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Louis,
You need to understand more of the basics of AFL, as the problem
you're mainly having now has nothing to do with linear regression but
rather just AFL language structure.
A function is a separate entity to the main code, and you need to
"return" the result from the function to use it elsewhere. Also, your
main code needs to call the function for it to do anything.
That function is not a correlation function, but one that calculates
the linear regression slope and intercept (if I interpreted the
formula correctly). The array 'b' is the slope, and 'a' the intercept.
What I called "line" and you've renamed to "slope" is the formula for
the resulting line in array form.
The parameters 'x' and 'y' are the generic coordinates for the data
points. In this case, 'x' would be the bar number (BarIndex()) and 'y'
the price (Close), giving a slope in dollars per bar.
So to finish the function correctly, assuming you just want the slope,
add the statement:
return b;
The "a + b*x" line can be removed, as it was only to indicate the
formula for the resulting line.
Then in the main code, add the lines:
range = HHVBars(High, 20);
slope = CorrV(BarIndex(), Close, range);
and use that "slope" variable in the arc-tangent formula to get an
array of slopes in degrees.
Tomasz indicated elsewhere that the LinRegSlope function can take an
array as the period, so you probably don't need your own function.
Also try:
slope = LinRegSlope(Close, HHVBars(High,20));
Regards,
GP
--- In amibroker@xxxxxxxxxxxxxxx, "Louis P." <rockprog80@xxx> wrote:
>
> Hi again,
>
> I did a small mistake in the last code. Here it is:
>
> range = HHVBars(High, 20);
> function CorrV( x, y, range )
> {
>
> avX = MA(x, range);
> avY = MA(y, range);
> avXY = MA(x*y, range);
> axXX = MA(x*x, range);
>
> b = (avXY - avX*avY) / (avXX - avX*avX);
> a = axY - b*avX;
>
> slope = a + b*x;
>
> }
>
> PI=3.14159265358979;
>
> Deg = atan(slope)*180/PI;
>
> Plot (deg,"deg",colorRed);
>
> I think I need to somehow "close" the function, but I don't know how
to do
> it. This code does not seem to work.
>
> Louis
>
> 2008/9/23 Louis P. <rockprog80@xxx>
>
> > Hi again,
> >
> > Here is a code I tested:
> >
> > range = HHVBars(High, 20);
> > function CorrV( x, y, range )
> > {
> >
> > avX = MA(x, range);
> > avY = MA(y, range);
> > avXY = MA(x*y, range);
> > axXX = MA(x*x, range);
> >
> > b = (avXY - avX*avY) / (avXX - avX*avX);
> > a = axY - b*avX;
> >
> > slope = a + b*x;
> >
> > }
> >
> > PI=3.14159265358979;
> >
> > Deg = atan(range)*180/PI;
> >
> > Plot (deg,"deg",colorRed);
> >
> >
> > Unfortunately it does not seem to work. Degree is very often
between 75
> > and 85 where I see a slope that is not that steep! What could be
wrong?
> >
> > Thanks,
> >
> > Louis
> >
> > 2008/9/23 Louis P. <rockprog80@xxx>
> >
> > Hi,
> >>
> >> First of all, thank you very much for your help. I have been
struggling
> >> with this for the last month, and I feel like I am close to get to
> >> something.
> >>
> >> @Barry: I did use the following
> >>
> >> vHHV = HHV(c,20);
> >> hhvValue = SelectedValue(Ref(vHHV, BarsSince(vHHV)));
> >> printf("HHV Value " + NumToStr(HHVValue, 1.2));
> >>
> >> but it does not seem to work, becasue I get a blank screen when
using it.
> >> And I feel (but I may be wrong) that the "Selectedvalue" might
still be a
> >> problem, because I want a formula that can be applied to any bar
on the
> >> chart, without having selected any one (something backtestable).
> >>
> >>
> >> @Tony: Before using that formula
> >>
> >> PI=3.14159265358979;
> >>
> >> Deg = atan(Slope)*180/PI;
> >>
> >>
> >> if I understand correctly I'd need to figure out what will be
"slope" and
> >> that "slope" should be the Linear Regression as gp_sydney said.
So I am
> >> still stuck as I have trouble establishing the LR outside of the
> >> selectedvalue.
> >>
> >> I understand what you say about the concept: it seems weird to
want to
> >> draw lines from each HHV to each bar; of course it would look
like a fan and
> >> would be useless. But this is not what I want; I don't want to
use it as a
> >> chart, but rather has something that can be backtested. I want
each bar to
> >> consider only its "own line" starting from its own HHV (c,X)
going to its
> >> own Close. The problem with Selectedvalue is that it consider
where I am
> >> on the chart, but as I see things I want to backtest the data so
I'd need
> >> something that can be used without seeing the chart.
> >>
> >> I'll try to explain again in different works.
> >>
> >> For any particular bar in the array (and without looking at the
chart; I
> >> want the whole thing to be backtestable) I want to get its own Linear
> >> Regression and then calculate the slope of that LR to get the
angle and the
> >> degree. Then, maybe I could use Rsquared to check if the data
fits the
> >> angle or if there are bars very high over the LR and very low
under that
> >> same LR?
> >>
> >> @gp_sydney:
> >>
> >> The formula you posted is this:
> >>
> >> n = HHVBars(High, periods);
> >> avX = MA(x, n);
> >> avY = MA(y, n);
> >> avXY = MA(x*y, n);
> >> axXX = MA(x*x, n);
> >>
> >> b = (avXY - avX*avY) / (avXX - avX*avX);
> >> a = axY - b*avX;
> >>
> >> line = a + b*x;
> >>
> >> Please forgive my ignorance (I have really trouble to work with this
> >> stuff), but I tried to use it but I got messages saying that x
variable had
> >> not been initialized, etc.
> >>
> >> But if I understand correctly, if I can make this thing work I could
> >> simply use the "line" and put it in place of "slope" in Tony's
formula,
> >> right?
> >>
> >>
> >> Thank you very much for your help!
> >>
> >> Louis
> >>
> >>
> >>
> >>
> >> 2008/9/22 gp_sydney <gp.investment@xxx>
> >>
> >> Louis,
> >>>
> >>> All I can suggest is using the LinRegSlope function to get the slope
> >>> array (or calculate it yourself using the other formula I
posted), and
> >>> then the arc-tangent formula to convert it to degrees. That will
give
> >>> you an array where for every bar, the value will be the slope of the
> >>> regression line that (I think) you want in degrees.
> >>>
> >>> Regards,
> >>> GP
> >>>
> >>> --- In amibroker@xxxxxxxxxxxxxxx <amibroker%40yahoogroups.com>,
"Louis
> >>> P." <rockprog80@> wrote:
> >>> >
> >>> > Hi,
> >>> >
> >>> > @Tony Grimes: I don't want a straight line between two points; I
> >>> want the LR
> >>> > from the HHV to each point. Let's say that bar 1 is the HHV, then
> >>> I'd like
> >>> > LR for bar 3, for bar 4, for bar 5, etc. until there is a new
HHV. The
> >>> > problem with the code I use is that it uses "selectedvalue" which
> >>> mean it
> >>> > can only draw the line from where I am on the chart; I'd like that
> >>> for each
> >>> > and every bar so it can be backtested! BTW, how would you use
that:
> >>> Deg =
> >>> > atan(Slope)*180/PI;? I am not sure how to set the variables
> >>> straight so it
> >>> > can work... Thanks!
> >>> >
> >>> > @gp_sydney: I know the 1$/bar idea is a bit stupid, but this
is the
> >>> only
> >>> > thing I have found for now. I asked support for help on this, but
> >>> of course
> >>> > they don't do coding but they told me I had to be able to set a Y
> >>> and an X
> >>> > that is not relative so I could draw an angle from this. Can you
> >>> think of
> >>> > something else to do the trick?
> >>> >
> >>> > Thanks,
> >>> >
> >>> > Louis
> >>> >
> >>> > 2008/9/20 gp_sydney <gp.investment@>
> >>> >
> >>> > > Louis,
> >>> > >
> >>> > > You need to explain how you define steepness. The regression
formula
> >>> > > gives the slope in dollars per bar. What units do you want
it in?
> >>> > >
> >>> > > When you talk of the slope in degrees, how do you define
that? For
> >>> > > example, 45 degrees is delta Y (price) and delta X (bars)
being the
> >>> > > same, or $1 / bar. That seems like a pretty useless figure to me
> >>> > > though, as it would be shallow on the graph of a $200 stock
but very
> >>> > > steep on the graph of a 20 cent stock (which is why I'd
probably be
> >>> > > using semi-log and trying to do linear regression based on a log
> >>> scale
> >>> > > with percentage gain per bar - but even then, what would 45
degrees
> >>> > > be, 1% per bar?).
> >>> > >
> >>> > >
> >>> > > GP
> >>> > >
> >>> > > --- In amibroker@xxxxxxxxxxxxxxx
<amibroker%40yahoogroups.com><amibroker%
> >>> 40yahoogroups.com>,
> >>> "Louis P."
> >>> > > <rockprog80@> wrote:
> >>> > > >
> >>> > > > Hi,
> >>> > > >
> >>> > > > @Tony:Sorry I thought gradient was the good word. Maybe I
meant
> >>> > > angle, or
> >>> > > > degree? I want a LR slope of approx. 30-50 degrees. How
can I do
> >>> that?
> >>> > > > @gp_sydney: I need to go to work, but will check this
tomorrow or
> >>> > > Monday. I
> >>> > > > looked very quickly and was wondering where in that
formula can I
> >>> > > determine
> >>> > > > how steep the slope will be. That would help a lot!
> >>> > > >
> >>> > > > Thank you you two a lot!
> >>> > > >
> >>> > > > Louis
> >>> > > >
> >>> > > >
> >>> > > >
> >>> > > > 2008/9/20 gp_sydney <gp.investment@>
> >>> > > >
> >>> > > > > Sorry, a couple of typos in the formula. It should be:
> >>> > > > >
> >>> > > > >
> >>> > > > > n = HHVBars(High, periods);
> >>> > > > > avX = MA(x, n);
> >>> > > > > avY = MA(y, n);
> >>> > > > > avXY = MA(x*y, n);
> >>> > > > > avXX = MA(x*x, n);
> >>> > > > >
> >>> > > > > b = (avXY - avX*avY) / (avXX - avX*avX);
> >>> > > > > a = avY - b*avX;
> >>> > > > >
> >>> > > > > GP
> >>> > > > >
> >>> > > > >
> >>> > > > > --- In amibroker@xxxxxxxxxxxxxxx
<amibroker%40yahoogroups.com>
> >>> <amibroker%40yahoogroups.com><amibroker%
> >>>
> >>> > > 40yahoogroups.com>,
> >>> > >
> >>> > > > > "gp_sydney" <gp.investment@> wrote:
> >>> > > > > >
> >>> > > > > > Louis,
> >>> > > > > >
> >>> > > > > > I gather you want to use regression to get a best-fit line
> >>> from the
> >>> > > > > > most-recent HHV, not just draw a line between the end
points.
> >>> > > From my
> >>> > > > > > understanding of regression though, which isn't a lot, I
> >>> gather that
> >>> > > > > > such a line may not actually pass through either end
point, so
> >>> > > may not
> >>> > > > > > go "from the HHV" exactly or end at the current bar
exactly
> >>> (to get
> >>> > > > > > that, just draw a line between those two points as
previously
> >>> > > > > > mentioned). Once you have the line though, you could
offset it
> >>> > > > > > vertically to get it to pass exactly through the HHV or
> >>> current bar
> >>> > > > > > (but not both of course).
> >>> > > > > >
> >>> > > > > > From a response Tomasz gave you in another thread
about how to
> >>> > > > > > calculate Rsquared without a loop:
> >>> > > > > >
> >>> > > > > >
http://finance.groups.yahoo.com/group/amibroker/message/129392
> >>> > > > > >
> >>> > > > > > you can similarly do the regression itself. According to
> >>> Tomasz,
> >>> > > > > > LinRegSlope already accepts variable periods (ie. arrays),
> >>> so to get
> >>> > > > > > the slope you should be able to just use that
function. Or to
> >>> > > > > > calculate both coefficients yourself, then something
like this
> >>> > > (based
> >>> > > > > > on the formula given at
http://www.educalc.net/2104083.page):
> >>> > > > > >
> >>> > > > > > n = HHVBars(High, periods);
> >>> > > > > > avX = MA(x, n);
> >>> > > > > > avY = MA(y, n);
> >>> > > > > > avXY = MA(x*y, n);
> >>> > > > > > axXX = MA(x*x, n);
> >>> > > > > >
> >>> > > > > > b = (avXY - avX*avY) / (avXX - avX*avX);
> >>> > > > > > a = axY - b*avX;
> >>> > > > > >
> >>> > > > > > Hopefully I've got that right (I haven't tested it).
The line
> >>> > > formula
> >>> > > > > > is then:
> >>> > > > > >
> >>> > > > > > line = a + b*x;
> >>> > > > > >
> >>> > > > > > The slope is 'b' and the intercept 'a'. The slope is in
> >>> dollars per
> >>> > > > > > bar. I'm not sure what you want by a percentage slope
> >>> though, as the
> >>> > > > > > 'x' and 'y' axes are in completely different units.
> >>> > > > > >
> >>> > > > > > Also note that the calculated lines would only be
straight on a
> >>> > > linear
> >>> > > > > > price scale (if they were plotted). If you use semi-log,
> >>> then the
> >>> > > > > > formula for a straight line is different - and how to do
> >>> regression
> >>> > > > > > for it would take some figuring out.
> >>> > > > > >
> >>> > > > > > Hope this helps.
> >>> > > > > >
> >>> > > > > > Regards,
> >>> > > > > > GP
> >>> > > > > >
> >>> > > > > >
> >>> > > > > > --- In amibroker@xxxxxxxxxxxxxxx
<amibroker%40yahoogroups.com>
> >>> <amibroker%40yahoogroups.com><amibroker%
> >>>
> >>> > > 40yahoogroups.com>,
> >>> > >
> >>> > > "Louis
> >>> > > > > P." <rockprog80@> wrote:
> >>> > > > > > >
> >>> > > > > > > Hi,
> >>> > > > > > >
> >>> > > > > > > I need the gradient of the slope, and for each bar. This
> >>> is where
> >>> > > > > it is
> >>> > > > > > > difficult...
> >>> > > > > > >
> >>> > > > > > > Thanks,
> >>> > > > > > >
> >>> > > > > > > Louis
> >>> > > > > > >
> >>> > > > > > > 2008/9/19 Tony Grimes <Tonez.Email@>
> >>> > > > > > >
> >>> > > > > > > > Louis,
> >>> > > > > > > >
> >>> > > > > > > > If your looking for the slope & difference between
HHV of
> >>> 20
> >>> > > bars
> >>> > > > > > & the
> >>> > > > > > > > current close, all you should need is this:
> >>> > > > > > > >
> >>> > > > > > > > Pds=20;
> >>> > > > > > > >
> >>> > > > > > > > LastHighBar = HHVBars(High, Pds);
> >>> > > > > > > > LastHighVal = HHV(High, Pds);
> >>> > > > > > > >
> >>> > > > > > > > Slope = IIf(LastHighBar,(Close - LastHighVal) /
> >>> LastHighBar,0);
> >>> > > > > > > > CloseDiff = Close - Ref(Close, -LastHighBar);
> >>> > > > > > > >
> >>> > > > > > > >
> >>> > > > > > > > On Fri, Sep 19, 2008 at 4:51 PM, Louis P.
<rockprog80@>
> >>> wrote:
> >>> > > > > > > >
> >>> > > > > > > >> Hi Tony,
> >>> > > > > > > >>
> >>> > > > > > > >> Thank you a lot for your response. I'm still very
weak
> >>> with
> >>> > > > > > loops. Last
> >>> > > > > > > >> time experimented with one, I had to reboot my
> >>> computer! :) So
> >>> > > > > > do you know
> >>> > > > > > > >> how such a loop could work? And if I run, let's say 2
> >>> minutes
> >>> > > > > > bar for one
> >>> > > > > > > >> year, wouldn't that be really really long to deal
with?
> >>> I have
> >>> > > > > > PIV with 1.5
> >>> > > > > > > >> GHz ram.
> >>> > > > > > > >>
> >>> > > > > > > >> I am looking for a line that ends at each bar and
that
> >>> starts
> >>> > > > > > from the HHV
> >>> > > > > > > >> of 20 bars, and I want to do things with this
bar: e.g.
> >>> compare
> >>> > > > > > the closes
> >>> > > > > > > >> between current bar and the HHV to the bar and
> >>> establish the
> >>> > > > > > gradient of
> >>> > > > > > > >> that linear regression bar for each bar.
> >>> > > > > > > >>
> >>> > > > > > > >> Thanks a lot!
> >>> > > > > > > >>
> >>> > > > > > > >> Louis
> >>> > > > > > > >>
> >>> > > > > > > >> 2008/9/19 Tony Grimes <Tonez.Email@>
> >>> > > > > > > >>
> >>> > > > > > > >>> Hi Louis,
> >>> > > > > > > >>>
> >>> > > > > > > >>> A loop will work, but how slow - it depends
(Speed of
> >>> your
> >>> > > > > computer,
> >>> > > > > > > >>> number of bars loaded, how many loops your using
etc...).
> >>> > > > > > Without seeing
> >>> > > > > > > >>> what your actually looking for (The end result), I
> >>> think you
> >>> > > > > > could do it
> >>> > > > > > > >>> with one loop, with only one pass through the loop.
> >>> The speed
> >>> > > > > > should be OK.
> >>> > > > > > > >>>
> >>> > > > > > > >>> Good Luck.
> >>> > > > > > > >>>
> >>> > > > > > > >>> Tony
> >>> > > > > > > >>>
> >>> > > > > > > >>> On Fri, Sep 19, 2008 at 2:47 PM, Louis P.
> >>> <rockprog80@> wrote:
> >>> > > > > > > >>>
> >>> > > > > > > >>>> Hi Tony,
> >>> > > > > > > >>>>
> >>> > > > > > > >>>> Thanks for the tips. Basically, I'd need a loop and
> >>> use it on
> >>> > > > > > each and
> >>> > > > > > > >>>> every bar of the array to determine the LR, right?
> >>> > > > > > > >>>>
> >>> > > > > > > >>>> That will slow down my computer a lot, don't
you think?
> >>> > > > > > > >>>>
> >>> > > > > > > >>>> Thanks,
> >>> > > > > > > >>>>
> >>> > > > > > > >>>> Louis
> >>> > > > > > > >>>>
> >>> > > > > > > >>>> 2008/9/16 Tony Grimes <Tonez.Email@>
> >>> > > > > > > >>>>
> >>> > > > > > > >>>>> SelectedValue takes an array ( of numbers) and
returns
> >>> a
> >>> > > > > single
> >>> > > > > > > >>>>> number based on the bar that is selected in
the chart.
> >>> > > > > > > >>>>>
> >>> > > > > > > >>>>> The first formula worked because SelectedValue was
> >>> > > giving you a
> >>> > > > > > > >>>>> number.
> >>> > > > > > > >>>>>
> >>> > > > > > > >>>>> Look at it this way: Array --> SelectedValue --->
> >>> Number.
> >>> > > > > > > >>>>>
> >>> > > > > > > >>>>> Remove SelectedValue: Array---->Array.
> >>> > > > > > > >>>>>
> >>> > > > > > > >>>>> You can draw a line with single numbers, but not
> >>> arrays.
> >>> > > > > > > >>>>>
> >>> > > > > > > >>>>> You can always use a loop.
> >>> > > > > > > >>>>>
> >>> > > > > > > >>>>> You might want to read: Understanding how AFL
works,
> >>> in the
> >>> > > > > > Amibroker
> >>> > > > > > > >>>>> users guide. Until you really understand AFL &
array
> >>> > > > > > processing, you are
> >>> > > > > > > >>>>> going to keep running into these problems, which
> >>> will just
> >>> > > > > > slow you down.
> >>> > > > > > > >>>>>
> >>> > > > > > > >>>>>
> >>> > > > > > > >>>>> On Tue, Sep 16, 2008 at 10:34 PM, Louis P.
> >>> > > <rockprog80@>wrote:
> >>> > > > > > > >>>>>
> >>> > > > > > > >>>>>> Hi Tony,
> >>> > > > > > > >>>>>>
> >>> > > > > > > >>>>>> Why was the first formula working (the one with
> >>> > > > > > selectedvalue) and not
> >>> > > > > > > >>>>>> the second one? Why simply deleting the
> >>> "selectedvalue"
> >>> > > > > > makes it an array
> >>> > > > > > > >>>>>> that will not be accept in "linearray"?
> >>> > > > > > > >>>>>>
> >>> > > > > > > >>>>>> Is there any way to draw a line without using
> >>> lastvalue or
> >>> > > > > > > >>>>>> selectedvalue? Do I need to use a loop?
> >>> > > > > > > >>>>>>
> >>> > > > > > > >>>>>> Thanks,
> >>> > > > > > > >>>>>>
> >>> > > > > > > >>>>>> Louis
> >>> > > > > > > >>>>>>
> >>> > > > > > > >>>>>> 2008/9/16 Tony Grimes <Tonez.Email@>
> >>> > > > > > > >>>>>>
> >>> > > > > > > >>>>>>> Louis,
> >>> > > > > > > >>>>>>>
> >>> > > > > > > >>>>>>> All of the variables you are creating for the
> >>> LineArray
> >>> > > > > > function are
> >>> > > > > > > >>>>>>> arrays themselves. Although LineArray
generates an
> >>> > > array, it
> >>> > > > > > does not accept
> >>> > > > > > > >>>>>>> any arrays as inputs. Additionally, your error
> >>> message was
> >>> > > > > > probably
> >>> > > > > > > >>>>>>> different. It probably went from complaining
about
> >>> > > argument
> >>> > > > > > #4 having the
> >>> > > > > > > >>>>>>> incorrect type (which you corrected) to argument
> >>> #3 having
> >>> > > > > > the incorrect
> >>> > > > > > > >>>>>>> type.
> >>> > > > > > > >>>>>>>
> >>> > > > > > > >>>>>>> On Tue, Sep 16, 2008 at 9:10 PM, Louis P.
> >>> > > <rockprog80@>wrote:
> >>> > > > > > > >>>>>>>
> >>> > > > > > > >>>>>>>> Hi,
> >>> > > > > > > >>>>>>>>
> >>> > > > > > > >>>>>>>> Thank you for your help.
> >>> > > > > > > >>>>>>>>
> >>> > > > > > > >>>>>>>> @Ara:
> >>> > > > > > > >>>>>>>>
> >>> > > > > > > >>>>>>>> If in
> >>> > > > > > > >>>>>>>>
> >>> > > > > > > >>>>>>>> barhh1 = HHVBars( High, Periods ) ;
> >>> > > > > > > >>>>>>>> bi1 = BarIndex();
> >>> > > > > > > >>>>>>>> y11 = LinearReg( C, barhh1 ) ;
> >>> > > > > > > >>>>>>>> y01 = LinRegIntercept( C, barhh1 ) ;
> >>> > > > > > > >>>>>>>> sl1 = LineArray( bi1-barhh1+0, y01, bi1,
y11, 0,
> >>> True );
> >>> > > > > > > >>>>>>>>
> >>> > > > > > > >>>>>>>> I replace
> >>> > > > > > > >>>>>>>>
> >>> > > > > > > >>>>>>>> sl1 = LineArray( bi1-barhh1+0, y01, bi1,
y11, 0,
> >>> True );
> >>> > > > > > > >>>>>>>>
> >>> > > > > > > >>>>>>>> by
> >>> > > > > > > >>>>>>>>
> >>> > > > > > > >>>>>>>> sl1 = LineArray( bi1-barhh1+0, y01, bi1,
> >>> > > LastValue(y11), 0,
> >>> > > > > > True
> >>> > > > > > > >>>>>>>> );
> >>> > > > > > > >>>>>>>>
> >>> > > > > > > >>>>>>>> I still have the same error message. I
don't know
> >>> from
> >>> > > > > > where it is
> >>> > > > > > > >>>>>>>> coming.. unfortunately!
> >>> > > > > > > >>>>>>>>
> >>> > > > > > > >>>>>>>>
> >>> > > > > > > >>>>>>>> @gp_sydney:
> >>> > > > > > > >>>>>>>>
> >>> > > > > > > >>>>>>>> That was a typo, you are right; I arranged
that by
> >>> > > adding a
> >>> > > > > > 1. But
> >>> > > > > > > >>>>>>>> still, the problem remains: the last line does
> >>> not work.
> >>> > > > > > > >>>>>>>>
> >>> > > > > > > >>>>>>>> One day, I asked support if I needed a loop
to do
> >>> such LR
> >>> > > > > > and they
> >>> > > > > > > >>>>>>>> said I should not need one.
> >>> > > > > > > >>>>>>>>
> >>> > > > > > > >>>>>>>> Here is the original code:
> >>> > > > > > > >>>>>>>>
> >>> > > > > > > >>>>>>>> barhh = SelectedValue( HHVBars( High,
Periods ) );
> >>> > > > > > > >>>>>>>> bi = SelectedValue( BarIndex() );
> >>> > > > > > > >>>>>>>> y1 = SelectedValue( LinearReg( C, barhh ) );
> >>> > > > > > > >>>>>>>> y0 = SelectedValue( LinRegIntercept( C,
barhh ) );
> >>> > > > > > > >>>>>>>> sl = LineArray( bi-barhh+0, y0, bi, y1, 0,
True );
> >>> > > > > > > >>>>>>>>
> >>> > > > > > > >>>>>>>> What I want to do is simply eliminate the
> >>> "selectedvalue"
> >>> > > > > > part and
> >>> > > > > > > >>>>>>>> use the code not only for the selected data but
> >>> for the
> >>> > > > > > whole data. I want
> >>> > > > > > > >>>>>>>> to be able to draw a line from each HHV to each
> >>> bar and
> >>> > > > > > then work with the
> >>> > > > > > > >>>>>>>> result.
> >>> > > > > > > >>>>>>>>
> >>> > > > > > > >>>>>>>> If it can't be done without a loop, I feel like
> >>> I'll be
> >>> > > > > > lost in time
> >>> > > > > > > >>>>>>>> again; last time I tried to run a loop on my
> >>> computer it
> >>> > > > > > freezed and after 2
> >>> > > > > > > >>>>>>>> minutes I decided to shut down AB...
> >>> > > > > > > >>>>>>>>
> >>> > > > > > > >>>>>>>>
> >>> > > > > > > >>>>>>>> Thanks for the help,
> >>> > > > > > > >>>>>>>>
> >>> > > > > > > >>>>>>>> Louis
> >>> > > > > > > >>>>>>>>
> >>> > > > > > > >>>>>>>>
> >>> > > > > > > >>>>>>>>
> >>> > > > > > > >>>>>>>>
> >>> > > > > > > >>>>>>>>
> >>> > > > > > > >>>>>>>>
> >>> > > > > > > >>>>>>>>
> >>> > > > > > > >>>>>>>> 2008/9/16 gp_sydney <gp.investment@>
> >>> > > > > > > >>>>>>>>
> >>> > > > > > > >>>>>>>>> As Ara said, in the shown code snippet you
don't
> >>> have
> >>> > > > > > "barhh"
> >>> > > > > > > >>>>>>>>> defined,
> >>> > > > > > > >>>>>>>>> only "barhh1".
> >>> > > > > > > >>>>>>>>>
> >>> > > > > > > >>>>>>>>> Beyond that, you have the same issue I
mentioned
> >>> > > > > > originally, that
> >>> > > > > > > >>>>>>>>> the
> >>> > > > > > > >>>>>>>>> linear regression functions and LineArray
> >>> function take
> >>> > > > > scalar
> >>> > > > > > > >>>>>>>>> values
> >>> > > > > > > >>>>>>>>> (ie. single numbers) as parameters, not
arrays.
> >>> > > > > > > >>>>>>>>>
> >>> > > > > > > >>>>>>>>> I gather you're trying to create a line
from the
> >>> > > > > > most-recent HHV
> >>> > > > > > > >>>>>>>>> value
> >>> > > > > > > >>>>>>>>> using the subsequent close data for every
bar on
> >>> the
> >>> > > > > > chart. As I
> >>> > > > > > > >>>>>>>>> don't
> >>> > > > > > > >>>>>>>>> think the linear regression functions can take
> >>> arrays
> >>> > > > > for the
> >>> > > > > > > >>>>>>>>> period,
> >>> > > > > > > >>>>>>>>> I think you'd need to use a loop and do
the linear
> >>> > > > > regression
> >>> > > > > > > >>>>>>>>> yourself
> >>> > > > > > > >>>>>>>>> at each bar (you could call the array
functions
> >>> > > within the
> >>> > > > > > loop,
> >>> > > > > > > >>>>>>>>> but
> >>> > > > > > > >>>>>>>>> since they fill a whole array each time, they
> >>> would do a
> >>> > > > > > lot of
> >>> > > > > > > >>>>>>>>> unnecessary work). If you do that yourself
> >>> inside the
> >>> > > > > > loop, then at
> >>> > > > > > > >>>>>>>>> each bar you'd have scalar 'x' and 'y'
values to
> >>> > > calculate
> >>> > > > > > the line
> >>> > > > > > > >>>>>>>>> slope and so on.
> >>> > > > > > > >>>>>>>>>
> >>> > > > > > > >>>>>>>>> For what it's worth, the BarIndex function
> >>> simply gives
> >>> > > > > > you the bar
> >>> > > > > > > >>>>>>>>> number. It provides a way of using the
current bar
> >>> > > number
> >>> > > > > > in array
> >>> > > > > > > >>>>>>>>> formula.
> >>> > > > > > > >>>>>>>>>
> >>> > > > > > > >>>>>>>>> Regards,
> >>> > > > > > > >>>>>>>>> GP
> >>> > > > > > > >>>>>>>>>
> >>> > > > > > > >>>>>>>>>
> >>> > > > > > > >>>>>>>>> --- In
> >>> > > amibroker@xxxxxxxxxxxxxxx <amibroker%40yahoogroups.com>
<amibroker%
> >>> 40yahoogroups.com><amibroker%
> >>> > > 40yahoogroups.com>
> >>> > > > > > <amibroker%40yahoogroups.com>,
> >>> > > > > > > >>>>>>>>> "Louis P." <rockprog80@> wrote:
> >>> > > > > > > >>>>>>>>> >
> >>> > > > > > > >>>>>>>>> > Hi,
> >>> > > > > > > >>>>>>>>> >
> >>> > > > > > > >>>>>>>>> > Sorry, You can replace "periods" by 50
if you
> >>> wish. I
> >>> > > > > > just forgot
> >>> > > > > > > >>>>>>>>> to
> >>> > > > > > > >>>>>>>>> > include that.
> >>> > > > > > > >>>>>>>>> >
> >>> > > > > > > >>>>>>>>> > barhh1 = HHVBars( High, *50* ) ;
> >>> > > > > > > >>>>>>>>> > bi1 = BarIndex() ;
> >>> > > > > > > >>>>>>>>> > y11 = LinearReg( C, barhh ) ;
> >>> > > > > > > >>>>>>>>> > y01 = LinRegIntercept( C, barhh ) ;
> >>> > > > > > > >>>>>>>>> > sl1 = LineArray( bi1-barhh1+0, y01, bi1,
y11, 0,
> >>> > > True );
> >>> > > > > > > >>>>>>>>> >
> >>> > > > > > > >>>>>>>>> > Still, it is not working, even if barhh1 is
> >>> defined...
> >>> > > > > > > >>>>>>>>> >
> >>> > > > > > > >>>>>>>>> > Louis
> >>> > > > > > > >>>>>>>>> >
> >>> > > > > > > >>>>>>>>> > 2008/9/16 Ara Kaloustian <ara1@>
> >>> > > > > > > >>>>>>>>> >
> >>> > > > > > > >>>>>>>>> > > y11 and y01 use "barhh" which is not
defined.
> >>> > > > > > > >>>>>>>>> > >
> >>> > > > > > > >>>>>>>>> > > You have defined "barhh1"
> >>> > > > > > > >>>>>>>>> > >
> >>> > > > > > > >>>>>>>>> > >
> >>> > > > > > > >>>>>>>>> > >
> >>> > > > > > > >>>>>>>>> > > ----- Original Message -----
> >>> > > > > > > >>>>>>>>> > > *From:* Louis P. <rockprog80@>
> >>> > > > > > > >>>>>>>>> > > *To:*
> >>> > > amibroker@xxxxxxxxxxxxxxx <amibroker%40yahoogroups.com>
<amibroker%
> >>> 40yahoogroups.com><amibroker%
> >>> > > 40yahoogroups.com>
> >>> > > > > > <amibroker%40yahoogroups.com>
> >>> > > > > > > >>>>>>>>> > > *Sent:* Tuesday, September 16, 2008
2:46 PM
> >>> > > > > > > >>>>>>>>> > > *Subject:* [amibroker] What is wrong?
> >>> > > > > > > >>>>>>>>> > >
> >>> > > > > > > >>>>>>>>> > > Hi,
> >>> > > > > > > >>>>>>>>> > >
> >>> > > > > > > >>>>>>>>> > > What is wrong in the following formula?
> >>> > > > > > > >>>>>>>>> > >
> >>> > > > > > > >>>>>>>>> > > barhh1 = HHVBars( High, Periods ) ;
> >>> > > > > > > >>>>>>>>> > > bi1 = BarIndex() ;
> >>> > > > > > > >>>>>>>>> > > y11 = LinearReg( C, barhh ) ;
> >>> > > > > > > >>>>>>>>> > > y01 = LinRegIntercept( C, barhh ) ;
> >>> > > > > > > >>>>>>>>> > > sl1 = LineArray( bi1-barhh1+0, y01,
bi1, y11,
> >>> 0,
> >>> > > True );
> >>> > > > > > > >>>>>>>>> > >
> >>> > > > > > > >>>>>>>>> > >
> >>> > > > > > > >>>>>>>>> > > Thanks,
> >>> > > > > > > >>>>>>>>> > >
> >>> > > > > > > >>>>>>>>> > > Louis
> >>> > > > > > > >>>>>>>>> > >
> >>> > > > > > > >>>>>>>>> > > p.s. There was "Selectedvalue" in the
first
> >>> four
> >>> > > lines
> >>> > > > > > but I
> >>> > > > > > > >>>>>>>>> don't
> >>> > > > > > > >>>>>>>>> want to
> >>> > > > > > > >>>>>>>>> > > plot it on the chart based on where I
am on
> >>> that
> >>> > > > > > chart, but
> >>> > > > > > > >>>>>>>>> simply
> >>> > > > > > > >>>>>>>>> set the
> >>> > > > > > > >>>>>>>>> > > variable so I can use the stuff later.
> >>> > > > > > > >>>>>>>>> > >
> >>> > > > > > > >>>>>>>>> > >
> >>> > > > > > > >>>>>>>>> > >
> >>> > > > > > > >>>>>>>>> >
> >>> > > > > > > >>>>>>>>>
> >>> > > > > > > >>>>>>>>>
> >>> > > > > > > >>>>>>>>
> >>> > > > > > > >>>>>>>
> >>> > > > > > > >>>>>>
> >>> > > > > > > >>>>>
> >>> > > > > > > >>>>
> >>> > > > > > > >>>
> >>> > > > > > > >>
> >>> > > > > > > >
> >>> > > > > > > >
> >>> > > > > > >
> >>> > > > > >
> >>> > > > >
> >>> > > > >
> >>> > > > >
> >>> > > >
> >>> > >
> >>> > >
> >>> > >
> >>> >
> >>>
> >>>
> >>>
> >>
> >>
> >
>
------------------------------------
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