSample[Range[100000], 50000];

          Part[arr, Delete[Range@Length[arr], inds]]; // AbsoluteTiming
          (* {0.006371, Null} *)





          share|improve this answer












          Exploiting the fact that Delete works fine on packed arrays, we can first construct an index vector, delete the unneeded indices, then finally use the remaining ones to index into the main array.



          arr = Developer`FromPackedArray@Range[100000];
          inds = List /@ RandomSample[Range[100000], 50000];

          Part[arr, Delete[Range@Length[arr], inds]]; // AbsoluteTiming
          (* {0.006371, Null} *)






          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Dec 3 at 9:42









          Szabolcs

          158k13432926




          158k13432926








          • 1




            Embarrassing for Delete that this method which ought to have terrible time complexity is so much faster... Have you tested some of the other things like delete on unpacked arrays?
            – b3m2a1
            Dec 3 at 9:52








          • 3




            @b3m2a1 I don't think that this has larger complexity... Still it is pretty bad that Delete is not clever enough to do that automatically. I'd suggest to inform Wolfram Support.
            – Henrik Schumacher
            Dec 3 at 10:00
















          • 1




            Embarrassing for Delete that this method which ought to have terrible time complexity is so much faster... Have you tested some of the other things like delete on unpacked arrays?
            – b3m2a1
            Dec 3 at 9:52








          • 3




            @b3m2a1 I don't think that this has larger complexity... Still it is pretty bad that Delete is not clever enough to do that automatically. I'd suggest to inform Wolfram Support.
            – Henrik Schumacher
            Dec 3 at 10:00










          1




          1




          Embarrassing for Delete that this method which ought to have terrible time complexity is so much faster... Have you tested some of the other things like delete on unpacked arrays?
          – b3m2a1
          Dec 3 at 9:52






          Embarrassing for Delete that this method which ought to have terrible time complexity is so much faster... Have you tested some of the other things like delete on unpacked arrays?
          – b3m2a1
          Dec 3 at 9:52






          3




          3




          @b3m2a1 I don't think that this has larger complexity... Still it is pretty bad that Delete is not clever enough to do that automatically. I'd suggest to inform Wolfram Support.
          – Henrik Schumacher
          Dec 3 at 10:00






          @b3m2a1 I don't think that this has larger complexity... Still it is pretty bad that Delete is not clever enough to do that automatically. I'd suggest to inform Wolfram Support.
          – Henrik Schumacher
          Dec 3 at 10:00












          up vote
          7
          down vote













          Why not use Part assignment (to Sequence) instead?



          arr = Developer`FromPackedArray@Range[100000];
          inds = List /@ RandomSample[Range[100000],50000];

          r1 = Part[arr, Delete[Range@Length[arr], inds]]; //RepeatedTiming
          (r2 = arr; r2[[Flatten @ inds]] = Sequence;) //RepeatedTiming

          r1 === r2



          {0.0059, Null}



          {0.0019, Null}



          True







          share|improve this answer





















          • Nothing is slightly faster than Sequence on my laptop.
            – Sjoerd C. de Vries
            Dec 3 at 20:11












          • @SjoerdC.deVries and Carl, to fully reduce the array the procedure needs to be followed by r2;. Both Sequence and Nothing will now produce same timings as Part+Delete. p.s. to see what I mean try this: r = {1, 2}; r[[1]] = Nothing; Information@r
            – Kuba
            Dec 4 at 7:28

















          up vote
          7
          down vote













          Why not use Part assignment (to Sequence) instead?



          arr = Developer`FromPackedArray@Range[100000];
          inds = List /@ RandomSample[Range[100000],50000];

          r1 = Part[arr, Delete[Range@Length[arr], inds]]; //RepeatedTiming
          (r2 = arr; r2[[Flatten @ inds]] = Sequence;) //RepeatedTiming

          r1 === r2



          {0.0059, Null}



          {0.0019, Null}



          True







          share|improve this answer





















          • Nothing is slightly faster than Sequence on my laptop.
            – Sjoerd C. de Vries
            Dec 3 at 20:11












          • @SjoerdC.deVries and Carl, to fully reduce the array the procedure needs to be followed by r2;. Both Sequence and Nothing will now produce same timings as Part+Delete. p.s. to see what I mean try this: r = {1, 2}; r[[1]] = Nothing; Information@r
            – Kuba
            Dec 4 at 7:28















          up vote
          7
          down vote










          up vote
          7
          down vote









          Why not use Part assignment (to Sequence) instead?



          arr = Developer`FromPackedArray@Range[100000];
          inds = List /@ RandomSample[Range[100000],50000];

          r1 = Part[arr, Delete[Range@Length[arr], inds]]; //RepeatedTiming
          (r2 = arr; r2[[Flatten @ inds]] = Sequence;) //RepeatedTiming

          r1 === r2



          {0.0059, Null}



          {0.0019, Null}



          True







          share|improve this answer












          Why not use Part assignment (to Sequence) instead?



          arr = Developer`FromPackedArray@Range[100000];
          inds = List /@ RandomSample[Range[100000],50000];

          r1 = Part[arr, Delete[Range@Length[arr], inds]]; //RepeatedTiming
          (r2 = arr; r2[[Flatten @ inds]] = Sequence;) //RepeatedTiming

          r1 === r2



          {0.0059, Null}



          {0.0019, Null}



          True








          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Dec 3 at 19:23









          Carl Woll

          66.7k385174




          66.7k385174












          • Nothing is slightly faster than Sequence on my laptop.
            – Sjoerd C. de Vries
            Dec 3 at 20:11












          • @SjoerdC.deVries and Carl, to fully reduce the array the procedure needs to be followed by r2;. Both Sequence and Nothing will now produce same timings as Part+Delete. p.s. to see what I mean try this: r = {1, 2}; r[[1]] = Nothing; Information@r
            – Kuba
            Dec 4 at 7:28




















          • Nothing is slightly faster than Sequence on my laptop.
            – Sjoerd C. de Vries
            Dec 3 at 20:11












          • @SjoerdC.deVries and Carl, to fully reduce the array the procedure needs to be followed by r2;. Both Sequence and Nothing will now produce same timings as Part+Delete. p.s. to see what I mean try this: r = {1, 2}; r[[1]] = Nothing; Information@r
            – Kuba
            Dec 4 at 7:28


















          Nothing is slightly faster than Sequence on my laptop.
          – Sjoerd C. de Vries
          Dec 3 at 20:11






          Nothing is slightly faster than Sequence on my laptop.
          – Sjoerd C. de Vries
          Dec 3 at 20:11














          @SjoerdC.deVries and Carl, to fully reduce the array the procedure needs to be followed by r2;. Both Sequence and Nothing will now produce same timings as Part+Delete. p.s. to see what I mean try this: r = {1, 2}; r[[1]] = Nothing; Information@r
          – Kuba
          Dec 4 at 7:28






          @SjoerdC.deVries and Carl, to fully reduce the array the procedure needs to be followed by r2;. Both Sequence and Nothing will now produce same timings as Part+Delete. p.s. to see what I mean try this: r = {1, 2}; r[[1]] = Nothing; Information@r
          – Kuba
          Dec 4 at 7:28




















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