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Gary,
The credit goes to Johan, I would not have thought
to do this in this manner. I was stumped.
Glen
<BLOCKQUOTE
>
----- Original Message -----
<DIV
>From:
Gary
A. Serkhoshian
To: <A title=amibroker@xxxxxxxxxxxxxxx
href="">amibroker@xxxxxxxxxxxxxxx
Sent: Sunday, March 21, 2004 8:29
PM
Subject: Re: [amibroker] Re: Help with
referencing previous prices
Glen,
I just read through all the posts on this thread, and I'm not exactly
sure what you are after but from my impressions your solution is
straightforward.
First, an explaination of the Error 3. IF, For, While only allow
for numbers NOT arrays. So, you can NOT do this:
If (C > 5)
{
blah;
}
That is why you are getting an Error 3. You could, however, use the
LastValue function to convert an array to a number. The problem is it
only returns the last value of the calculation.
You can do this:
x = 1;
If (x)
{
blahblah;
}
Tomasz has created this IIF function for these situations.
IIF(Sum(C > Ref(C, -1), 3) == 3, 1, -1);
Hope this helps as I'm not sure if you are trying to gate your code with
the condition (which IF will do barring the restriction to a number not an
array) or are just trying to generate some numbers from the results of
the condition (which IIF provides).
Kind Regards,
Gary
Glen Haponek <ghaponek@xxxxxxx> wrote:
<BLOCKQUOTE class=replbq
>
Johan,
So you are defining the relationship between
the three bars as a mathematical function of the element subscript, and the
condition is identified as true when the numeric value of y equals 3 ?
The first iteration through this is : x =
C [ i - k ] >C [ i -k -1 ];
x = C [ 3 - 0 ] > C [ 3 - 0 -1];
<FONT face=Arial
size=2>
= C [ 3 ] > C [ 2 ];
<FONT face=Arial
size=2> y
= 0 + 1 = 1
The second
:
x = C [ 4 - 1] > C [ 4 - 1 - 1]
<FONT face=Arial
size=2> =
C [ 3 ] > C [ 2 ];
<FONT face=Arial
size=2>
y = y + x = 1 + 1 = 2
The
third:
x = C [5 -2] > C [ 5 - 2 - 1] ;
<FONT face=Arial
size=2>
= C [ 3 ] > C [ 2] ;
<FONT face=Arial
size=2>
y = y + x = 2 + 1 = 3
and Cond [ i ] or Cond [ 3 ] == 3, and is
true, except that it looks like only C [ 3 ] > C [ 2 ] has been
tested.
Is this what you have performed here?
I initially described the relationship as
: C [ 3 ] > C [ 2 ] and C [ 2 ] > C [1 ]
and C [ 1 ] > C [ 0 ] , which isn't exactly what I think you have
done, but the specific price relationship wasn't the point of my
question. Thank you for your insight in this.
If I my interpretation is correct,
what would be changed to perform the above?
Might this work?
for ( i=3; i<BarCount ; i++){k = 0
;y = 0 ;
m = 0 ;
while( k <= 2
){x=C[i-k-m]>C[i-k-n-1];y=y+x;k++;
m = m++;
Cond[i]=y==3;}}
First iteration : x = C
[i-k-m]>C[i-k-m-1];
<FONT face=Arial
size=2> =
C [3-0-0] > C [3 -0 -0 -1]
<FONT face=Arial
size=2>
= C [ 3 ] > C [ 2 ]
Second
: x = C [i-k-m] >
C[i-k-n];
<FONT face=Arial
size=2> =
C [4 -1 -1] > C [ 4 - 1 - 1 -1]
<FONT face=Arial
size=2>
= C [ 2 ] > C [ 1 ]
Third
: x = C
[i-k-m] > C[i-k-n];
<FONT face=Arial
size=2> =
C [ 5 -2 -2] > C [ 5 - 2 -2 -1]
<FONT face=Arial
size=2>
= C [ 1 ] > C [ 0 ]
If the initial relationship were : C [ 3 ]
> C [ 2 ] and C [ 2 ] > C [ 1 ] and C [ 1 ]
< C [ 0 ]
How would this be addressed? Wouldn't the
algorithm have to be split into 2 parts?
Glen<FONT face=Arial
size=2>
<FONT
size=2>
<BLOCKQUOTE
>
----- Original Message -----
<DIV
>From:
johsun
To: <A
title=amibroker@xxxxxxxxxxxxxxx
href="">amibroker@xxxxxxxxxxxxxxx
Sent: Sunday, March 21, 2004 2:34
PM
Subject: [amibroker] Re: Help with
referencing previous prices
OK, guess I didn't read your message properly.Does
this help you in any
way:for(i=3;i<BarCount;i++){k=0;y=0;while(k<=2){x=C[i-k]>C[i-k-1];y=y+x;k++;Cond[i]=y==3;}}Plot(C,"",colorBlack,styleCandle);Plot(Cond,"",colorBlue,styleHistogram|styleOwnScale);JohanSend
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