I can not reproduce your timings. The following is what I got:
EMA=:((2 % [:>: #) $: ]) : (4 :'(((x * [) + (-.x) * ])/\.)&.|. y')
EMA1=:((<a:;1)&{)@:((([EMAIL PROTECTED] , (*/@[) + (1- [EMAIL PROTECTED]) *
{:@])/\.)&.|.)@:,.
recur=:[ (] * [:+/\ [ % ]) [: */\ ]
EMA2=:([EMAIL PROTECTED] , [ * [EMAIL PROTECTED]) recur (1, (<:@[EMAIL
PROTECTED]) $ [EMAIL PROTECTED])
ts=: 6!:2 , 7!:[EMAIL PROTECTED]
ts '0.1 EMA i.1e6'
3.74642 1.61362e8
ts '0.1 EMA1 i.1e6'
4.75599 6.29162e7
ts '0.1 EMA2 i.1e6'
2.41342 4.61384e7
EMA2 is faster, but not by a factor of 10.
----- Original Message -----
From: Xu Zuoqian <[EMAIL PROTECTED]>
Date: Saturday, March 8, 2008 20:34
Subject: Re: [Jprogramming] Is there a tacit version EMA?
To: Programming forum <[email protected]>
> Oh,Mike,your solution is so beautiful!
>
> recur=:[ (] * [:+/\ [ % ]) [: */\ ]
> EMA2=:([EMAIL PROTECTED] , [ * [EMAIL PROTECTED]) recur (1, (<:@[EMAIL
> PROTECTED]) $ [EMAIL PROTECTED])
>
> (0.1&EMA -: 0.1&EMA2) p:i.10
> 1
>
> 6!:2 '0.1&EMA i.1e6'
> 2.80876
>
> 6!:2 '0.1&EMA1 i.1e6'
> 3.781998
>
> 6!:2 '0.1&EMA2 i.1e6'
> 0.314285
>
> 2008/3/9, Mike Powell <[EMAIL PROTECTED]>:
> > Xu,
> >
> > Your EMA question is related to algorithms for linear
> recurrence> relations. There is a much loved algorithm
> (probably because it's
> > usually found to be lean and mean) that I first
> encountered at IP
> > Sharp in a newsletter quiz. We used it all over the place,
> > particularly wherever a present value or discounted cash
> flow was in
> > play. Where it came from before that, I know not. It
> goes like this.
> >
> > If you have a linear recurrence relation of the form:
> >
> > r[0] is m[0] times a[0]
> > r[i] is a[i] + m[i] times r[i-
> 1] (i >0)
> >
> > then this can be calculated in vector languages in the
> following way:
> >
> > a recur m;t
> > t is timesscan m
> > r is t times plusscan a div t
> >
> > I'd be most interested to see a J algorithm for this,
> particularly a
> > tacit one. I note that we have a reflexive piece (t) and
> that there
> > may be an opportunity for an inverse (* and %).
> >
> > As far as the EMA is concerned, I suggest Xu's original
> requirement is
> > solved with:
> >
> > (y[0] , k times 1 drop y) recur (length y) replicate 1 - k
> >
> >
> > Mike Powell
----------------------------------------------------------------------
For information about J forums see http://www.jsoftware.com/forums.htm
