List Interval Sum











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3
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I am doing a large data-set computation. Among those computational steps, in one step I need to do a sum with a pattern: Sum the elements with same interval.



For example, for a list from 1 to 9; With the interval being set to 3 manually. So it could be other values in different cases.



And the list would be calculated as:



1 + 4 + 7 = 12;
2 + 5 + 8 = 15;
3 + 6 + 9 = 18;


So for list = Range[1,9],the final desired result would be {12,15,18} in this example. I attached an illustration for a further elaboration: sum the element with the same color when interval = 3:



The example illustration, sum same color with interval equals to 3 in this case



Thanks for @Chris's Answer, the above case could be solved by:



 Total[Partition[Range[9], 3]]


Edit my original question from here:



But what I actually want to do is only sum "N" numbers in a time. N is settled and when N = 3 in below example:



enter image description here



It should be computed by :



1 + 4 + 7 = 12;
2 + 5 + 8 = 15;
3 + 6 + 9 = 18;
10 + 13 + 16 = 39;
11 + 14 + 17 = 42;
12 + 15 + 18 = 45;


Hereby the result would be {12,15,18,39,42,45}



I think this might be not hard, but I just can't think it very clearly when I want to utilize the parallelization characteristics of MMA and trying to avoid Unpacked Array results.










share|improve this question
























  • is the length of the input list always a multiple of n? If not, what is the desired output for inputs Range[19] and Range[20]?
    – kglr
    Dec 5 at 3:09












  • The length of list will always be Mod[Length[list],N] = 0, so no worry about corner cases
    – cj9435042
    Dec 5 at 3:13

















up vote
3
down vote

favorite












I am doing a large data-set computation. Among those computational steps, in one step I need to do a sum with a pattern: Sum the elements with same interval.



For example, for a list from 1 to 9; With the interval being set to 3 manually. So it could be other values in different cases.



And the list would be calculated as:



1 + 4 + 7 = 12;
2 + 5 + 8 = 15;
3 + 6 + 9 = 18;


So for list = Range[1,9],the final desired result would be {12,15,18} in this example. I attached an illustration for a further elaboration: sum the element with the same color when interval = 3:



The example illustration, sum same color with interval equals to 3 in this case



Thanks for @Chris's Answer, the above case could be solved by:



 Total[Partition[Range[9], 3]]


Edit my original question from here:



But what I actually want to do is only sum "N" numbers in a time. N is settled and when N = 3 in below example:



enter image description here



It should be computed by :



1 + 4 + 7 = 12;
2 + 5 + 8 = 15;
3 + 6 + 9 = 18;
10 + 13 + 16 = 39;
11 + 14 + 17 = 42;
12 + 15 + 18 = 45;


Hereby the result would be {12,15,18,39,42,45}



I think this might be not hard, but I just can't think it very clearly when I want to utilize the parallelization characteristics of MMA and trying to avoid Unpacked Array results.










share|improve this question
























  • is the length of the input list always a multiple of n? If not, what is the desired output for inputs Range[19] and Range[20]?
    – kglr
    Dec 5 at 3:09












  • The length of list will always be Mod[Length[list],N] = 0, so no worry about corner cases
    – cj9435042
    Dec 5 at 3:13















up vote
3
down vote

favorite









up vote
3
down vote

favorite











I am doing a large data-set computation. Among those computational steps, in one step I need to do a sum with a pattern: Sum the elements with same interval.



For example, for a list from 1 to 9; With the interval being set to 3 manually. So it could be other values in different cases.



And the list would be calculated as:



1 + 4 + 7 = 12;
2 + 5 + 8 = 15;
3 + 6 + 9 = 18;


So for list = Range[1,9],the final desired result would be {12,15,18} in this example. I attached an illustration for a further elaboration: sum the element with the same color when interval = 3:



The example illustration, sum same color with interval equals to 3 in this case



Thanks for @Chris's Answer, the above case could be solved by:



 Total[Partition[Range[9], 3]]


Edit my original question from here:



But what I actually want to do is only sum "N" numbers in a time. N is settled and when N = 3 in below example:



enter image description here



It should be computed by :



1 + 4 + 7 = 12;
2 + 5 + 8 = 15;
3 + 6 + 9 = 18;
10 + 13 + 16 = 39;
11 + 14 + 17 = 42;
12 + 15 + 18 = 45;


Hereby the result would be {12,15,18,39,42,45}



I think this might be not hard, but I just can't think it very clearly when I want to utilize the parallelization characteristics of MMA and trying to avoid Unpacked Array results.










share|improve this question















I am doing a large data-set computation. Among those computational steps, in one step I need to do a sum with a pattern: Sum the elements with same interval.



For example, for a list from 1 to 9; With the interval being set to 3 manually. So it could be other values in different cases.



And the list would be calculated as:



1 + 4 + 7 = 12;
2 + 5 + 8 = 15;
3 + 6 + 9 = 18;


So for list = Range[1,9],the final desired result would be {12,15,18} in this example. I attached an illustration for a further elaboration: sum the element with the same color when interval = 3:



The example illustration, sum same color with interval equals to 3 in this case



Thanks for @Chris's Answer, the above case could be solved by:



 Total[Partition[Range[9], 3]]


Edit my original question from here:



But what I actually want to do is only sum "N" numbers in a time. N is settled and when N = 3 in below example:



enter image description here



It should be computed by :



1 + 4 + 7 = 12;
2 + 5 + 8 = 15;
3 + 6 + 9 = 18;
10 + 13 + 16 = 39;
11 + 14 + 17 = 42;
12 + 15 + 18 = 45;


Hereby the result would be {12,15,18,39,42,45}



I think this might be not hard, but I just can't think it very clearly when I want to utilize the parallelization characteristics of MMA and trying to avoid Unpacked Array results.







list-manipulation






share|improve this question















share|improve this question













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share|improve this question








edited Dec 5 at 1:29

























asked Dec 5 at 0:04









cj9435042

34216




34216












  • is the length of the input list always a multiple of n? If not, what is the desired output for inputs Range[19] and Range[20]?
    – kglr
    Dec 5 at 3:09












  • The length of list will always be Mod[Length[list],N] = 0, so no worry about corner cases
    – cj9435042
    Dec 5 at 3:13




















  • is the length of the input list always a multiple of n? If not, what is the desired output for inputs Range[19] and Range[20]?
    – kglr
    Dec 5 at 3:09












  • The length of list will always be Mod[Length[list],N] = 0, so no worry about corner cases
    – cj9435042
    Dec 5 at 3:13


















is the length of the input list always a multiple of n? If not, what is the desired output for inputs Range[19] and Range[20]?
– kglr
Dec 5 at 3:09






is the length of the input list always a multiple of n? If not, what is the desired output for inputs Range[19] and Range[20]?
– kglr
Dec 5 at 3:09














The length of list will always be Mod[Length[list],N] = 0, so no worry about corner cases
– cj9435042
Dec 5 at 3:13






The length of list will always be Mod[Length[list],N] = 0, so no worry about corner cases
– cj9435042
Dec 5 at 3:13












3 Answers
3






active

oldest

votes

















up vote
4
down vote













Total[Partition[Range[9], 3]]



{12, 15, 18}




Update for revised question:



r = Range[18]    

Total /@ Flatten[Partition[#, 3] & /@ {r[[1 ;; ;; 3]], r[[2 ;; ;; 3]], r[[3 ;; ;; 3]]}, 1]





share|improve this answer























  • Hi Chris, sorry I wasn't clarify the problem clearly. I just updated my question would you still interest to help?
    – cj9435042
    Dec 5 at 1:30


















up vote
3
down vote













Using the six-argument form of Partition:



Join @@ Partition[Partition[Range[9], 3], 3, 3, {1, 1}, {}, Plus]



{12, 15, 18}




Join @@ Partition[Partition[Range[18], 3], 3, 3, {1, 1}, {}, Plus]



{12, 15, 18, 39, 42, 45}




More generally,



ClearAll[partsums]
partsums[lst_List, n_Integer] := Join@@Partition[Partition[lst, n], n, n, {1,1}, {}, Plus]


Examples:



partsums[Range[18], 3]



{12, 15, 18, 39, 42, 45}




Grid[Prepend[Table[{i, Column[i Range[7]], Column[partsums[Range@#, i] & /@ 
(i Range[7])]}, {i, {3, 4, 5}}], {"n", "Length@list" , "f[list, n]"}],
Alignment -> Center, Dividers -> All] // TeXForm



$smallbegin{array}{|c|c|c|}
hline
text{n} & text{Length@list} & text{f[list, n]} \
hline
3 &
begin{array}{l}
3 \
6 \
9 \
12 \
15 \
18 \
21 \
end{array}
&
begin{array}{l}
{1,2,3} \
{5,7,9} \
{12,15,18} \
{12,15,18,10,11,12} \
{12,15,18,23,25,27} \
{12,15,18,39,42,45} \
{12,15,18,39,42,45,19,20,21} \
end{array}
\
hline
4 &
begin{array}{l}
4 \
8 \
12 \
16 \
20 \
24 \
28 \
end{array}
&
begin{array}{l}
{1,2,3,4} \
{6,8,10,12} \
{15,18,21,24} \
{28,32,36,40} \
{28,32,36,40,17,18,19,20} \
{28,32,36,40,38,40,42,44} \
{28,32,36,40,63,66,69,72} \
end{array}
\
hline
5 &
begin{array}{l}
5 \
10 \
15 \
20 \
25 \
30 \
35 \
end{array}
&
begin{array}{l}
{1,2,3,4,5} \
{7,9,11,13,15} \
{18,21,24,27,30} \
{34,38,42,46,50} \
{55,60,65,70,75} \
{55,60,65,70,75,26,27,28,29,30} \
{55,60,65,70,75,57,59,61,63,65} \
end{array}
\
hline
end{array}$







share|improve this answer






























    up vote
    2
    down vote













    Total@Take[Range@9, {#, -1, 3}] & /@ Range@3    



    {12, 15, 18}




    or..



    Total /@ Transpose@Partition[Range@9, 3]   



    {12, 15, 18}







    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








      up vote
      4
      down vote













      Total[Partition[Range[9], 3]]



      {12, 15, 18}




      Update for revised question:



      r = Range[18]    

      Total /@ Flatten[Partition[#, 3] & /@ {r[[1 ;; ;; 3]], r[[2 ;; ;; 3]], r[[3 ;; ;; 3]]}, 1]





      share|improve this answer























      • Hi Chris, sorry I wasn't clarify the problem clearly. I just updated my question would you still interest to help?
        – cj9435042
        Dec 5 at 1:30















      up vote
      4
      down vote













      Total[Partition[Range[9], 3]]



      {12, 15, 18}




      Update for revised question:



      r = Range[18]    

      Total /@ Flatten[Partition[#, 3] & /@ {r[[1 ;; ;; 3]], r[[2 ;; ;; 3]], r[[3 ;; ;; 3]]}, 1]





      share|improve this answer























      • Hi Chris, sorry I wasn't clarify the problem clearly. I just updated my question would you still interest to help?
        – cj9435042
        Dec 5 at 1:30













      up vote
      4
      down vote










      up vote
      4
      down vote









      Total[Partition[Range[9], 3]]



      {12, 15, 18}




      Update for revised question:



      r = Range[18]    

      Total /@ Flatten[Partition[#, 3] & /@ {r[[1 ;; ;; 3]], r[[2 ;; ;; 3]], r[[3 ;; ;; 3]]}, 1]





      share|improve this answer














      Total[Partition[Range[9], 3]]



      {12, 15, 18}




      Update for revised question:



      r = Range[18]    

      Total /@ Flatten[Partition[#, 3] & /@ {r[[1 ;; ;; 3]], r[[2 ;; ;; 3]], r[[3 ;; ;; 3]]}, 1]






      share|improve this answer














      share|improve this answer



      share|improve this answer








      edited Dec 5 at 2:16

























      answered Dec 5 at 0:10









      Chris

      54116




      54116












      • Hi Chris, sorry I wasn't clarify the problem clearly. I just updated my question would you still interest to help?
        – cj9435042
        Dec 5 at 1:30


















      • Hi Chris, sorry I wasn't clarify the problem clearly. I just updated my question would you still interest to help?
        – cj9435042
        Dec 5 at 1:30
















      Hi Chris, sorry I wasn't clarify the problem clearly. I just updated my question would you still interest to help?
      – cj9435042
      Dec 5 at 1:30




      Hi Chris, sorry I wasn't clarify the problem clearly. I just updated my question would you still interest to help?
      – cj9435042
      Dec 5 at 1:30










      up vote
      3
      down vote













      Using the six-argument form of Partition:



      Join @@ Partition[Partition[Range[9], 3], 3, 3, {1, 1}, {}, Plus]



      {12, 15, 18}




      Join @@ Partition[Partition[Range[18], 3], 3, 3, {1, 1}, {}, Plus]



      {12, 15, 18, 39, 42, 45}




      More generally,



      ClearAll[partsums]
      partsums[lst_List, n_Integer] := Join@@Partition[Partition[lst, n], n, n, {1,1}, {}, Plus]


      Examples:



      partsums[Range[18], 3]



      {12, 15, 18, 39, 42, 45}




      Grid[Prepend[Table[{i, Column[i Range[7]], Column[partsums[Range@#, i] & /@ 
      (i Range[7])]}, {i, {3, 4, 5}}], {"n", "Length@list" , "f[list, n]"}],
      Alignment -> Center, Dividers -> All] // TeXForm



      $smallbegin{array}{|c|c|c|}
      hline
      text{n} & text{Length@list} & text{f[list, n]} \
      hline
      3 &
      begin{array}{l}
      3 \
      6 \
      9 \
      12 \
      15 \
      18 \
      21 \
      end{array}
      &
      begin{array}{l}
      {1,2,3} \
      {5,7,9} \
      {12,15,18} \
      {12,15,18,10,11,12} \
      {12,15,18,23,25,27} \
      {12,15,18,39,42,45} \
      {12,15,18,39,42,45,19,20,21} \
      end{array}
      \
      hline
      4 &
      begin{array}{l}
      4 \
      8 \
      12 \
      16 \
      20 \
      24 \
      28 \
      end{array}
      &
      begin{array}{l}
      {1,2,3,4} \
      {6,8,10,12} \
      {15,18,21,24} \
      {28,32,36,40} \
      {28,32,36,40,17,18,19,20} \
      {28,32,36,40,38,40,42,44} \
      {28,32,36,40,63,66,69,72} \
      end{array}
      \
      hline
      5 &
      begin{array}{l}
      5 \
      10 \
      15 \
      20 \
      25 \
      30 \
      35 \
      end{array}
      &
      begin{array}{l}
      {1,2,3,4,5} \
      {7,9,11,13,15} \
      {18,21,24,27,30} \
      {34,38,42,46,50} \
      {55,60,65,70,75} \
      {55,60,65,70,75,26,27,28,29,30} \
      {55,60,65,70,75,57,59,61,63,65} \
      end{array}
      \
      hline
      end{array}$







      share|improve this answer



























        up vote
        3
        down vote













        Using the six-argument form of Partition:



        Join @@ Partition[Partition[Range[9], 3], 3, 3, {1, 1}, {}, Plus]



        {12, 15, 18}




        Join @@ Partition[Partition[Range[18], 3], 3, 3, {1, 1}, {}, Plus]



        {12, 15, 18, 39, 42, 45}




        More generally,



        ClearAll[partsums]
        partsums[lst_List, n_Integer] := Join@@Partition[Partition[lst, n], n, n, {1,1}, {}, Plus]


        Examples:



        partsums[Range[18], 3]



        {12, 15, 18, 39, 42, 45}




        Grid[Prepend[Table[{i, Column[i Range[7]], Column[partsums[Range@#, i] & /@ 
        (i Range[7])]}, {i, {3, 4, 5}}], {"n", "Length@list" , "f[list, n]"}],
        Alignment -> Center, Dividers -> All] // TeXForm



        $smallbegin{array}{|c|c|c|}
        hline
        text{n} & text{Length@list} & text{f[list, n]} \
        hline
        3 &
        begin{array}{l}
        3 \
        6 \
        9 \
        12 \
        15 \
        18 \
        21 \
        end{array}
        &
        begin{array}{l}
        {1,2,3} \
        {5,7,9} \
        {12,15,18} \
        {12,15,18,10,11,12} \
        {12,15,18,23,25,27} \
        {12,15,18,39,42,45} \
        {12,15,18,39,42,45,19,20,21} \
        end{array}
        \
        hline
        4 &
        begin{array}{l}
        4 \
        8 \
        12 \
        16 \
        20 \
        24 \
        28 \
        end{array}
        &
        begin{array}{l}
        {1,2,3,4} \
        {6,8,10,12} \
        {15,18,21,24} \
        {28,32,36,40} \
        {28,32,36,40,17,18,19,20} \
        {28,32,36,40,38,40,42,44} \
        {28,32,36,40,63,66,69,72} \
        end{array}
        \
        hline
        5 &
        begin{array}{l}
        5 \
        10 \
        15 \
        20 \
        25 \
        30 \
        35 \
        end{array}
        &
        begin{array}{l}
        {1,2,3,4,5} \
        {7,9,11,13,15} \
        {18,21,24,27,30} \
        {34,38,42,46,50} \
        {55,60,65,70,75} \
        {55,60,65,70,75,26,27,28,29,30} \
        {55,60,65,70,75,57,59,61,63,65} \
        end{array}
        \
        hline
        end{array}$







        share|improve this answer

























          up vote
          3
          down vote










          up vote
          3
          down vote









          Using the six-argument form of Partition:



          Join @@ Partition[Partition[Range[9], 3], 3, 3, {1, 1}, {}, Plus]



          {12, 15, 18}




          Join @@ Partition[Partition[Range[18], 3], 3, 3, {1, 1}, {}, Plus]



          {12, 15, 18, 39, 42, 45}




          More generally,



          ClearAll[partsums]
          partsums[lst_List, n_Integer] := Join@@Partition[Partition[lst, n], n, n, {1,1}, {}, Plus]


          Examples:



          partsums[Range[18], 3]



          {12, 15, 18, 39, 42, 45}




          Grid[Prepend[Table[{i, Column[i Range[7]], Column[partsums[Range@#, i] & /@ 
          (i Range[7])]}, {i, {3, 4, 5}}], {"n", "Length@list" , "f[list, n]"}],
          Alignment -> Center, Dividers -> All] // TeXForm



          $smallbegin{array}{|c|c|c|}
          hline
          text{n} & text{Length@list} & text{f[list, n]} \
          hline
          3 &
          begin{array}{l}
          3 \
          6 \
          9 \
          12 \
          15 \
          18 \
          21 \
          end{array}
          &
          begin{array}{l}
          {1,2,3} \
          {5,7,9} \
          {12,15,18} \
          {12,15,18,10,11,12} \
          {12,15,18,23,25,27} \
          {12,15,18,39,42,45} \
          {12,15,18,39,42,45,19,20,21} \
          end{array}
          \
          hline
          4 &
          begin{array}{l}
          4 \
          8 \
          12 \
          16 \
          20 \
          24 \
          28 \
          end{array}
          &
          begin{array}{l}
          {1,2,3,4} \
          {6,8,10,12} \
          {15,18,21,24} \
          {28,32,36,40} \
          {28,32,36,40,17,18,19,20} \
          {28,32,36,40,38,40,42,44} \
          {28,32,36,40,63,66,69,72} \
          end{array}
          \
          hline
          5 &
          begin{array}{l}
          5 \
          10 \
          15 \
          20 \
          25 \
          30 \
          35 \
          end{array}
          &
          begin{array}{l}
          {1,2,3,4,5} \
          {7,9,11,13,15} \
          {18,21,24,27,30} \
          {34,38,42,46,50} \
          {55,60,65,70,75} \
          {55,60,65,70,75,26,27,28,29,30} \
          {55,60,65,70,75,57,59,61,63,65} \
          end{array}
          \
          hline
          end{array}$







          share|improve this answer














          Using the six-argument form of Partition:



          Join @@ Partition[Partition[Range[9], 3], 3, 3, {1, 1}, {}, Plus]



          {12, 15, 18}




          Join @@ Partition[Partition[Range[18], 3], 3, 3, {1, 1}, {}, Plus]



          {12, 15, 18, 39, 42, 45}




          More generally,



          ClearAll[partsums]
          partsums[lst_List, n_Integer] := Join@@Partition[Partition[lst, n], n, n, {1,1}, {}, Plus]


          Examples:



          partsums[Range[18], 3]



          {12, 15, 18, 39, 42, 45}




          Grid[Prepend[Table[{i, Column[i Range[7]], Column[partsums[Range@#, i] & /@ 
          (i Range[7])]}, {i, {3, 4, 5}}], {"n", "Length@list" , "f[list, n]"}],
          Alignment -> Center, Dividers -> All] // TeXForm



          $smallbegin{array}{|c|c|c|}
          hline
          text{n} & text{Length@list} & text{f[list, n]} \
          hline
          3 &
          begin{array}{l}
          3 \
          6 \
          9 \
          12 \
          15 \
          18 \
          21 \
          end{array}
          &
          begin{array}{l}
          {1,2,3} \
          {5,7,9} \
          {12,15,18} \
          {12,15,18,10,11,12} \
          {12,15,18,23,25,27} \
          {12,15,18,39,42,45} \
          {12,15,18,39,42,45,19,20,21} \
          end{array}
          \
          hline
          4 &
          begin{array}{l}
          4 \
          8 \
          12 \
          16 \
          20 \
          24 \
          28 \
          end{array}
          &
          begin{array}{l}
          {1,2,3,4} \
          {6,8,10,12} \
          {15,18,21,24} \
          {28,32,36,40} \
          {28,32,36,40,17,18,19,20} \
          {28,32,36,40,38,40,42,44} \
          {28,32,36,40,63,66,69,72} \
          end{array}
          \
          hline
          5 &
          begin{array}{l}
          5 \
          10 \
          15 \
          20 \
          25 \
          30 \
          35 \
          end{array}
          &
          begin{array}{l}
          {1,2,3,4,5} \
          {7,9,11,13,15} \
          {18,21,24,27,30} \
          {34,38,42,46,50} \
          {55,60,65,70,75} \
          {55,60,65,70,75,26,27,28,29,30} \
          {55,60,65,70,75,57,59,61,63,65} \
          end{array}
          \
          hline
          end{array}$








          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited Dec 5 at 6:33

























          answered Dec 5 at 4:44









          kglr

          175k9197402




          175k9197402






















              up vote
              2
              down vote













              Total@Take[Range@9, {#, -1, 3}] & /@ Range@3    



              {12, 15, 18}




              or..



              Total /@ Transpose@Partition[Range@9, 3]   



              {12, 15, 18}







              share|improve this answer

























                up vote
                2
                down vote













                Total@Take[Range@9, {#, -1, 3}] & /@ Range@3    



                {12, 15, 18}




                or..



                Total /@ Transpose@Partition[Range@9, 3]   



                {12, 15, 18}







                share|improve this answer























                  up vote
                  2
                  down vote










                  up vote
                  2
                  down vote









                  Total@Take[Range@9, {#, -1, 3}] & /@ Range@3    



                  {12, 15, 18}




                  or..



                  Total /@ Transpose@Partition[Range@9, 3]   



                  {12, 15, 18}







                  share|improve this answer












                  Total@Take[Range@9, {#, -1, 3}] & /@ Range@3    



                  {12, 15, 18}




                  or..



                  Total /@ Transpose@Partition[Range@9, 3]   



                  {12, 15, 18}








                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered Dec 5 at 0:08









                  J42161217

                  3,712220




                  3,712220






























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