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Both codes in one plot, Fred's in red, mine in blue. To make sure
Random() doesn't interfere I replaced it with a sturdy Cycle-array
based on a sine.
Can somebody at least acknowledge the difference? Fred? Tomasz?
vlanschot? cpescho?
// code start
Cycle = int( 30 + 20 * sin( C ) ) ;
Plot(Cycle,"Cycle",colorLightGrey,styleStaircase|styleOwnScale);
//Fred's code
//LV = 10;
//HV = 50;
//Length = int(Random() * (HV - LV) + LV);
Length = Cycle ;
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_Fred", colorRed);
// my code
function Randomize(a,b)
{ return Random(1)*(b-a)+a ; }
Cycle = int( Randomize(10,50) ) ;
n = BarIndex() + 1 ;
function StDvX2(X)
{ return sqrt(Cum ((X - Cum(X)/n) ^ 2) / n) ; }
function Cycleconstant(number)
{ return Cum(0) + Cycle[ number ] ; }
function X2(i)
{ return MA(C,Cycleconstant(i)) ; }
for (i = 0 ; i < BarCount ; i++ )
{
y = StDvX2( X2( i ) ) ;
FinalArray[ i ] = y[ i ] ;
}
Plot(FinalArray,"StDvX_treliff",colorBlue);
GraphZOrder = 1;
// code end
--- In amibroker@xxxxxxxxxxxxxxx, "Fred" <ftonetti@xxxx> wrote:
> 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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