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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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