ByteCount on Function










5












$begingroup$


I was comparing RAM and CPU efficiency for List, Rule, Association, Function. The following code



n=10^6; 
tbl=Table[2i,i,n]; ByteCount[tbl]
rls=Table[2i->i,i,n]; ByteCount[rls]
asc=Association@Table[2i->i,i,n]; ByteCount[asc]
Do[fnc[2i]=i,i,n]; ByteCount[fnc]
AbsoluteTiming[Position[tbl,2n][[1,1]]]
AbsoluteTiming[2n/.rls]
AbsoluteTiming[asc[2n]]
AbsoluteTiming[fnc[2n]]


gave the following results:




8000144



96000080



128382000



0



0.142235, 1000000



0.805691, 1000000



0.00001, 1000000



4.*10^-6, 1000000




Thus Function is the fastest, but how do I get its real memory requirement?










share|improve this question









$endgroup$
















    5












    $begingroup$


    I was comparing RAM and CPU efficiency for List, Rule, Association, Function. The following code



    n=10^6; 
    tbl=Table[2i,i,n]; ByteCount[tbl]
    rls=Table[2i->i,i,n]; ByteCount[rls]
    asc=Association@Table[2i->i,i,n]; ByteCount[asc]
    Do[fnc[2i]=i,i,n]; ByteCount[fnc]
    AbsoluteTiming[Position[tbl,2n][[1,1]]]
    AbsoluteTiming[2n/.rls]
    AbsoluteTiming[asc[2n]]
    AbsoluteTiming[fnc[2n]]


    gave the following results:




    8000144



    96000080



    128382000



    0



    0.142235, 1000000



    0.805691, 1000000



    0.00001, 1000000



    4.*10^-6, 1000000




    Thus Function is the fastest, but how do I get its real memory requirement?










    share|improve this question









    $endgroup$














      5












      5








      5





      $begingroup$


      I was comparing RAM and CPU efficiency for List, Rule, Association, Function. The following code



      n=10^6; 
      tbl=Table[2i,i,n]; ByteCount[tbl]
      rls=Table[2i->i,i,n]; ByteCount[rls]
      asc=Association@Table[2i->i,i,n]; ByteCount[asc]
      Do[fnc[2i]=i,i,n]; ByteCount[fnc]
      AbsoluteTiming[Position[tbl,2n][[1,1]]]
      AbsoluteTiming[2n/.rls]
      AbsoluteTiming[asc[2n]]
      AbsoluteTiming[fnc[2n]]


      gave the following results:




      8000144



      96000080



      128382000



      0



      0.142235, 1000000



      0.805691, 1000000



      0.00001, 1000000



      4.*10^-6, 1000000




      Thus Function is the fastest, but how do I get its real memory requirement?










      share|improve this question









      $endgroup$




      I was comparing RAM and CPU efficiency for List, Rule, Association, Function. The following code



      n=10^6; 
      tbl=Table[2i,i,n]; ByteCount[tbl]
      rls=Table[2i->i,i,n]; ByteCount[rls]
      asc=Association@Table[2i->i,i,n]; ByteCount[asc]
      Do[fnc[2i]=i,i,n]; ByteCount[fnc]
      AbsoluteTiming[Position[tbl,2n][[1,1]]]
      AbsoluteTiming[2n/.rls]
      AbsoluteTiming[asc[2n]]
      AbsoluteTiming[fnc[2n]]


      gave the following results:




      8000144



      96000080



      128382000



      0



      0.142235, 1000000



      0.805691, 1000000



      0.00001, 1000000



      4.*10^-6, 1000000




      Thus Function is the fastest, but how do I get its real memory requirement?







      functions table associations memory data-structures






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked Aug 25 '18 at 17:09









      LeonLeon

      376111




      376111




















          3 Answers
          3






          active

          oldest

          votes


















          3












          $begingroup$

          Since you are using positive integer labels with rather high density in their range you should also try a classical array as lookup table. A further variant that works well when the keys are integers or lists of integers with fixed length is by using a BSP tree like with Nearest:



          n = 10^6;
          tbl = Table[2 i, i, n]; ByteCount[tbl]
          rls = Table[2 i -> i, i, n]; ByteCount[rls]
          rls2 = Dispatch[rls]; ByteCount[rls2]
          asc = AssociationThread[Range[2, 2 n, 2], Range[n]]; ByteCount[asc]
          ClearAll[fnc];
          Do[fnc[2 i] = i, i, n]; ByteCount[DownValues[fnc]]
          lookuptable = ConstantArray[0, 2 n];
          lookuptable[[2 ;; ;; 2]] = Range[1, n]; ByteCount[lookuptable]
          nf = Nearest[Range[2, 2 n, 2] -> Automatic];values = Range[n];ByteCount[nf] + ByteCount[values]



          8000144



          96000080



          126473520



          126473432



          192000080



          16000144



          16000488




          a = 2 RandomInteger[1, n, 100000];
          RepeatedTiming[r2 = a /. rls2][[1]]
          RepeatedTiming[r3a = asc /@ a][[1]]
          RepeatedTiming[r3b = Lookup[asc, a]][[1]]
          RepeatedTiming[r4 = fnc /@ a][[1]]
          RepeatedTiming[r5 = lookuptable[[a]]][[1]]
          RepeatedTiming[r6 = values[[Flatten[nf[a, 1]]]]][[1]]
          r2 = r3a == r3b == r4 == r5 == r6



          0.907



          0.8451



          0.614



          1.0



          0.011



          0.11



          True




          Also notice that these methods behave very differently to each other when looking up invalid keys.






          share|improve this answer











          $endgroup$












          • $begingroup$
            What a nice overview of methods.
            $endgroup$
            – Mr.Wizard
            Sep 17 '18 at 7:49










          • $begingroup$
            Thanks! When you say that, I really appreciate it!
            $endgroup$
            – Henrik Schumacher
            Sep 17 '18 at 8:39


















          4












          $begingroup$

          DownValues should get close to the actual value I believe.



          ByteCount[DownValues[fnc]]



          192000080



          You could also use MaxMemoryUsed during the construction:



          ClearAll[fnc]

          MaxMemoryUsed[Do[fnc[2 i] = i, i, 10^6];]



          168327840






          share|improve this answer









          $endgroup$












          • $begingroup$
            DownValues returns way more, than only the right hand side. Is it a fair comparison? I guess all memory structures should be compared by DownValues then.
            $endgroup$
            – Johu
            Aug 25 '18 at 18:17











          • $begingroup$
            @Johu I think it's a good first order approximation at least. I did provide a second method that I presume is more accurate.
            $endgroup$
            – Mr.Wizard
            Aug 25 '18 at 18:18


















          4












          $begingroup$

          In your example the value of the argument is a part of the definition. Value of fnc[2 i], where i is a symbol, is not defined in your MWE.



          Total@Table[ByteCount[fnc[2 i]], i, n]



          16 000 000




          (Edit: note, that my solution only gives only the size of the right hand side and probably underestimates the real memory cost. See the other solution.)



          Note also, that your timing measurement lead to misleading results as you apply it to so simple operations. Compare to



          n2 = 10;
          AbsoluteTiming[Position[tbl, #] & /@ (2 RandomInteger[n, n2])] // First
          AbsoluteTiming[Replace[rls] /@ (2 RandomInteger[n, n2]);] // First



          2.19423 
          1.8028



          Both of these approaches are very slow, as they require going through the whole list to fine the element.



          These are much much faster:



          n2 = 100000;
          AbsoluteTiming[fnc /@ (2 RandomInteger[n, n2]);] // First
          AbsoluteTiming[asc /@ (2 RandomInteger[n, n2]);] // First
          AbsoluteTiming[Lookup[asc, 2 RandomInteger[n, n2]];] // First

          > 0.267781
          > 0.226777
          > 0.171752


          I think it is misleading to call fnc[i] a function, as a function usually is evaluated runtime. In your MWE you save a precomputed value. This technique is referred to as memoization.



          When you wonder which one you should use, I would always use what makes sense semantically, because the engineers behind the kernel and native commands have but a lot of effort into finding a balance between all the features one usually needs from a List, Rules, Associations and Symbols. Associations and Symbols are the ones, which require fast random access.






          share|improve this answer











          $endgroup$












          • $begingroup$
            Hmm, this is very low RAM usage, comparable to Table, yet faster than all of them. Why wouldn't one use Function instead of Table or Association?
            $endgroup$
            – Leon
            Aug 25 '18 at 17:21











          • $begingroup$
            Please see the updated answer.
            $endgroup$
            – Johu
            Aug 25 '18 at 17:43






          • 1




            $begingroup$
            Your problem is that foo /@ 2 RandomInteger[n, n2] is parsed as (foo /@ 2) RandomInteger[n, n2].
            $endgroup$
            – Carl Woll
            Aug 25 '18 at 20:42










          • $begingroup$
            @CarlWoll oh lord. how stupid.
            $endgroup$
            – Johu
            Aug 25 '18 at 22:36











          • $begingroup$
            I fixed it now and adjusted my conclusions. I did think it had to be impossible for the lookup from List to be that fast. Now it all makes sense.
            $endgroup$
            – Johu
            Aug 25 '18 at 22:46










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          3 Answers
          3






          active

          oldest

          votes








          3 Answers
          3






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          3












          $begingroup$

          Since you are using positive integer labels with rather high density in their range you should also try a classical array as lookup table. A further variant that works well when the keys are integers or lists of integers with fixed length is by using a BSP tree like with Nearest:



          n = 10^6;
          tbl = Table[2 i, i, n]; ByteCount[tbl]
          rls = Table[2 i -> i, i, n]; ByteCount[rls]
          rls2 = Dispatch[rls]; ByteCount[rls2]
          asc = AssociationThread[Range[2, 2 n, 2], Range[n]]; ByteCount[asc]
          ClearAll[fnc];
          Do[fnc[2 i] = i, i, n]; ByteCount[DownValues[fnc]]
          lookuptable = ConstantArray[0, 2 n];
          lookuptable[[2 ;; ;; 2]] = Range[1, n]; ByteCount[lookuptable]
          nf = Nearest[Range[2, 2 n, 2] -> Automatic];values = Range[n];ByteCount[nf] + ByteCount[values]



          8000144



          96000080



          126473520



          126473432



          192000080



          16000144



          16000488




          a = 2 RandomInteger[1, n, 100000];
          RepeatedTiming[r2 = a /. rls2][[1]]
          RepeatedTiming[r3a = asc /@ a][[1]]
          RepeatedTiming[r3b = Lookup[asc, a]][[1]]
          RepeatedTiming[r4 = fnc /@ a][[1]]
          RepeatedTiming[r5 = lookuptable[[a]]][[1]]
          RepeatedTiming[r6 = values[[Flatten[nf[a, 1]]]]][[1]]
          r2 = r3a == r3b == r4 == r5 == r6



          0.907



          0.8451



          0.614



          1.0



          0.011



          0.11



          True




          Also notice that these methods behave very differently to each other when looking up invalid keys.






          share|improve this answer











          $endgroup$












          • $begingroup$
            What a nice overview of methods.
            $endgroup$
            – Mr.Wizard
            Sep 17 '18 at 7:49










          • $begingroup$
            Thanks! When you say that, I really appreciate it!
            $endgroup$
            – Henrik Schumacher
            Sep 17 '18 at 8:39















          3












          $begingroup$

          Since you are using positive integer labels with rather high density in their range you should also try a classical array as lookup table. A further variant that works well when the keys are integers or lists of integers with fixed length is by using a BSP tree like with Nearest:



          n = 10^6;
          tbl = Table[2 i, i, n]; ByteCount[tbl]
          rls = Table[2 i -> i, i, n]; ByteCount[rls]
          rls2 = Dispatch[rls]; ByteCount[rls2]
          asc = AssociationThread[Range[2, 2 n, 2], Range[n]]; ByteCount[asc]
          ClearAll[fnc];
          Do[fnc[2 i] = i, i, n]; ByteCount[DownValues[fnc]]
          lookuptable = ConstantArray[0, 2 n];
          lookuptable[[2 ;; ;; 2]] = Range[1, n]; ByteCount[lookuptable]
          nf = Nearest[Range[2, 2 n, 2] -> Automatic];values = Range[n];ByteCount[nf] + ByteCount[values]



          8000144



          96000080



          126473520



          126473432



          192000080



          16000144



          16000488




          a = 2 RandomInteger[1, n, 100000];
          RepeatedTiming[r2 = a /. rls2][[1]]
          RepeatedTiming[r3a = asc /@ a][[1]]
          RepeatedTiming[r3b = Lookup[asc, a]][[1]]
          RepeatedTiming[r4 = fnc /@ a][[1]]
          RepeatedTiming[r5 = lookuptable[[a]]][[1]]
          RepeatedTiming[r6 = values[[Flatten[nf[a, 1]]]]][[1]]
          r2 = r3a == r3b == r4 == r5 == r6



          0.907



          0.8451



          0.614



          1.0



          0.011



          0.11



          True




          Also notice that these methods behave very differently to each other when looking up invalid keys.






          share|improve this answer











          $endgroup$












          • $begingroup$
            What a nice overview of methods.
            $endgroup$
            – Mr.Wizard
            Sep 17 '18 at 7:49










          • $begingroup$
            Thanks! When you say that, I really appreciate it!
            $endgroup$
            – Henrik Schumacher
            Sep 17 '18 at 8:39













          3












          3








          3





          $begingroup$

          Since you are using positive integer labels with rather high density in their range you should also try a classical array as lookup table. A further variant that works well when the keys are integers or lists of integers with fixed length is by using a BSP tree like with Nearest:



          n = 10^6;
          tbl = Table[2 i, i, n]; ByteCount[tbl]
          rls = Table[2 i -> i, i, n]; ByteCount[rls]
          rls2 = Dispatch[rls]; ByteCount[rls2]
          asc = AssociationThread[Range[2, 2 n, 2], Range[n]]; ByteCount[asc]
          ClearAll[fnc];
          Do[fnc[2 i] = i, i, n]; ByteCount[DownValues[fnc]]
          lookuptable = ConstantArray[0, 2 n];
          lookuptable[[2 ;; ;; 2]] = Range[1, n]; ByteCount[lookuptable]
          nf = Nearest[Range[2, 2 n, 2] -> Automatic];values = Range[n];ByteCount[nf] + ByteCount[values]



          8000144



          96000080



          126473520



          126473432



          192000080



          16000144



          16000488




          a = 2 RandomInteger[1, n, 100000];
          RepeatedTiming[r2 = a /. rls2][[1]]
          RepeatedTiming[r3a = asc /@ a][[1]]
          RepeatedTiming[r3b = Lookup[asc, a]][[1]]
          RepeatedTiming[r4 = fnc /@ a][[1]]
          RepeatedTiming[r5 = lookuptable[[a]]][[1]]
          RepeatedTiming[r6 = values[[Flatten[nf[a, 1]]]]][[1]]
          r2 = r3a == r3b == r4 == r5 == r6



          0.907



          0.8451



          0.614



          1.0



          0.011



          0.11



          True




          Also notice that these methods behave very differently to each other when looking up invalid keys.






          share|improve this answer











          $endgroup$



          Since you are using positive integer labels with rather high density in their range you should also try a classical array as lookup table. A further variant that works well when the keys are integers or lists of integers with fixed length is by using a BSP tree like with Nearest:



          n = 10^6;
          tbl = Table[2 i, i, n]; ByteCount[tbl]
          rls = Table[2 i -> i, i, n]; ByteCount[rls]
          rls2 = Dispatch[rls]; ByteCount[rls2]
          asc = AssociationThread[Range[2, 2 n, 2], Range[n]]; ByteCount[asc]
          ClearAll[fnc];
          Do[fnc[2 i] = i, i, n]; ByteCount[DownValues[fnc]]
          lookuptable = ConstantArray[0, 2 n];
          lookuptable[[2 ;; ;; 2]] = Range[1, n]; ByteCount[lookuptable]
          nf = Nearest[Range[2, 2 n, 2] -> Automatic];values = Range[n];ByteCount[nf] + ByteCount[values]



          8000144



          96000080



          126473520



          126473432



          192000080



          16000144



          16000488




          a = 2 RandomInteger[1, n, 100000];
          RepeatedTiming[r2 = a /. rls2][[1]]
          RepeatedTiming[r3a = asc /@ a][[1]]
          RepeatedTiming[r3b = Lookup[asc, a]][[1]]
          RepeatedTiming[r4 = fnc /@ a][[1]]
          RepeatedTiming[r5 = lookuptable[[a]]][[1]]
          RepeatedTiming[r6 = values[[Flatten[nf[a, 1]]]]][[1]]
          r2 = r3a == r3b == r4 == r5 == r6



          0.907



          0.8451



          0.614



          1.0



          0.011



          0.11



          True




          Also notice that these methods behave very differently to each other when looking up invalid keys.







          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited Sep 16 '18 at 11:46

























          answered Sep 16 '18 at 11:31









          Henrik SchumacherHenrik Schumacher

          50.3k469144




          50.3k469144











          • $begingroup$
            What a nice overview of methods.
            $endgroup$
            – Mr.Wizard
            Sep 17 '18 at 7:49










          • $begingroup$
            Thanks! When you say that, I really appreciate it!
            $endgroup$
            – Henrik Schumacher
            Sep 17 '18 at 8:39
















          • $begingroup$
            What a nice overview of methods.
            $endgroup$
            – Mr.Wizard
            Sep 17 '18 at 7:49










          • $begingroup$
            Thanks! When you say that, I really appreciate it!
            $endgroup$
            – Henrik Schumacher
            Sep 17 '18 at 8:39















          $begingroup$
          What a nice overview of methods.
          $endgroup$
          – Mr.Wizard
          Sep 17 '18 at 7:49




          $begingroup$
          What a nice overview of methods.
          $endgroup$
          – Mr.Wizard
          Sep 17 '18 at 7:49












          $begingroup$
          Thanks! When you say that, I really appreciate it!
          $endgroup$
          – Henrik Schumacher
          Sep 17 '18 at 8:39




          $begingroup$
          Thanks! When you say that, I really appreciate it!
          $endgroup$
          – Henrik Schumacher
          Sep 17 '18 at 8:39











          4












          $begingroup$

          DownValues should get close to the actual value I believe.



          ByteCount[DownValues[fnc]]



          192000080



          You could also use MaxMemoryUsed during the construction:



          ClearAll[fnc]

          MaxMemoryUsed[Do[fnc[2 i] = i, i, 10^6];]



          168327840






          share|improve this answer









          $endgroup$












          • $begingroup$
            DownValues returns way more, than only the right hand side. Is it a fair comparison? I guess all memory structures should be compared by DownValues then.
            $endgroup$
            – Johu
            Aug 25 '18 at 18:17











          • $begingroup$
            @Johu I think it's a good first order approximation at least. I did provide a second method that I presume is more accurate.
            $endgroup$
            – Mr.Wizard
            Aug 25 '18 at 18:18















          4












          $begingroup$

          DownValues should get close to the actual value I believe.



          ByteCount[DownValues[fnc]]



          192000080



          You could also use MaxMemoryUsed during the construction:



          ClearAll[fnc]

          MaxMemoryUsed[Do[fnc[2 i] = i, i, 10^6];]



          168327840






          share|improve this answer









          $endgroup$












          • $begingroup$
            DownValues returns way more, than only the right hand side. Is it a fair comparison? I guess all memory structures should be compared by DownValues then.
            $endgroup$
            – Johu
            Aug 25 '18 at 18:17











          • $begingroup$
            @Johu I think it's a good first order approximation at least. I did provide a second method that I presume is more accurate.
            $endgroup$
            – Mr.Wizard
            Aug 25 '18 at 18:18













          4












          4








          4





          $begingroup$

          DownValues should get close to the actual value I believe.



          ByteCount[DownValues[fnc]]



          192000080



          You could also use MaxMemoryUsed during the construction:



          ClearAll[fnc]

          MaxMemoryUsed[Do[fnc[2 i] = i, i, 10^6];]



          168327840






          share|improve this answer









          $endgroup$



          DownValues should get close to the actual value I believe.



          ByteCount[DownValues[fnc]]



          192000080



          You could also use MaxMemoryUsed during the construction:



          ClearAll[fnc]

          MaxMemoryUsed[Do[fnc[2 i] = i, i, 10^6];]



          168327840







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Aug 25 '18 at 18:10









          Mr.WizardMr.Wizard

          231k294741041




          231k294741041











          • $begingroup$
            DownValues returns way more, than only the right hand side. Is it a fair comparison? I guess all memory structures should be compared by DownValues then.
            $endgroup$
            – Johu
            Aug 25 '18 at 18:17











          • $begingroup$
            @Johu I think it's a good first order approximation at least. I did provide a second method that I presume is more accurate.
            $endgroup$
            – Mr.Wizard
            Aug 25 '18 at 18:18
















          • $begingroup$
            DownValues returns way more, than only the right hand side. Is it a fair comparison? I guess all memory structures should be compared by DownValues then.
            $endgroup$
            – Johu
            Aug 25 '18 at 18:17











          • $begingroup$
            @Johu I think it's a good first order approximation at least. I did provide a second method that I presume is more accurate.
            $endgroup$
            – Mr.Wizard
            Aug 25 '18 at 18:18















          $begingroup$
          DownValues returns way more, than only the right hand side. Is it a fair comparison? I guess all memory structures should be compared by DownValues then.
          $endgroup$
          – Johu
          Aug 25 '18 at 18:17





          $begingroup$
          DownValues returns way more, than only the right hand side. Is it a fair comparison? I guess all memory structures should be compared by DownValues then.
          $endgroup$
          – Johu
          Aug 25 '18 at 18:17













          $begingroup$
          @Johu I think it's a good first order approximation at least. I did provide a second method that I presume is more accurate.
          $endgroup$
          – Mr.Wizard
          Aug 25 '18 at 18:18




          $begingroup$
          @Johu I think it's a good first order approximation at least. I did provide a second method that I presume is more accurate.
          $endgroup$
          – Mr.Wizard
          Aug 25 '18 at 18:18











          4












          $begingroup$

          In your example the value of the argument is a part of the definition. Value of fnc[2 i], where i is a symbol, is not defined in your MWE.



          Total@Table[ByteCount[fnc[2 i]], i, n]



          16 000 000




          (Edit: note, that my solution only gives only the size of the right hand side and probably underestimates the real memory cost. See the other solution.)



          Note also, that your timing measurement lead to misleading results as you apply it to so simple operations. Compare to



          n2 = 10;
          AbsoluteTiming[Position[tbl, #] & /@ (2 RandomInteger[n, n2])] // First
          AbsoluteTiming[Replace[rls] /@ (2 RandomInteger[n, n2]);] // First



          2.19423 
          1.8028



          Both of these approaches are very slow, as they require going through the whole list to fine the element.



          These are much much faster:



          n2 = 100000;
          AbsoluteTiming[fnc /@ (2 RandomInteger[n, n2]);] // First
          AbsoluteTiming[asc /@ (2 RandomInteger[n, n2]);] // First
          AbsoluteTiming[Lookup[asc, 2 RandomInteger[n, n2]];] // First

          > 0.267781
          > 0.226777
          > 0.171752


          I think it is misleading to call fnc[i] a function, as a function usually is evaluated runtime. In your MWE you save a precomputed value. This technique is referred to as memoization.



          When you wonder which one you should use, I would always use what makes sense semantically, because the engineers behind the kernel and native commands have but a lot of effort into finding a balance between all the features one usually needs from a List, Rules, Associations and Symbols. Associations and Symbols are the ones, which require fast random access.






          share|improve this answer











          $endgroup$












          • $begingroup$
            Hmm, this is very low RAM usage, comparable to Table, yet faster than all of them. Why wouldn't one use Function instead of Table or Association?
            $endgroup$
            – Leon
            Aug 25 '18 at 17:21











          • $begingroup$
            Please see the updated answer.
            $endgroup$
            – Johu
            Aug 25 '18 at 17:43






          • 1




            $begingroup$
            Your problem is that foo /@ 2 RandomInteger[n, n2] is parsed as (foo /@ 2) RandomInteger[n, n2].
            $endgroup$
            – Carl Woll
            Aug 25 '18 at 20:42










          • $begingroup$
            @CarlWoll oh lord. how stupid.
            $endgroup$
            – Johu
            Aug 25 '18 at 22:36











          • $begingroup$
            I fixed it now and adjusted my conclusions. I did think it had to be impossible for the lookup from List to be that fast. Now it all makes sense.
            $endgroup$
            – Johu
            Aug 25 '18 at 22:46















          4












          $begingroup$

          In your example the value of the argument is a part of the definition. Value of fnc[2 i], where i is a symbol, is not defined in your MWE.



          Total@Table[ByteCount[fnc[2 i]], i, n]



          16 000 000




          (Edit: note, that my solution only gives only the size of the right hand side and probably underestimates the real memory cost. See the other solution.)



          Note also, that your timing measurement lead to misleading results as you apply it to so simple operations. Compare to



          n2 = 10;
          AbsoluteTiming[Position[tbl, #] & /@ (2 RandomInteger[n, n2])] // First
          AbsoluteTiming[Replace[rls] /@ (2 RandomInteger[n, n2]);] // First



          2.19423 
          1.8028



          Both of these approaches are very slow, as they require going through the whole list to fine the element.



          These are much much faster:



          n2 = 100000;
          AbsoluteTiming[fnc /@ (2 RandomInteger[n, n2]);] // First
          AbsoluteTiming[asc /@ (2 RandomInteger[n, n2]);] // First
          AbsoluteTiming[Lookup[asc, 2 RandomInteger[n, n2]];] // First

          > 0.267781
          > 0.226777
          > 0.171752


          I think it is misleading to call fnc[i] a function, as a function usually is evaluated runtime. In your MWE you save a precomputed value. This technique is referred to as memoization.



          When you wonder which one you should use, I would always use what makes sense semantically, because the engineers behind the kernel and native commands have but a lot of effort into finding a balance between all the features one usually needs from a List, Rules, Associations and Symbols. Associations and Symbols are the ones, which require fast random access.






          share|improve this answer











          $endgroup$












          • $begingroup$
            Hmm, this is very low RAM usage, comparable to Table, yet faster than all of them. Why wouldn't one use Function instead of Table or Association?
            $endgroup$
            – Leon
            Aug 25 '18 at 17:21











          • $begingroup$
            Please see the updated answer.
            $endgroup$
            – Johu
            Aug 25 '18 at 17:43






          • 1




            $begingroup$
            Your problem is that foo /@ 2 RandomInteger[n, n2] is parsed as (foo /@ 2) RandomInteger[n, n2].
            $endgroup$
            – Carl Woll
            Aug 25 '18 at 20:42










          • $begingroup$
            @CarlWoll oh lord. how stupid.
            $endgroup$
            – Johu
            Aug 25 '18 at 22:36











          • $begingroup$
            I fixed it now and adjusted my conclusions. I did think it had to be impossible for the lookup from List to be that fast. Now it all makes sense.
            $endgroup$
            – Johu
            Aug 25 '18 at 22:46













          4












          4








          4





          $begingroup$

          In your example the value of the argument is a part of the definition. Value of fnc[2 i], where i is a symbol, is not defined in your MWE.



          Total@Table[ByteCount[fnc[2 i]], i, n]



          16 000 000




          (Edit: note, that my solution only gives only the size of the right hand side and probably underestimates the real memory cost. See the other solution.)



          Note also, that your timing measurement lead to misleading results as you apply it to so simple operations. Compare to



          n2 = 10;
          AbsoluteTiming[Position[tbl, #] & /@ (2 RandomInteger[n, n2])] // First
          AbsoluteTiming[Replace[rls] /@ (2 RandomInteger[n, n2]);] // First



          2.19423 
          1.8028



          Both of these approaches are very slow, as they require going through the whole list to fine the element.



          These are much much faster:



          n2 = 100000;
          AbsoluteTiming[fnc /@ (2 RandomInteger[n, n2]);] // First
          AbsoluteTiming[asc /@ (2 RandomInteger[n, n2]);] // First
          AbsoluteTiming[Lookup[asc, 2 RandomInteger[n, n2]];] // First

          > 0.267781
          > 0.226777
          > 0.171752


          I think it is misleading to call fnc[i] a function, as a function usually is evaluated runtime. In your MWE you save a precomputed value. This technique is referred to as memoization.



          When you wonder which one you should use, I would always use what makes sense semantically, because the engineers behind the kernel and native commands have but a lot of effort into finding a balance between all the features one usually needs from a List, Rules, Associations and Symbols. Associations and Symbols are the ones, which require fast random access.






          share|improve this answer











          $endgroup$



          In your example the value of the argument is a part of the definition. Value of fnc[2 i], where i is a symbol, is not defined in your MWE.



          Total@Table[ByteCount[fnc[2 i]], i, n]



          16 000 000




          (Edit: note, that my solution only gives only the size of the right hand side and probably underestimates the real memory cost. See the other solution.)



          Note also, that your timing measurement lead to misleading results as you apply it to so simple operations. Compare to



          n2 = 10;
          AbsoluteTiming[Position[tbl, #] & /@ (2 RandomInteger[n, n2])] // First
          AbsoluteTiming[Replace[rls] /@ (2 RandomInteger[n, n2]);] // First



          2.19423 
          1.8028



          Both of these approaches are very slow, as they require going through the whole list to fine the element.



          These are much much faster:



          n2 = 100000;
          AbsoluteTiming[fnc /@ (2 RandomInteger[n, n2]);] // First
          AbsoluteTiming[asc /@ (2 RandomInteger[n, n2]);] // First
          AbsoluteTiming[Lookup[asc, 2 RandomInteger[n, n2]];] // First

          > 0.267781
          > 0.226777
          > 0.171752


          I think it is misleading to call fnc[i] a function, as a function usually is evaluated runtime. In your MWE you save a precomputed value. This technique is referred to as memoization.



          When you wonder which one you should use, I would always use what makes sense semantically, because the engineers behind the kernel and native commands have but a lot of effort into finding a balance between all the features one usually needs from a List, Rules, Associations and Symbols. Associations and Symbols are the ones, which require fast random access.







          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited Aug 25 '18 at 22:45

























          answered Aug 25 '18 at 17:15









          JohuJohu

          3,6681037




          3,6681037











          • $begingroup$
            Hmm, this is very low RAM usage, comparable to Table, yet faster than all of them. Why wouldn't one use Function instead of Table or Association?
            $endgroup$
            – Leon
            Aug 25 '18 at 17:21











          • $begingroup$
            Please see the updated answer.
            $endgroup$
            – Johu
            Aug 25 '18 at 17:43






          • 1




            $begingroup$
            Your problem is that foo /@ 2 RandomInteger[n, n2] is parsed as (foo /@ 2) RandomInteger[n, n2].
            $endgroup$
            – Carl Woll
            Aug 25 '18 at 20:42










          • $begingroup$
            @CarlWoll oh lord. how stupid.
            $endgroup$
            – Johu
            Aug 25 '18 at 22:36











          • $begingroup$
            I fixed it now and adjusted my conclusions. I did think it had to be impossible for the lookup from List to be that fast. Now it all makes sense.
            $endgroup$
            – Johu
            Aug 25 '18 at 22:46
















          • $begingroup$
            Hmm, this is very low RAM usage, comparable to Table, yet faster than all of them. Why wouldn't one use Function instead of Table or Association?
            $endgroup$
            – Leon
            Aug 25 '18 at 17:21











          • $begingroup$
            Please see the updated answer.
            $endgroup$
            – Johu
            Aug 25 '18 at 17:43






          • 1




            $begingroup$
            Your problem is that foo /@ 2 RandomInteger[n, n2] is parsed as (foo /@ 2) RandomInteger[n, n2].
            $endgroup$
            – Carl Woll
            Aug 25 '18 at 20:42










          • $begingroup$
            @CarlWoll oh lord. how stupid.
            $endgroup$
            – Johu
            Aug 25 '18 at 22:36











          • $begingroup$
            I fixed it now and adjusted my conclusions. I did think it had to be impossible for the lookup from List to be that fast. Now it all makes sense.
            $endgroup$
            – Johu
            Aug 25 '18 at 22:46















          $begingroup$
          Hmm, this is very low RAM usage, comparable to Table, yet faster than all of them. Why wouldn't one use Function instead of Table or Association?
          $endgroup$
          – Leon
          Aug 25 '18 at 17:21





          $begingroup$
          Hmm, this is very low RAM usage, comparable to Table, yet faster than all of them. Why wouldn't one use Function instead of Table or Association?
          $endgroup$
          – Leon
          Aug 25 '18 at 17:21













          $begingroup$
          Please see the updated answer.
          $endgroup$
          – Johu
          Aug 25 '18 at 17:43




          $begingroup$
          Please see the updated answer.
          $endgroup$
          – Johu
          Aug 25 '18 at 17:43




          1




          1




          $begingroup$
          Your problem is that foo /@ 2 RandomInteger[n, n2] is parsed as (foo /@ 2) RandomInteger[n, n2].
          $endgroup$
          – Carl Woll
          Aug 25 '18 at 20:42




          $begingroup$
          Your problem is that foo /@ 2 RandomInteger[n, n2] is parsed as (foo /@ 2) RandomInteger[n, n2].
          $endgroup$
          – Carl Woll
          Aug 25 '18 at 20:42












          $begingroup$
          @CarlWoll oh lord. how stupid.
          $endgroup$
          – Johu
          Aug 25 '18 at 22:36





          $begingroup$
          @CarlWoll oh lord. how stupid.
          $endgroup$
          – Johu
          Aug 25 '18 at 22:36













          $begingroup$
          I fixed it now and adjusted my conclusions. I did think it had to be impossible for the lookup from List to be that fast. Now it all makes sense.
          $endgroup$
          – Johu
          Aug 25 '18 at 22:46




          $begingroup$
          I fixed it now and adjusted my conclusions. I did think it had to be impossible for the lookup from List to be that fast. Now it all makes sense.
          $endgroup$
          – Johu
          Aug 25 '18 at 22:46

















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