List Interval Sum











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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
    7 mins ago












  • The length of list will always be Mod[Length[list],N] = 0, so no worry about corner cases
    – cj9435042
    4 mins ago

















up vote
2
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
    7 mins ago












  • The length of list will always be Mod[Length[list],N] = 0, so no worry about corner cases
    – cj9435042
    4 mins ago















up vote
2
down vote

favorite









up vote
2
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






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edited 1 hour ago

























asked 3 hours ago









cj9435042

33716




33716












  • 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
    7 mins ago












  • The length of list will always be Mod[Length[list],N] = 0, so no worry about corner cases
    – cj9435042
    4 mins ago




















  • 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
    7 mins ago












  • The length of list will always be Mod[Length[list],N] = 0, so no worry about corner cases
    – cj9435042
    4 mins ago


















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
7 mins ago






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
7 mins ago














The length of list will always be Mod[Length[list],N] = 0, so no worry about corner cases
– cj9435042
4 mins ago






The length of list will always be Mod[Length[list],N] = 0, so no worry about corner cases
– cj9435042
4 mins ago












2 Answers
2






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up vote
3
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
    1 hour ago


















up vote
1
down vote













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



{12, 15, 18}




or..



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



{12, 15, 18}







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    2 Answers
    2






    active

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    2 Answers
    2






    active

    oldest

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    active

    oldest

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    up vote
    3
    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
      1 hour ago















    up vote
    3
    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
      1 hour ago













    up vote
    3
    down vote










    up vote
    3
    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 1 hour ago

























    answered 3 hours ago









    Chris

    52116




    52116












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


















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
















    Hi Chris, sorry I wasn't clarify the problem clearly. I just updated my question would you still interest to help?
    – cj9435042
    1 hour ago




    Hi Chris, sorry I wasn't clarify the problem clearly. I just updated my question would you still interest to help?
    – cj9435042
    1 hour ago










    up vote
    1
    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
      1
      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
        1
        down vote










        up vote
        1
        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 3 hours ago









        J42161217

        3,597220




        3,597220






























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