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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. 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) or
so.)
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 ] ; }
Can the StDvX definition create the FinalArray without a loop ?
--- 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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