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To use your example oscillator "X" of an MA with a randomly generated
Length between 10 and 50 ...
LV = 10;
HV = 50;
Length = int(Random() * (HV - LV) + LV);
X = MA(C, Length);
n = BarIndex() + 1;
CumX = Cum(X);
MeanX = CumX / n;
XMean = X - MeanX;
Mean2 = XMean ^ 2;
CumM2 = Cum(Mean2);
nCumM = CumM2 / n;
StDvX = sqrt(nCumM);
Plot(StDvX, "StDvX", colorWhite);
Filter = 1;
AddColumn(C, "Close", 1.5);
AddColumn(Length, "Length", 1.0);
AddColumn(X, "X", 1.5);
AddColumn(n, "n", 1.0);
AddColumn(CumX, "CumX", 1.5);
AddColumn(MeanX, "MeanX", 1.5);
AddColumn(XMean, "XMeanX", 1.5);
AddColumn(Mean2, "Mean2", 1.5);
AddColumn(CumM2, "CumM2", 1.5);
AddColumn(nCumM, "nCumX", 1.5);
AddColumn(StDvX, "StDevX", 1.5);
--- In amibroker@xxxxxxxxxxxxxxx, "Fred" <ftonetti@xxxx> wrote:
> What you need to do for starters so that you understand is
ELIMANATE
> ALL THE FUNCTIONS ...
>
> There's really no need for them ... This is straightline code.
>
> You may think you are simplifying things but instead you are
instead
> needlessly doing calculations FOR EVERY BAR that only need be done
> once ...
>
> PS ... I'm on EDT as well
>
> --- In amibroker@xxxxxxxxxxxxxxx, "treliff" <treliff@xxxx> wrote:
> > 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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