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Hi, did 2 changes, it passes syntax-checking now, but the result is
always 2.
regards
enzo
r=log(C/Ref(C,-1));
n=100; //n--period
i=0;
arun=1;
brun=1;
do
{
x=IIf(Ref(r,-i)>0,1,0);
y=IIf(Ref(r,-i-1)>0,1,0);
t=IIf(x-y==0,1,0); // t=IIf(x-y=0,1,0);
arun=arun+t;
maxr=Max(arun,brun);
brun=maxr;
IIf(t==0,arun=1,arun=arun); // if(t==0) arun=1;
i=i+1;
} while (i<n);
Plot(brun,"",1,1);
--- In amibroker@xxxxxxxxxxxxxxx, "liugq_99" <gliu@xxxx> wrote:
>
> Sorry too late back on this. Thanks for giving suggestions.
However,
> the result is not quite I'm looking for. I have written VBscript
> codes to try the job. The codes are shown as below. Again, thanks.
>
> liugq_99
>
>
> EnableScript("vbscript");
>
> ix=34;
>
> //----------------------------------
>
> r=log(c/Ref(c,-1))*100;
>
> <%
> rr=afl("r")
> ix=afl("ix")
> nsize=ubound(rr)
>
> arun=0;
> brun=0;
>
> for i=0 to ix
> xx=rr(i)
> yy=rr(i-1)
> arun=1
> if (xx == yy) arun=arun+1
> maxr=Max(arun,brun)
> brun=maxr
> i=i+1;
> next i
>
> afl(xy")=brun
> %>
>
> xy=brun;
>
> Plot(xy,"",1,1);
>
> --- In amibroker@xxxxxxxxxxxxxxx, "Christoper" <turkey@xxxx> wrote:
> >
> > I tried looking at your code and coudln' figure out why AB was
> > thinking i or n were arrays and not just single numbers.
> >
> > Having said that... I think you might be able to achiece what
you
> want
> > without loops. Take a look at this and let me know if this is
waht
> > you are looking for.
> >
> > Also, how are you using this?
> >
> > ----------start-----------
> > Lookback = 100;
> >
> > Runup=HHV(BarsSince(Cross(C, Ref(C, -1))) * (C > Ref(C,-1))+1,
> Lookback );
> > Rundown=HHV(BarsSince(Cross(Ref(C, -1), C)) * (Ref(C,-1) > C)+1,
> > Lookback );
> >
> > Plot(Runup,"Max Runup", colorGreen, styleLine);
> > Plot(rundown,"Max Rundown", colorRed, styleLine);
> > ----------end-------------
> >
> > --- In amibroker@xxxxxxxxxxxxxxx, "liugq_99" <gliu@xxxx> wrote:
> > >
> > > Hi,
> > >
> > > A maximum run is defined as the maximum number of days (daily
> data)
> > > consecutive runing up or running down for a certain time of
> period.
> > > For example, a 100 day maximum run being 5 means the maximum
> number
> > > of days consecutively running up (or down) is 5.
> > >
> > > I need a help to write a AFL code for this. Thanks.
> > >
> > > liugq_99
> > >
> > > My AFL code below has a "if" error:
> > >
> > >
> > > r=log(C/Ref(C,-1));
> > > n=100; //n--period
> > >
> > > i=0;
> > > arun=1;
> > > brun=1;
> > > do
> > > {
> > > x=IIf(Ref(r,-i)>0,1,0);
> > > y=IIf(Ref(r,-i-1)>0,1,0);
> > > t=IIf(x-y=0,1,0);
> > > arun=arun+t;
> > > maxr=Max(arun,brun);
> > > brun=maxr;
> > > if(t==0) arun=1;
> > > i=i+1;} while (i<n);
> > >
> > > Plot(brun,"",1,1);
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