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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:
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];
= C [ 3 ] > C [ 2 ];
y = 0 + 1 = 1
The second : x = C [ 4 - 1] > C [ 4 - 1 - 1]
= C [ 3 ] > C [ 2 ];
y = y + x = 1 + 1 = 2
The third: x = C [5 -2] > C [ 5 - 2 - 1] ;
= C [ 3 ] > C [ 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];
= C [3-0-0] > C [3 -0 -0 -1]
= C [ 3 ] > C [ 2 ]
Second : x = C [i-k-m] > C[i-k-n];
= C [4 -1 -1] > C [ 4 - 1 - 1 -1]
= C [ 2 ] > C [ 1 ]
Third : x = C [i-k-m] > C[i-k-n];
= C [ 5 -2 -2] > C [ 5 - 2 -2 -1]
= 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
----- Original Message -----
From: johsun
To: 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 BUG REPORTS to bugs@xxxxxxxxxxxxxSend SUGGESTIONS to suggest@xxxxxxxxxxxxx-----------------------------------------Post AmiQuote-related messages ONLY to: amiquote@xxxxxxxxxxxxxxx (Web page: http://groups.yahoo.com/group/amiquote/messages/)--------------------------------------------Check group FAQ at: http://groups.yahoo.com/group/amibroker/files/groupfaq.html Send BUG REPORTS
to bugs@xxxxxxxxxxxxxSend SUGGESTIONS to suggest@xxxxxxxxxxxxx-----------------------------------------Post AmiQuote-related messages ONLY to: amiquote@xxxxxxxxxxxxxxx (Web page: http://groups.yahoo.com/group/amiquote/messages/)--------------------------------------------Check group FAQ at: http://groups.yahoo.com/group/amibroker/files/groupfaq.html Do you Yahoo!?
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