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Dennis,
The 4 last peaks of H [for per=10] are
per=10;
//Plot(H,"H",colorBlack,1);Plot(Zig(H,per),"",colorYellow,1);
Title="";
for(J=1;J<5;J++)
{
X=ValueWhen(PeakBars(H,per)==0,H,J);
Title=Title+"\nH"+WriteVal(j,1.0)+"="+WriteVal(x);
}
Their average will be
per=5;av=0;j0=4;
Plot(H,"H",colorBlack,1);Plot(Zig(H,per),"",colorYellow,1);
Title="";
for(J=1;J<=4;J++)
{
X=ValueWhen(PeakBars(H,per)==0,H,J);
av=av+x;
Title=Title+"\nH"+WriteVal(j,1.0)+"="+WriteVal(x);
}
Title=Title+"\nHav="+WriteVal(av/j0);
Dimitris
--- In amibroker@xxxxxxxxxxxxxxx, "conn_dennis" <dennisconn@xxxx>
wrote:
>
> So was the question that stupid, or hasn't anyone ever attempted it?
>
> Anyone???
>
> --- In amibroker@xxxxxxxxxxxxxxx, "conn_dennis" <dennisconn@xxxx>
> wrote:
> >
> > I know that the median function will return the median value of
the
> > entire array, but how would one go about finding the median of
only
> > some of the values in an array?
> >
> > So far, I've managed only to identify the peaks in an oscillator
> I've
> > coded, but I'm trying to have the program find the median of ONLY
> the
> > four highest peaks within a user-defined lookback range (likewise
> > with the four lowest troughs below zero).
> >
> > I think I can stumble around and figure out how to code what I
need
> > once I get past this median problem. I've spent two days on this
> > trying to figure it out (I feel like Thomas Edison - I've
> discovered
> > a thousand ways to NOT find the median). : )
> >
> > Has anyone had a similar problem and found a somewhat simple
> > solution, or perhaps be willing to point me in some direction?
> >
> > DC
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