Fastest way to go from linear index to grid index












3














I'm sure this has been asked before but I'm interested in going from a position in a vector to the index in the grid version of the vector with given strides, for example, say I have the vector:



vec = {58, 94, 19, 68, 54, 77, 1, 18, 49, 20, 90, 44, 91, 89, 15, 0, 

   60, 18, 19, 44, 87, 5, 8, 42, 51, 55, 87, 71, 83, 68, 53, 58, 27, 

   17, 8, 14, 33, 58, 86, 3, 91, 66, 3, 16, 98, 84, 72, 98, 9, 30, 90,

    99, 15, 0, 82, 76, 86, 58, 77, 58};


And say I have strides {5, 4, 3}, the position 35 in the vector would correspond to the index {3, 4, 2}:



vec[[35]]



4



ArrayReshape[vec, {5, 4, 3}][[3, 4, 2]]



4


How can I get this index fast and in a vectorized fashion because I will have potentially many positions to extract?






list-manipulation performance-tuning







share|improve this question



























    3














    I'm sure this has been asked before but I'm interested in going from a position in a vector to the index in the grid version of the vector with given strides, for example, say I have the vector:



    vec = {58, 94, 19, 68, 54, 77, 1, 18, 49, 20, 90, 44, 91, 89, 15, 0, 
    
   60, 18, 19, 44, 87, 5, 8, 42, 51, 55, 87, 71, 83, 68, 53, 58, 27, 
    
   17, 8, 14, 33, 58, 86, 3, 91, 66, 3, 16, 98, 84, 72, 98, 9, 30, 90,
    
    99, 15, 0, 82, 76, 86, 58, 77, 58};


    And say I have strides {5, 4, 3}, the position 35 in the vector would correspond to the index {3, 4, 2}:



    vec[[35]]
    

    
4
    

    
ArrayReshape[vec, {5, 4, 3}][[3, 4, 2]]
    

    
4


    How can I get this index fast and in a vectorized fashion because I will have potentially many positions to extract?






    list-manipulation performance-tuning







    share|improve this question

























      3












      3








      3







      I'm sure this has been asked before but I'm interested in going from a position in a vector to the index in the grid version of the vector with given strides, for example, say I have the vector:



      vec = {58, 94, 19, 68, 54, 77, 1, 18, 49, 20, 90, 44, 91, 89, 15, 0, 
      
   60, 18, 19, 44, 87, 5, 8, 42, 51, 55, 87, 71, 83, 68, 53, 58, 27, 
      
   17, 8, 14, 33, 58, 86, 3, 91, 66, 3, 16, 98, 84, 72, 98, 9, 30, 90,
      
    99, 15, 0, 82, 76, 86, 58, 77, 58};


      And say I have strides {5, 4, 3}, the position 35 in the vector would correspond to the index {3, 4, 2}:



      vec[[35]]
      

      
4
      

      
ArrayReshape[vec, {5, 4, 3}][[3, 4, 2]]
      

      
4


      How can I get this index fast and in a vectorized fashion because I will have potentially many positions to extract?






      list-manipulation performance-tuning







      share|improve this question













      I'm sure this has been asked before but I'm interested in going from a position in a vector to the index in the grid version of the vector with given strides, for example, say I have the vector:



      vec = {58, 94, 19, 68, 54, 77, 1, 18, 49, 20, 90, 44, 91, 89, 15, 0, 
      
   60, 18, 19, 44, 87, 5, 8, 42, 51, 55, 87, 71, 83, 68, 53, 58, 27, 
      
   17, 8, 14, 33, 58, 86, 3, 91, 66, 3, 16, 98, 84, 72, 98, 9, 30, 90,
      
    99, 15, 0, 82, 76, 86, 58, 77, 58};


      And say I have strides {5, 4, 3}, the position 35 in the vector would correspond to the index {3, 4, 2}:



      vec[[35]]
      

      
4
      

      
ArrayReshape[vec, {5, 4, 3}][[3, 4, 2]]
      

      
4


      How can I get this index fast and in a vectorized fashion because I will have potentially many positions to extract?






      list-manipulation performance-tuning




      list-manipulation performance-tuning






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked 1 hour ago









      b3m2a1

      26.7k257154




      26.7k257154






















          2 Answers
          2






          active

          oldest

          votes


















          2














          I think this is what you want:



          IntegerDigits[35 - 1, MixedRadix[{5, 4, 3}], 3] + 1


          In general:



          gridIndex[n_Integer, shape_List] := 
          
 IntegerDigits[n - 1, MixedRadix[shape], Length@shape] + 1





          share|improve this answer



















          • 1




            @C.E. You're right, it just needs dimension length specification
            – swish
            45 mins ago






          • 1




            This is a very nice solution, +1
            – C. E.
            43 mins ago



















          1














          Here's what I came up with:



          getSubindex[index_, stride_] := {
          
  Mod[index, stride, 1],
          
  Ceiling[index/stride]
          
  }
          
getIndex[index_, strides_] := 
          
 Reverse@FoldPairList[getSubindex, index, Reverse@strides]


          This is comparable to swish's solution speed-wise:



          gridIndex[1000, {3, 5, 4, 6}] // RepeatedTiming



          {0.000061, {3, 2, 3, 4}}




          getIndex[1000, {3, 5, 4, 6}] // RepeatedTiming



          {0.000052, {3, 2, 3, 4}}







          share|improve this answer























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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            2














            I think this is what you want:



            IntegerDigits[35 - 1, MixedRadix[{5, 4, 3}], 3] + 1


            In general:



            gridIndex[n_Integer, shape_List] := 
            
 IntegerDigits[n - 1, MixedRadix[shape], Length@shape] + 1





            share|improve this answer



















            • 1




              @C.E. You're right, it just needs dimension length specification
              – swish
              45 mins ago






            • 1




              This is a very nice solution, +1
              – C. E.
              43 mins ago
















            2














            I think this is what you want:



            IntegerDigits[35 - 1, MixedRadix[{5, 4, 3}], 3] + 1


            In general:



            gridIndex[n_Integer, shape_List] := 
            
 IntegerDigits[n - 1, MixedRadix[shape], Length@shape] + 1





            share|improve this answer



















            • 1




              @C.E. You're right, it just needs dimension length specification
              – swish
              45 mins ago






            • 1




              This is a very nice solution, +1
              – C. E.
              43 mins ago














            2












            2








            2






            I think this is what you want:



            IntegerDigits[35 - 1, MixedRadix[{5, 4, 3}], 3] + 1


            In general:



            gridIndex[n_Integer, shape_List] := 
            
 IntegerDigits[n - 1, MixedRadix[shape], Length@shape] + 1





            share|improve this answer














            I think this is what you want:



            IntegerDigits[35 - 1, MixedRadix[{5, 4, 3}], 3] + 1


            In general:



            gridIndex[n_Integer, shape_List] := 
            
 IntegerDigits[n - 1, MixedRadix[shape], Length@shape] + 1






            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited 42 mins ago

























            answered 53 mins ago









            swish

            3,9611534




            3,9611534








            • 1




              @C.E. You're right, it just needs dimension length specification
              – swish
              45 mins ago






            • 1




              This is a very nice solution, +1
              – C. E.
              43 mins ago














            • 1




              @C.E. You're right, it just needs dimension length specification
              – swish
              45 mins ago






            • 1




              This is a very nice solution, +1
              – C. E.
              43 mins ago








            1




            1




            @C.E. You're right, it just needs dimension length specification
            – swish
            45 mins ago




            @C.E. You're right, it just needs dimension length specification
            – swish
            45 mins ago




            1




            1




            This is a very nice solution, +1
            – C. E.
            43 mins ago




            This is a very nice solution, +1
            – C. E.
            43 mins ago











            1














            Here's what I came up with:



            getSubindex[index_, stride_] := {
            
  Mod[index, stride, 1],
            
  Ceiling[index/stride]
            
  }
            
getIndex[index_, strides_] := 
            
 Reverse@FoldPairList[getSubindex, index, Reverse@strides]


            This is comparable to swish's solution speed-wise:



            gridIndex[1000, {3, 5, 4, 6}] // RepeatedTiming



            {0.000061, {3, 2, 3, 4}}




            getIndex[1000, {3, 5, 4, 6}] // RepeatedTiming



            {0.000052, {3, 2, 3, 4}}







            share|improve this answer




























              1














              Here's what I came up with:



              getSubindex[index_, stride_] := {
              
  Mod[index, stride, 1],
              
  Ceiling[index/stride]
              
  }
              
getIndex[index_, strides_] := 
              
 Reverse@FoldPairList[getSubindex, index, Reverse@strides]


              This is comparable to swish's solution speed-wise:



              gridIndex[1000, {3, 5, 4, 6}] // RepeatedTiming



              {0.000061, {3, 2, 3, 4}}




              getIndex[1000, {3, 5, 4, 6}] // RepeatedTiming



              {0.000052, {3, 2, 3, 4}}







              share|improve this answer


























                1












                1








                1






                Here's what I came up with:



                getSubindex[index_, stride_] := {
                
  Mod[index, stride, 1],
                
  Ceiling[index/stride]
                
  }
                
getIndex[index_, strides_] := 
                
 Reverse@FoldPairList[getSubindex, index, Reverse@strides]


                This is comparable to swish's solution speed-wise:



                gridIndex[1000, {3, 5, 4, 6}] // RepeatedTiming



                {0.000061, {3, 2, 3, 4}}




                getIndex[1000, {3, 5, 4, 6}] // RepeatedTiming



                {0.000052, {3, 2, 3, 4}}







                share|improve this answer














                Here's what I came up with:



                getSubindex[index_, stride_] := {
                
  Mod[index, stride, 1],
                
  Ceiling[index/stride]
                
  }
                
getIndex[index_, strides_] := 
                
 Reverse@FoldPairList[getSubindex, index, Reverse@strides]


                This is comparable to swish's solution speed-wise:



                gridIndex[1000, {3, 5, 4, 6}] // RepeatedTiming



                {0.000061, {3, 2, 3, 4}}




                getIndex[1000, {3, 5, 4, 6}] // RepeatedTiming



                {0.000052, {3, 2, 3, 4}}








                share|improve this answer














                share|improve this answer



                share|improve this answer








                edited 24 mins ago

























                answered 45 mins ago









                C. E.

                49.9k397202




                49.9k397202






























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