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



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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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