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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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