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how about a simple loop, I
will just type one here so no guarantee it is completely
correct
Cond =
WhatYourCondIs;//make this binary ie 0 or 1
<FONT
face=Arial>Result=0;
for (
k=25;k>=10;k--)
{ <SPAN
class=564423400-20022004>if ( sum(Cond,k)>0) {Result = k;}
<FONT
face=Arial>}
FinalResult =
Result>0;
Cheers,Graham<A
href="">http://e-wire.net.au/~eb_kavan/
<FONT
face=Tahoma size=2>-----Original Message-----From: mrdavis9
[mailto:mrdavis9@xxxxxxxxxx] Sent: Friday, 20 February 2004 7:48
AMTo: amibroker@xxxxxxxxxxxxxxxSubject: Re: [amibroker]
Re: IsNotThere(Array or Ref(Array,-1) or Ref(Array,-1) or
I am going to code an exploration that
looks back over the last 25 daily bars to see if an Array condition has
been met on ALL of the last 25 bars. If it fails to find that
ALL of the last 25 bars are true, it will then look for 24 true bars,
then 23 true bars, and so on till it finally quits looking at about 10 bars
back. I can code this with lots of repititons of coding,
but I would like to find a more elegant way to code it.
I was unable to divise a way to do this with Sum() or
Cum(). Is there any other simple way to code this?
If loops is the way to go, then I would sure appreciate
seeing this coded with loops. I learn best by seeing a working example
of something that I am attempting to do. TIA Ron
D
<BLOCKQUOTE
>
----- Original Message -----
<DIV
>From:
johsun
To: <A title=amibroker@xxxxxxxxxxxxxxx
href="">amibroker@xxxxxxxxxxxxxxx
Sent: Wednesday, February 18, 2004
11:42 AM
Subject: [amibroker] Re:
IsNotThere(Array or Ref(Array,-1) or Ref(Array,-1) or
cond=Your conditionx=Sum(cond,4)>=1;this will
return True if cond has occurred in the last four days including today.
Is that what you wanted?JS--- In <A
href="">amibroker@xxxxxxxxxxxxxxx,
"mrdavis9" <mrdavis9@x...>
wrote:> Below is coding that allows the Seq10e2up2e10 OR
SeqR14toR14e15 conditions to have occurred upto three days before the
RSIEMA2cross occurs. Is there a lot more elegant and simple way to
do this?> > An IsNotThere(Array) would do what I
want. Is there one that I missed that can be made to work. I
still have not been able to grasp doing loops. If looping is a
good way to do this, I would sure appreciate seeing the below IIF coding
made into a looping function. Then, by knowing what it is designed
to do, I should be able to finally understand how looping works.
Ron D> > Day3a=IIf(Ref(Seq10e2up2e10 OR
SeqR14toR14e15,-3),1,0);> > Day3=IIf(Ref(Seq10e2up2e10 OR
SeqR14toR14e15,-3),1,0);> > Day2a=IIf(Ref(Seq10e2up2e10 OR
SeqR14toR14e15,-1),1,day3a);> > Day2=IIf(Ref(Seq10e2up2e10 OR
SeqR14toR14e15,-2),Day3,0);> > Day1a=IIf(Ref(Seq10e2up2e10 OR
SeqR14toR14e15,-1),Day2,Day2a);> > Day1=IIf(Ref(Seq10e2up2e10
OR SeqR14toR14e15,-1),Day2,0);> > Day0a=IIf(Seq10e2up2e10 OR
SeqR14toR14e15,Day1,Day1a);> >
Day0=IIf(RSIEMA2cross,Day0a,0);Send BUG REPORTS to <A
href="">bugs@xxxxxxxxxxxxxSend SUGGESTIONS
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