Tukukkk

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:

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:

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.

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]

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

{12, 15, 18}

or..

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

{12, 15, 18}