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AutoScale makes things wrong.
The slope value is good.
You need to get 1$ = 1 Unit time
Imagine to do the works on a millimeter paper sheet...
Y start from zero and X axis one millimeter equals Y axis one
millimeter !
e.g
1$ = 1 minute = 1 mm on X and Y
BR
--- In amibroker@xxxxxxxxxxxxxxx, "Louis P." <rockprog80@xxx> wrote:
>
> 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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