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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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