Re: [amibroker] Re: What is wrong?, AmiBroker Email List Archive

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@xxxxxxxxx>

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, "Louis P." <rockprog80@xxx> 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@xxx>

>
> > 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>,
"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>,
> >
> > > > "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>,
> >
> > "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>,
> > > > > > >>>>>>>>> "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>
> > > > > > >>>>>>>>> > > *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.
> > > > > > >>>>>>>>> > >
> > > > > > >>>>>>>>> > >
> > > > > > >>>>>>>>> > >
> > > > > > >>>>>>>>> >
> > > > > > >>>>>>>>>
> > > > > > >>>>>>>>>
> > > > > > >>>>>>>>
> > > > > > >>>>>>>
> > > > > > >>>>>>
> > > > > > >>>>>
> > > > > > >>>>
> > > > > > >>>
> > > > > > >>
> > > > > > >
> > > > > > >
> > > > > >
> > > > >
> > > >
> > > >
> > > >
> > >
> >
> >
> >
>