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Fred, I don't doubt your expertise nor my own inexperience and this
line
"your oscillator regardless of what it is comprised of is no
different."
sure makes sense. I just have to put this aside now and check back
tomorrow (I'm ET).
Meanwhile I simplified the code, replaced the mysterious Oscillator
(nothing secret, it will just clutter the code further) with a simple
MA and used your StDvX definition (but turned it into a function to
save space).
Now a lot may be wrong (could be improved) in this code, it may
actually be total and complete Nonsense..... but one thing makes it
unique (so far): it gives the correct result.
I enjoy chewing on your pointers but the bottom line of course is to
find a code without loop (or at least much faster) that gives me
exactly the same result (plot) as this one.
No need (yet) to post the end result, but are you sure you can do
it?
Sure do appreciate your time & patience so far!!
// code start
function Randomize(a,b)
{ return Random(1)*(b-a)+a ; }
Cycle = int( Randomize(10,50) ) ;
n = BarIndex() + 1 ;
function StDvX(X)
{ return sqrt(Cum ((X - Cum(X)/n) ^ 2) / n) ; }
function Cycleconstant(number)
{ return Cum(0) + Cycle[ number ] ; }
function X(i)
{ return MA(C,Cycleconstant(i)) ; }
for (i = 0 ; i < BarCount ; i++ )
{
y = StDvX( X( i ) ) ;
FinalArray[ i ] = y[ i ] ;
}
Plot(FinalArray,"FinalArray",colorBlack);
// code end
--- In amibroker@xxxxxxxxxxxxxxx, "Fred" <ftonetti@xxxx> wrote:
> For example one of the things that StDev is a function of is how
many
> bars i.e. "n" ... right ?
>
> But in our example "n" is constantly changing ... on bar 1 it's
1 ...
> on bar 100 it's 100 ...
>
> Which is why I wrote ...
>
> n = BarIndex() + 1;
>
> n is used several ways ...
>
> It is used to get our Mean at each bar ...
>
> Mean = Cum(X) / n ...
>
> In the above calculation Cum(X) is an array ... so is Mean ... AND
SO
> IS "n" ...
>
> I don't think you are seeing this ... your oscillator regardless of
> what it is comprised of is no different.
>
> --- In amibroker@xxxxxxxxxxxxxxx, "Fred" <ftonetti@xxxx> wrote:
> > Nonsense ... you still don't get it
> >
> > --- In amibroker@xxxxxxxxxxxxxxx, "treliff" <treliff@xxxx> wrote:
> > > 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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