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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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