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[amibroker] Re: some looping help needed .......



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You removed the random part from the code I wrote ... but not from 
your code ...

--- In amibroker@xxxxxxxxxxxxxxx, "treliff" <treliff@xxxx> wrote:
> Both codes in one plot, Fred's in red, mine in blue. To make sure 
> Random() doesn't interfere I replaced it with a sturdy Cycle-array 
> based on a sine.  
> 
> Can somebody at least acknowledge the difference? Fred? Tomasz? 
> vlanschot? cpescho?
> 
> // code start
> 
> Cycle = int( 30 + 20 * sin( C ) ) ;
> Plot(Cycle,"Cycle",colorLightGrey,styleStaircase|styleOwnScale);
> 
> //Fred's code
> 
> //LV = 10;
> //HV = 50;
> //Length = int(Random() * (HV - LV) + LV);
> 
> Length = Cycle ;
> 
> X = MA(C, Length);
> 
> n = BarIndex() + 1;
> CumX = Cum(X);
> MeanX = CumX / n;
> XMean = X - MeanX;
> Mean2 = XMean ^ 2;
> CumM2 = Cum(Mean2);
> nCumM = CumM2 / n;
> StDvX = sqrt(nCumM);
> 
> Plot(StDvX, "StDvX_Fred", colorRed);
> 
> // my code
> 
> function Randomize(a,b)
> { return Random(1)*(b-a)+a ; }
> 
> Cycle = int( Randomize(10,50) ) ;
> 
> n = BarIndex() + 1 ;
> 
> function StDvX2(X)
> { return sqrt(Cum ((X - Cum(X)/n) ^ 2) / n) ; }
> 
> function Cycleconstant(number)
> { return Cum(0) + Cycle[ number ] ; }
> 
> function X2(i) 
> { return MA(C,Cycleconstant(i)) ; } 
> 
> for (i = 0 ; i < BarCount ; i++ )
> {
> y = StDvX2( X2( i ) ) ; 
> FinalArray[ i ] = y[ i ] ; 
> }
> 
> Plot(FinalArray,"StDvX_treliff",colorBlue);
> GraphZOrder = 1;
> 
> // code end
> 
> 
> --- In amibroker@xxxxxxxxxxxxxxx, "Fred" <ftonetti@xxxx> wrote:
> > To use your example oscillator "X" of an MA with a randomly 
> generated 
> > Length between 10 and 50 ...
> > 
> > LV = 10;
> > HV = 50;
> > 
> > Length = int(Random() * (HV - LV) + LV);
> > 
> > X     = MA(C, Length);
> > 
> > n     = BarIndex() + 1;
> > CumX  = Cum(X);
> > MeanX = CumX / n;
> > XMean = X - MeanX;
> > Mean2 = XMean ^ 2;
> > CumM2 = Cum(Mean2);
> > nCumM = CumM2 / n;
> > StDvX = sqrt(nCumM);
> > 
> > Plot(StDvX, "StDvX", colorWhite);
> > 
> > Filter = 1;
> > 
> > AddColumn(C,      "Close",  1.5); 
> > AddColumn(Length, "Length", 1.0); 
> > AddColumn(X,      "X",      1.5); 
> > 
> > AddColumn(n,      "n",      1.0); 
> > AddColumn(CumX,   "CumX",   1.5);  
> > AddColumn(MeanX,  "MeanX",  1.5);  
> > AddColumn(XMean,  "XMeanX", 1.5);  
> > AddColumn(Mean2,  "Mean2",  1.5);  
> > AddColumn(CumM2,  "CumM2",  1.5);  
> > AddColumn(nCumM,  "nCumX",  1.5);  
> > AddColumn(StDvX,  "StDevX", 1.5);  
> > 
> > 
> > 
> > --- In amibroker@xxxxxxxxxxxxxxx, "Fred" <ftonetti@xxxx> wrote:
> > > What you need to do for starters so that you understand is 
> > ELIMANATE 
> > > ALL THE FUNCTIONS ...
> > > 
> > > There's really no need for them ... This is straightline code.
> > > 
> > > You may think you are simplifying things but instead you are 
> > instead 
> > > needlessly doing calculations FOR EVERY BAR that only need be 
> done 
> > > once ...
> > > 
> > > PS ... I'm on EDT as well
> > > 
> > > --- In amibroker@xxxxxxxxxxxxxxx, "treliff" <treliff@xxxx> 
wrote:
> > > > Fred, I don't doubt your expertise nor my own inexperience 
and 
> > this 
> > > > line 
> > > > 
> > > > "your oscillator regardless of what it is comprised of is no 
> > > > different."
> > > > 
> > > > sure makes sense. I just have to put this aside now and 
check 
> > back 
> > > > tomorrow (I'm ET). 
> > > > 
> > > > Meanwhile I simplified the code, replaced the mysterious 
> > Oscillator 
> > > > (nothing secret, it will just clutter the code further) with 
a 
> > > simple 
> > > > MA and used your StDvX definition (but turned it into a 
> function 
> > to 
> > > > save space).
> > > > 
> > > > Now a lot may be wrong (could be improved) in this code, it 
may 
> > > > actually be total and complete Nonsense..... but one thing 
> makes 
> > it 
> > > > unique (so far): it gives the correct result.
> > > > 
> > > > I enjoy chewing on your pointers but the bottom line of 
course 
> is 
> > > to 
> > > > find a code without loop (or at least much faster) that 
gives 
> me 
> > > > exactly the same result (plot) as this one.  
> > > > 
> > > > No need (yet) to post the end result, but are you sure you 
can 
> do 
> > > > it?  
> > > > 
> > > > Sure do appreciate your time & patience so far!!
> > > > 
> > > > // code start
> > > > 
> > > > function Randomize(a,b)
> > > > { return Random(1)*(b-a)+a ; }
> > > > 
> > > > Cycle = int( Randomize(10,50) ) ;
> > > > 
> > > > n = BarIndex() + 1 ;
> > > > 
> > > > function StDvX(X) 
> > > > { return sqrt(Cum ((X - Cum(X)/n) ^ 2) / n) ; }
> > > > 
> > > > function Cycleconstant(number)
> > > > { return Cum(0) + Cycle[ number ] ; }
> > > > 
> > > > function X(i) 
> > > > { return MA(C,Cycleconstant(i)) ; } 
> > > > 
> > > > for (i = 0 ; i < BarCount ; i++ )
> > > > {
> > > > y = StDvX( X( i ) ) ; 
> > > > FinalArray[ i ] = y[ i ] ; 
> > > > }
> > > > 
> > > > Plot(FinalArray,"FinalArray",colorBlack);
> > > > 
> > > > // code end
> > > > 
> > > > 
> > > > --- In amibroker@xxxxxxxxxxxxxxx, "Fred" <ftonetti@xxxx> 
wrote:
> > > > > For example one of the things that StDev is a function of 
is 
> > how 
> > > > many 
> > > > > bars i.e. "n" ... right ?
> > > > > 
> > > > > But in our example "n" is constantly changing ... on bar 1 
> it's 
> > > > 1 ... 
> > > > > on bar 100 it's 100 ...
> > > > > 
> > > > > Which is why I wrote ... 
> > > > > 
> > > > > n = BarIndex() + 1;
> > > > > 
> > > > > n is used several ways ... 
> > > > > 
> > > > > It is used to get our Mean at each bar ...
> > > > > 
> > > > > Mean = Cum(X) / n ...
> > > > > 
> > > > > In the above calculation Cum(X) is an array ... so is 
> Mean ... 
> > > AND 
> > > > SO 
> > > > > IS "n" ...
> > > > > 
> > > > > I don't think you are seeing this ... your oscillator 
> > regardless 
> > > of 
> > > > > what it is comprised of is no different.
> > > > > 
> > > > > --- In amibroker@xxxxxxxxxxxxxxx, "Fred" <ftonetti@xxxx> 
> wrote:
> > > > > > Nonsense ... you still don't get it
> > > > > > 
> > > > > > --- In amibroker@xxxxxxxxxxxxxxx, "treliff" 
<treliff@xxxx> 
> > > wrote:
> > > > > > > Oscillator fits into a single dimension array, but it 
is 
> a 
> > > > > FUNCTION 
> > > > > > > of, among others, the bar number i, or better, it is a 
> > > function 
> > > > > of 
> > > > > > > Cycle[i].
> > > > > > > 
> > > > > > > Because Cycle varies from 10 to 50 we in fact have 41 
> > > different 
> > > > > > > Oscillator arrays:
> > > > > > > 
> > > > > > > Oscillator(10)
> > > > > > > Oscillator(11)
> > > > > > > .
> > > > > > > .
> > > > > > > Oscillator(50)
> > > > > > > 
> > > > > > > Could just as well be:
> > > > > > > 
> > > > > > > MA(C,10)
> > > > > > > .
> > > > > > > .
> > > > > > > .
> > > > > > > MA(C,50)
> > > > > > > 
> > > > > > > (well, MA doesn't really oscillate around zero but 
that 
> > > doesn't 
> > > > > > > matter) 
> > > > > > > 
> > > > > > > Now we arrive at bar 300 with Cycle[300] is, say, 27.
> > > > > > > Then I want FinalArray[300] to contain 
> > > > > > > 
> > > > > > > StDvX( MA(C,27) ) [300]
> > > > > > > 
> > > > > > > (this is not good code but just indicates: the 300th 
> array 
> > > > > element 
> > > > > > of 
> > > > > > > StDvX( MA(C,27) )
> > > > > > > 
> > > > > > > Next bar 301 has Cycle[301] which is 49.
> > > > > > > So FinalArray[301] should contain 
> > > > > > > 
> > > > > > > StDvX( MA(C,49) ) [301]
> > > > > > > 
> > > > > > > etc.
> > > > > > > 
> > > > > > > 
> > > > > > > --- In amibroker@xxxxxxxxxxxxxxx, "Fred" 
<ftonetti@xxxx> 
> > > wrote:
> > > > > > > > Wrong ... You can not have the equivalent of doubly 
> > > > dimensioned 
> > > > > > > > arrays i.e. tables in AFL ... one dimension is all 
you 
> > > > get ... 
> > > > > or 
> > > > > > > at 
> > > > > > > > least not without using something external i.e. 
Osaka 
> or 
> > > > ABTool 
> > > > > > > > plugins ... 
> > > > > > > > 
> > > > > > > > If you can get your "oscillator" into an array "X" 
then 
> > the 
> > > > AFL 
> > > > > I 
> > > > > > > > wrote will give you the standard deviation at each 
bar 
> > > using 
> > > > > all 
> > > > > > > the 
> > > > > > > > prior elements of X ( your osciallator ) ... That is 
> what 
> > > you 
> > > > > > were 
> > > > > > > > looking for, wasn't it ?  
> > > > > > > > 
> > > > > > > > Is there something about your osciallator that 
doesn't 
> > > allow 
> > > > it 
> > > > > > to 
> > > > > > > > fit into a single dimension array ?!  
> > > > > > > > 
> > > > > > > > --- In amibroker@xxxxxxxxxxxxxxx, "treliff" 
> > <treliff@xxxx> 
> > > > > wrote:
> > > > > > > > > More food for thought... I have to chew on all 
that 
> but 
> > > one 
> > > > > > thing 
> > > > > > > > > right away:
> > > > > > > > > 
> > > > > > > > > "X(i) is an ELEMENT of the array X."
> > > > > > > > > 
> > > > > > > > > NO: each X(i) is a separate array (otherwise it 
would 
> > be X
> > > > [i] 
> > > > > > > > right? 
> > > > > > > > > Or wrong?)
> > > > > > > > > 
> > > > > > > > > In my code:
> > > > > > > > > 
> > > > > > > > > Cycle is an array
> > > > > > > > > bar i has Cycle[i] 
> > > > > > > > > Cycleconstant(i) = Cum(0)+Cycle[i] is an array 
> > > > (a "constant" 
> > > > > > > array)
> > > > > > > > > X(i) = Oscillator(Cycleconstant(i))is an **ARRAY** 
> for 
> > > each 
> > > > i.
> > > > > > > > > 
> > > > > > > > > 
> > > > > > > > > --- In amibroker@xxxxxxxxxxxxxxx, "Fred" 
> > <ftonetti@xxxx> 
> > > > > wrote:
> > > > > > > > > > It appears you don't understand the array .vs. 
> > element 
> > > of 
> > > > > an 
> > > > > > > > array 
> > > > > > > > > > concept ...
> > > > > > > > > > 
> > > > > > > > > > Is the equation
> > > > > > > > > > 
> > > > > > > > > > X(i) = BarIndex() + i 
> > > > > > > > > > 
> > > > > > > > > > even meaningful ?  X(i) is an ELEMENT of the 
array 
> > X.  
> > > > > > BarIndex
> > > > > > > () 
> > > > > > > > > is 
> > > > > > > > > > an ARRAY.  How does one equate an ELEMENT of an 
> array 
> > > > i.e. X
> > > > > > (i) 
> > > > > > > > to 
> > > > > > > > > > the entire contents of another array i.e. 
BarIndex
> () 
> > + 
> > > a 
> > > > > > > modifier 
> > > > > > > > > i ?
> > > > > > > > > > 
> > > > > > > > > > Not doable, is it ?.  Further more why do you 
think 
> > you 
> > > > > need 
> > > > > > or 
> > > > > > > > > want 
> > > > > > > > > > to do this ?  With regards to ...
> > > > > > > > > > 
> > > > > > > > > > "Note however that in real life the X(i)'s are 
> > > > > independent.  
> > > > > > > > There 
> > > > > > > > > is 
> > > > > > > > > > no way to express X(i) in terms of X(i-1)"
> > > > > > > > > > 
> > > > > > > > > > Nor is there a need to ...
> > > > > > > > > > 
> > > > > > > > > > "Can the StDvX definition create the FinalArray 
> > without 
> > > a 
> > > > > > > loop ?"
> > > > > > > > > > 
> > > > > > > > > > Did you play with the code ? Look at the 
> > results ? ... 
> > > > > > Doesn't 
> > > > > > > it 
> > > > > > > > > do 
> > > > > > > > > > precisely that ?  It matters not what is in the 
> array 
> > > of 
> > > > X 
> > > > > in 
> > > > > > > > terms 
> > > > > > > > > > of being able to calc the StDev of it's elements 
> from 
> > > the 
> > > > > > first 
> > > > > > > > one 
> > > > > > > > > > to each bar along the way.  In fact that code 
with 
> > > minor 
> > > > > mods 
> > > > > > > > could 
> > > > > > > > > > be used to calc a variable length StDev based on 
a 
> > > > changing 
> > > > > > > value 
> > > > > > > > > of 
> > > > > > > > > > n where n was an array of elements based on 
> whatever 
> > > calc 
> > > > > one 
> > > > > > > > > wanted.
> > > > > > > > > > 
> > > > > > > > > > --- In amibroker@xxxxxxxxxxxxxxx, "treliff" 
> > > > <treliff@xxxx> 
> > > > > > > wrote:
> > > > > > > > > > > So far so good, but now suppose that the array 
in 
> > > > > question, 
> > > > > > > the 
> > > > > > > > > one 
> > > > > > > > > > > we need to calculate the standard deviation 
over, 
> > > > changes 
> > > > > > > with 
> > > > > > > > > each 
> > > > > > > > > > > bar. In other words, there is not one array 
> > > > > > > > > > > 
> > > > > > > > > > > X = BarIndex() + 100;
> > > > > > > > > > > 
> > > > > > > > > > > but there are different arrays like for example
> > > > > > > > > > > 
> > > > > > > > > > > X(i) = BarIndex() + i;
> > > > > > > > > > > 
> > > > > > > > > > > (In my code this would be Oscillator
(Cycleconstant
> > > (i)) 
> > > > > but 
> > > > > > > that 
> > > > > > > > > is 
> > > > > > > > > > > not of the essence. > 
> > > > > > > > > > > In my opinion this now is the remaining 
problem 
> and 
> > > the 
> > > > > > real 
> > > > > > > > time-
> > > > > > > > > > > consumer: 
> > > > > > > > > > > 
> > > > > > > > > > > for (i = 0 ; i < BarCount ; i++ )
> > > > > > > > > > > { y = StDvX( X( i ) ) ) ; 
> > > > > > > > > > >   FinalArray[ i ] = y[ i ] ; } 
> > > > > > > > > > > 
> > > > > > > > > > > > 
> > > > > > > > > > > 
> > > > > > > > > > > --- In amibroker@xxxxxxxxxxxxxxx, "Fred" 
> > > > <ftonetti@xxxx> 
> > > > > > > wrote:
> > > > > > > > > > > > The question simplifies to ... how do I 
> calculate 
> > > > > > standard 
> > > > > > > > > > > deviation 
> > > > > > > > > > > > at the current bar for all past values of 
some 
> > > array 
> > > > > > > without 
> > > > > > > > > > using 
> > > > > > > > > > > a 
> > > > > > > > > > > > loop, thereby eliminating the innermost loop 
> and 
> > > > > leaving 
> > > > > > > only 
> > > > > > > > > the 
> > > > > > > > > > > > outer one.
> > > > > > > > > > > > 
> > > > > > > > > > > > When looking at most problems like this 
where 
> the 
> > > > > > solution 
> > > > > > > > may 
> > > > > > > > > > not 
> > > > > > > > > > > be 
> > > > > > > > > > > > immediately obvious, the simplest way is to 
> break 
> > > the 
> > > > > > > problem 
> > > > > > > > > > down 
> > > > > > > > > > > > into its individual components and use 
EXPLORE 
> to 
> > > see 
> > > > > > that 
> > > > > > > > each 
> > > > > > > > > > > > calculation is doing what it's supposed to 
and 
> > from 
> > > > the 
> > > > > > > > > > perspective 
> > > > > > > > > > > > of speed it won't be any slower to do it 
this 
> > way, 
> > > in 
> > > > > > some 
> > > > > > > > > cases 
> > > > > > > > > > it 
> > > > > > > > > > > > may actually be faster i.e. here's the way 
most 
> > > > people 
> > > > > > > write 
> > > > > > > > a 
> > > > > > > > > > > > stochastic calc ...
> > > > > > > > > > > > 
> > > > > > > > > > > > Sto = (C - LLV(C, Length)) / (HHV(C, 
Length) - 
> LLV
> > > (C, 
> > > > > > > > Length));
> > > > > > > > > > > > 
> > > > > > > > > > > > The problem of course is that one has done 
the 
> > calc 
> > > > LLV
> > > > > > (C, 
> > > > > > > > > > Length) 
> > > > > > > > > > > > twice ... Simpler and of course faster is ...
> > > > > > > > > > > > 
> > > > > > > > > > > > LLVX = LLV(C, Length)
> > > > > > > > > > > > Sto = (C - LLVX) / (HHV(C, Length) - LLVX); 
> > > > > > > > > > > > 
> > > > > > > > > > > > Back to your problem ... 
> > > > > > > > > > > > 
> > > > > > > > > > > > StDevX = Sqrt(Cum ((X - Average(X)) ^ 2) / n)
> > > > > > > > > > > > 
> > > > > > > > > > > > Let's assume we want to see how to get the 
> > > > calculations 
> > > > > > > > correct 
> > > > > > > > > > at 
> > > > > > > > > > > > BarIndex() == 10 ( The 11th Bar ) without 
using 
> a 
> > > > loop 
> > > > > > and 
> > > > > > > > for 
> > > > > > > > > > the 
> > > > > > > > > > > > moment we won't care if the calc is correct 
at 
> BI
> > () 
> > > = 
> > > > 9 
> > > > > > or 
> > > > > > > 11 
> > > > > > > > > > > because 
> > > > > > > > > > > > we know we can always write a loop to go 
around 
> > all 
> > > > of 
> > > > > > this 
> > > > > > > > if 
> > > > > > > > > we 
> > > > > > > > > > > > need to ...
> > > > > > > > > > > > 
> > > > > > > > > > > > // Let's generate some simple dummy data "X" 
> to  
> > > > > > > > > > > > // use where we can easily eyeball the 
results
> > > > > > > > > > > > // "X" can always be replaced by something 
real
> > > > > > > > > > > > 
> > > > > > > > > > > > X = BarIndex() + 100;
> > > > > > > > > > > > 
> > > > > > > > > > > > // The components
> > > > > > > > > > > > 
> > > > > > > > > > > > n = BarIndex() + 1;
> > > > > > > > > > > > CumX  = Cum(X);
> > > > > > > > > > > > MeanX = CumX / n;
> > > > > > > > > > > > XMean = X - MeanX;
> > > > > > > > > > > > Mean2 = XMean ^ 2;
> > > > > > > > > > > > CumM2 = Cum(Mean2);
> > > > > > > > > > > > nCumM = CumM2 / n;
> > > > > > > > > > > > StDvX = sqrt(nCumM);
> > > > > > > > > > > > 
> > > > > > > > > > > > Filter = BarIndex() <= 10;
> > > > > > > > > > > > 
> > > > > > > > > > > > AddColumn(X, "X", 1.0);
> > > > > > > > > > > > AddColumn(n, "n", 1.0);
> > > > > > > > > > > > AddColumn(CumX,  "CumX", 1.0);
> > > > > > > > > > > > AddColumn(MeanX, "MeanX", 1.2);
> > > > > > > > > > > > AddColumn(XMean, "X-MeanX", 1.2);
> > > > > > > > > > > > AddColumn(Mean2, "Mean2", 1.2);
> > > > > > > > > > > > AddColumn(CumM2, "CumM2", 1.2);
> > > > > > > > > > > > AddColumn(nCumM, "nCumX", 1.2);
> > > > > > > > > > > > AddColumn(StDvX, "StDevX", 1.2);
> > > > > > > > > > > > 
> > > > > > > > > > > > Try taking what's above and running it as an 
> > > > > EXPLORE ... 
> > > > > > > see 
> > > > > > > > > the 
> > > > > > > > > > > > columns (below "hopefully") it shows i.e. 
one 
> for 
> > > > each 
> > > > > > > > > component 
> > > > > > > > > > > > including the data "X" ... It would appear 
that 
> > > > StDevX 
> > > > > is 
> > > > > > > > > correct 
> > > > > > > > > > > not 
> > > > > > > > > > > > only for BI() == 10 but for ALL the other 
bars 
> as 
> > > > well 
> > > > > > > > without 
> > > > > > > > > > ANY 
> > > > > > > > > > > > loops.
> > > > > > > > > > > > 
> > > > > > > > > > > > X	n	CumX	MeanX	X-MeanX	Mean2
	CumM2
> > > > 	nCumX
> > > > > > > > 	StDevX
> > > > > > > > > > > > 100	1	100	100.00	0.00	0.00
> 	0.00
> > > > > 	0.00
> > > > > > > > > 	0.00
> > > > > > > > > > > > 101	2	201	100.50	0.50	0.25
> 	0.25
> > > > > 	0.13
> > > > > > > > > 	0.35
> > > > > > > > > > > > 102	3	303	101.00	1.00	1.00
> 	1.25
> > > > > 	0.42
> > > > > > > > > 	0.65
> > > > > > > > > > > > 103	4	406	101.50	1.50	2.25
> 	3.50
> > > > > 	0.88
> > > > > > > > > 	0.94
> > > > > > > > > > > > 104	5	510	102.00	2.00	4.00
> 	7.50
> > > > > 	1.50
> > > > > > > > > 	1.22
> > > > > > > > > > > > 105	6	615	102.50	2.50	6.25
> 	13.75
> > > > > 	2.29
> > > > > > > > > 	1.51
> > > > > > > > > > > > 106	7	721	103.00	3.00	9.00
> 	22.75
> > > > > 	3.25
> > > > > > > > > 	1.80
> > > > > > > > > > > > 107	8	828	103.50	3.50	12.25
> 	35.00
> > > > > 	4.38
> > > > > > > > > 	2.09
> > > > > > > > > > > > 108	9	936	104.00	4.00	16.00
> 	51.00
> > > > > 	5.67
> > > > > > > > > 	2.38
> > > > > > > > > > > > 109	10	1045	104.50	4.50	20.25
> 	71.25
> > > > > 	7.13
> > > > > > > > > 	2.67
> > > > > > > > > > > > 110	11	1155	105.00	5.00	25.00
> 	96.25
> > > > > 	8.75
> > > > > > > > > 	2.96
> > > > > > > > > > > > 
> > > > > > > > > > > > Since the rest of your AFL doesn't require 
any 
> > > loops, 
> > > > > one 
> > > > > > > > would 
> > > > > > > > > > > > conclude that your AFL really needs NO loops 
at 
> > all.




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