exponential moving average function for Q

[Wind]

I have found the ema function from the kxforums.com. But there is
some error. Maybe because of the difference between K and Q. Can
anybody help change the function for Q?

/ the original function for exponential moving average
[code:1]\ema:{a:2%x+1;b:1-a;c:(+/x#y)%x;((x-1)#0n),c{(a*y)+b*x}\x _ y}
[/code:1]

I have changed it to the following format, but the "/" maybe should
also be replaced:
ema:{[x;y] a:2%x+1;b:1-a;c:(+/x#y)%x;((x-1)#0n),c{(a*y)+b*x}\x _ y};

Thanks.

Wind

[Niall Dalton]

Untested, but something like ema:{[x;y]b:1-a:2%x+1;c:(sum x#y)%x;
((x-1)#0n),c,c{[a;b;c;d](a*d)+b*c}[a;b]\x _y} should provide a
starting point.

[Aaron Davies]

On Apr 10, 2009, at 11:10 AM, Wind wrote:

> ema:{a:2%x+1;b:1-a;c:(+/x#y)%x;((x-1)#0n),c{(a*y)+b*x}\x _ y}

here's a direct port from to k4:

k)ema:{a:2%x+1;b:1-a;c:(+/x#y)%x;((x-1)#0n),c{[a;b;x;y](a*y)+b*x}[a;b]
\x _ y}

k3 must have made locals available to nested lambdas

[Wind]

I copied Nial's ema function directly. It works well and is exactly
what I need.
Thanks Niall and Aaron.

[Attila Vrabecz]

m:{((x-1)#0n),i,{z+x*y}\[i:avg x#y;1-a;(x _y)*a:2%1+x]}
is a bit faster

q)x:1000000?10f
q)\t ema[100;x]
750
q)\t m[100;x]
531
Attila

[Xi Chen]

if just want the first one as intial value, one can  use

ewma:{{y+x*z-y}[x:2%1+x]\[y]}

Xi

[Wind]

I tried Xi and Attila's methods. All are perfect q codes.
Thanks again.

On Apr 10, 11:26 pm, Xi Chen <heydic...@gmail.com> wrote:
> if just want the first one as intial value, one can  use
> ewma:{{y+x*z-y}[x:2%1+x]\[y]}
> Xi
>

> On Fri, Apr 10, 2009 at 9:30 AM, Attila Vrabecz <attila.vrab...@gmail.com>wrote:
>
>
>
> > m:{((x-1)#0n),i,{z+x*y}\[i:avg x#y;1-a;(x _y)*a:2%1+x]}
> > is a bit faster
>
> > q)x:1000000?10f
> > q)\t ema[100;x]
> > 750
> > q)\t m[100;x]
> > 531
> >  Attila
>