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From: "Fred"
> LOL ... Ok, if you want to deem that to be a multidimensional array,
> fine ...
>
I don't find it funny. Rude ? Yes.
Best regards,
Tomasz Janeczko
amibroker.com
----- Original Message -----
From: "Fred" <ftonetti@xxxxxxxxxxxxx>
To: <amibroker@xxxxxxxxxxxxxxx>
Sent: Monday, August 29, 2005 6:44 PM
Subject: [amibroker] Re: some looping help needed .......
> LOL ... Ok, if you want to deem that to be a multidimensional array,
> fine ...
>
> --- In amibroker@xxxxxxxxxxxxxxx, "Tomasz Janeczko" <amibroker@xxxx>
> wrote:
>> Hello,
>>
>> > Wrong ... You can not have the equivalent of doubly dimensioned
>> > arrays i.e. tables in AFL
>>
>> Actually.... you can. Via VarGet/VarSet
>> http://www.amibroker.com/f?varset
>> http://www.amibroker.com/f?varget
>>
>> you can create equivalent of arrays of any dimension simply using
> dynamic variables.
>>
>>
>> Best regards,
>> Tomasz Janeczko
>> amibroker.com
>> ----- Original Message -----
>> From: "Fred" <ftonetti@xxxx>
>> To: <amibroker@xxxxxxxxxxxxxxx>
>> Sent: Monday, August 29, 2005 5:11 AM
>> Subject: [amibroker] Re: some looping help needed .......
>>
>>
>> > 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.
>> >
>> >
>> >
>> >
>> >
>> > Please note that this group is for discussion between users only.
>> >
>> > To get support from AmiBroker please send an e-mail directly to
>> > SUPPORT {at} amibroker.com
>> >
>> > For other support material please check also:
>> > http://www.amibroker.com/support.html
>> >
>> >
>> > Yahoo! Groups Links
>> >
>> >
>> >
>> >
>> >
>> >
>> >
>
>
>
>
>
>
> Please note that this group is for discussion between users only.
>
> To get support from AmiBroker please send an e-mail directly to
> SUPPORT {at} amibroker.com
>
> For other support material please check also:
> http://www.amibroker.com/support.html
>
>
> Yahoo! Groups Links
>
>
>
>
>
>
>
>
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