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Dicken's avatar
Dicken
Post Prodigy
2 months ago
Solved

Rolling accumulation

Hi, I have been trying various methods of a rolling accumultion or bucketiing;  so  buckets of 3  =  {1..10} =  {  {1}, {1,2}, { 1,2,3}, { 2,3,4} , {3,4,5}, {4,5,6}, {5,,6.7}, { 6,7,8} , { 7,8,9}, ...
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  • pcoley's avatar
    pcoley
    2 months ago

    Dicken please try with: 

    let
    alist = {1 .. 15},
    window = 3,
    pad = List.Repeat({0}, window - 1),           // {0, 0}
    padded = List.Combine({pad, alist}),          // {0,0,1,2,...,15}
    positions = {0 .. List.Count(alist) - 1},     // 0-based indices for final result
    buckets = List.Transform(
        positions,
        each List.Range(padded, _, window)        // always take exactly 'window' items
    )
in
    buckets
  • pcoley's avatar
    pcoley
    2 months ago

    Dicken Another option could be with List.Accumulate: 

    let
    alist = {1 .. 15},
    window = 3,
    buckets = List.Accumulate(
        {1 .. List.Count(alist)},
        {},
        (state, current) => 
            state & {
                if current <= window 
                then List.Range(alist, 0, current)
                else List.Range(alist, current - window, window)
            }
    )
in
    buckets
slorin's avatar
slorin
Super User
2 months ago

Hi
Another solution

= List.Transform(
   {1..List.Count(alist)},
   (i)=> List.Transform(
      List.Numbers(List.Max({0, i-window}), window),
      each alist{_}))


Stéphane