Forum Discussion
Accumulated average
I want to accumulte the but as ans average, so : { 1, 1, 1 ,1} = 1 / 1 , 2 /2 , 3/3 , 4/ 4
so here the average is calculated in select;
let
alist =
let
alist = List.Numbers(1, 12, 1)
in
alist,
result = List.Generate(
() => [x = 0, y = alist{0}, z = y / (x + 1)],
each [x] < List.Count(alist),
each [x = [x] + 1, y = alist{x}, z = [z] + y],
each [z] / ([x] + 1)
)
in
result
but can i get the average in the function, i have tried various such as ;
= List.Generate
()=> [ x = 1 , y = alist {0} , z = y / x ] ,
each [x] < List.Count( alist ),
each [ x = [x] + 1, y = alist {x} , z = List.Sum( { [z], y} ) / x ],
each [z] )
i have played around with various x = 0 , x = 1 etc. but cannot get one to work?
I want the cuumulative average, the example i gave shows a correct result.
7 Replies
Not sure what you want for a result.
What result would you like to see for your List.Numbers(1,12,1) list?
Your code gives the average where for each row n, the average of values alist{0..n}Perhaps if you gave an example other than {1,1,1,1}??
I want the cuumulative average, the example i gave shows a correct result.
o as is said, but done within the fuunction, not select part of generate so;
Still not sure exactly what you want.
If you want a standalone function that will do the computation, then try:(alist as list)=> List.Generate( ()=>[a=alist{0}, idx=0], each [idx] < List.Count(alist), each [idx=[idx]+1, a = List.Sum(List.Range(alist,0, idx+1))/(idx+1)], each [a])You could use like:
let alist = List.Numbers(1,12,1), avg = fnRunningAvg(alist), tbl = Table.FromColumns({alist, avg}, type table[Numbers=Int64.Type, Running Avg = number]) in tbl
posssibel method adapted from post prodigy;
let alist = {1..12}, Custom1 = List.Generate( ()=> 1, each _ < List.Count( alist ) , each _ + 1, each List.Average( List.Range( alist, 0,_ )) ) in Custom1
Hi
List.Transform(
{1..List.Count(alist)},
each List.Average(List.FirstN(alist, _))
)List.Generate(
()=>[i=1, x=alist{0}],
each [i]<=List.Count(alist),
each [i=[i]+1, x= ([x]*[i]+alist{[i]}) / ([i]+1) ],
each [x]
)Stéphane
