Date: Fri, 29 Dec 2006 12:46:28 -0500
Reply-To: "data _null_;" <datanull@GMAIL.COM>
Sender: "SAS(r) Discussion" <SAS-L@LISTSERV.UGA.EDU>
From: "data _null_;" <datanull@GMAIL.COM>
Subject: Re: legend for graphs
In-Reply-To: <1167411723.45954a0bd5758@ssl0.ovh.net>
Content-Type: text/plain; charset=ISO-8859-1; format=flowed
See this example from the support.sas.com
/* TITLE: Replaying multiple graphs to a page with a shared/common
TITLE and LEGEND */
http://ftp.sas.com/techsup/download/sample/graph/greplay-shlegend.html
Similar
http://support.sas.com/ctx/samples/index.jsp?sid=430&tab=code
On 12/29/06, Stéphane COLAS <scolas@datametric.fr> wrote:
> Hi all,
>
> I was wondering if I missed something about the legend for graphs created with a
> BY statement...
>
> If you generate several graphs with a BY Statement and where the differents
> graphs use the same legend, how do you produce only one 'box' that represents
> the legend?
>
> for instance, if you produce graphs with this code, You can see the same legend
> each time.
> The purpose on the work is to group them on a page with a proc greplay. I don't
> want to produce the same legend box in each image, I want to produce only one
> box.
>
> proc gplot data=sashelp.class uniform;
> plot height*weight=sex;
> by name;
> run;
> quit;
>
>
> If the gplot is not the good proc for that, which is the more appropriate ?
>
> TIA.
>
> Stéphane.
>
> Selon Dorian <d.noel@ICMACENTRE.RDG.AC.UK>:
>
> > Howard,
> >
> > Your code gives the correct solution to my problem and thus, you have a
> > fair understanding of the problem at hand.
> >
> > However, your solution may takes quite some time to run for a large
> > sample. I'm looking for a reasonably fast code beacuse of the size of
> > the dataset I'm working with (400K observations).
> >
> > Is there a faster way to getting the output?
> >
> > Thanks for your assistance.
> >
> > Dorian
> >
> >
> > "Howard Schreier <hs AT dc-sug DOT org>" wrote:
> > > Maybe a reflexive join ...
> > >
> > > data try;
> > > input
> > > Ohecode : $10. Time : time8. ohebs:$1. lo_price:8. B1pr:8. S1pr:8.;
> > > format time time8.;
> > > cards;
> > > DGAE709 08:07:48 B 122.50 122.50 123.00
> > > TANA05T 08:07:50 B 121.50 122.50 123.00
> > > SGTIA05T 08:07:52 S 122.75 123.50 123.75
> > > GJCXEUT 08:07:55 B 122.00 121.00 121.50
> > > DGAE709 08:08:00 B 122.50 122.75 123.00
> > > SGTIA05T 08:08:10 S 122.75 121.75 122.00
> > > GUOLS5T 08:08:15 S 123.00 122.75 123.00
> > > TANA05T 08:09:50 B 121.50 122.75 123.00
> > > ;
> > >
> > > proc sql;
> > > select Ohecode
> > > , ohebs
> > > , lo_price
> > > , entry
> > > , exit
> > > , Event_time
> > > , case when ohebs='B' then b1pr
> > > when ohebs='S' then s1pr
> > > else .
> > > end as Best_price
> > > from
> > > (select time as Event_time, b1pr, s1pr from try)
> > > natural join
> > > (select ohecode
> > > , ohebs
> > > , lo_price
> > > , min(time) as entry format=time8.
> > > , max(time) as exit format=time8.
> > > from try
> > > group by ohecode, ohebs, lo_price
> > > )
> > > where entry le Event_time le exit
> > > group by ohecode, ohebs, lo_price
> > > having case when ohebs='B' then b1pr=max(b1pr)
> > > when ohebs='S' then s1pr=min(s1pr)
> > > else .
> > > end
> > > ;
> > >
> > > Result is
> > >
> > > Ohecode ohebs lo_price entry exit Event_time Best_price
> > > -----------------------------------------------------------------------
> > > DGAE709 B 122.5 8:07:48 8:08:00 8:07:52 123.5
> > > GJCXEUT B 122 8:07:55 8:07:55 8:07:55 121
> > > GUOLS5T S 123 8:08:15 8:08:15 8:08:15 123
> > > SGTIA05T S 122.75 8:07:52 8:08:10 8:07:55 121.5
> > > TANA05T B 121.5 8:07:50 8:09:50 8:07:52 123.5
> > >
> > > On Thu, 28 Dec 2006 10:07:51 -0800, Dorian <d.noel@ICMACENTRE.RDG.AC.UK>
> > wrote:
> > >
> > > >Thanks for the assistance but I don't think your code will generate the
> > > >desired results.
> > > >
> > > >The solution requires that the price of an order is to be compared with
> > > >all prices from its time of entry to the time of exit from the sample.
> > > >
> > > >Let's take for example; DGAE709. The order entry is 08:07:48 and exit
> > > >on 08:08:00. Its price is 122.50 and its type is 'B'.
> > > >
> > > >What is required then is that its price 122.50 is to be compared with
> > > >prices in the column B1pr between 08:07:48 and 08:08:00, returning the
> > > >highest price value.
> > > >
> > > >For completeness:
> > > >
> > > > 08:07:48 Max( 122.50, 122.50) DGAE709's entry time
> > > > 08:07:50 Max( 122.50, 122.50)
> > > > 08:07:52 Max( 122.50, 123.50)
> > > > 08:07:55 Max( 123.50, 121.50)
> > > > 08:08:00 Max( 123.50, 122.75) DGAE709's exit time
> > > >
> > > >output dataset:
> > > >
> > > >Ohecode Time Best_price Event_time
> > > >
> > > >DGAE709 08:08:00 123.50 08:07:52
> > > >
> > > >
> > > >For order of the type 'S', the solution, using an example, is as
> > > >follows:
> > > >
> > > >Comparing the order's price with prices in the column S1pr:
> > > >
> > > >08:07:52 Min(122.75, 123.75) SGTIA05T's time of entry
> > > >08:07:55 Min(122.75, 121.50)
> > > >08:08:00 Min(121.50, 123.00)
> > > >08:08:10 Min(121.50, 122.00) SGTIA05T's time of exit
> > > >
> > > >output
> > > >
> > > >Ohecode Time Best_price Event_time
> > > >
> > > >DGAE709 08:08:00 123.50 08:07:52
> > > >SGTIA05T 08:08:10 121.50 08:07:55
> > > >
> > > >Thanks once again for your assistance.
> > > >
> > > >Dorian
> > > >
> > > >Gerhard Hellriegel wrote:
> > > >> Ok, maybe something like:
> > > >>
> > > >> data try;
> > > >> input
> > > >> Ohecode : $10. Time : time8. ohebs:$1. lo_price:8. B1pr:8. S1pr:8.;
> > > >> cards;
> > > >> DGAE709 08:07:48 B 122.50 122.50 123.00
> > > >> TANA05T 08:07:50 B 121.50 122.50 123.00
> > > >> SGTIA05T 08:07:52 S 122.75 123.50 123.75
> > > >> GJCXEUT 08:07:55 B 122.00 121.00 121.50
> > > >> DGAE709 08:08:00 B 122.50 122.75 123.00
> > > >> SGTIA05T 08:08:10 S 122.75 121.75 122.00
> > > >> GUOLS5T 08:08:15 S 123.00 122.75 123.00
> > > >> TANA05T 08:09:50 B 121.50 122.75 123.00
> > > >> ;
> > > >> run;
> > > >>
> > > >> proc sort;
> > > >> by ohecode time;
> > > >> run;
> > > >>
> > > >> data high low;
> > > >> retain eventtime;
> > > >> set try;
> > > >> by ohecode;
> > > >> if first.ohecode then eventtime=time;
> > > >> if ohebs="B" then do;
> > > >> bestprice=max(lo_price,b1pr);
> > > >> output low;
> > > >> end;
> > > >> else do;
> > > >> bestprice=min(lo_price,s1pr);
> > > >> output high;
> > > >> end;
> > > >> format eventtime time time10.;
> > > >> run;
> > > >>
> > > >> proc sort data=low;
> > > >> by ohecode descending bestprice;
> > > >> run;
> > > >>
> > > >> data low;
> > > >> set low;
> > > >> by ohecode;
> > > >> if first.ohecode;
> > > >> run;
> > > >>
> > > >> proc sort data=high;
> > > >> by ohecode bestprice;
> > > >> run;
> > > >>
> > > >> data high;
> > > >> set high;
> > > >> by ohecode;
> > > >> if first.ohecode;
> > > >> run;
> > > >>
> > > >> data all;
> > > >> set low high;
> > > >> by ohecode;
> > > >> keep ohecode bestprice ohebs time eventtime;
> > > >> run;
> > > >>
> > > >>
> > > >> ....with other results like you had.
> > > >> e.g.
> > > >>
> > > >> SGTIA05T 08:08:10 S 122.75 121.75 122.00
> > > >>
> > > >> You have best (lowest) price 121.75. You compared that with b1pr, but
> > should
> > > >> (?) be s1pr = 122.00
> > > >>
> > > >> Try it out, if the results are like you want them.
> > > >>
> > > >> Regards,
> > > >> Gerhard
> > > >>
> > > >>
> > > >>
> > > >>
> > > >>
> > > >>
> > > >>
> > > >> On Thu, 28 Dec 2006 05:03:35 -0800, Dorian <d.noel@ICMACENTRE.RDG.AC.UK>
> > > wrote:
> > > >>
> > > >> >Thanks for your interest and your comment is valid.
> > > >> >
> > > >> >I thought the logical expression will be suffice to explain the final
> > > >> >result I seek but it is confusing and need further clarification.
> > > >> >
> > > >> >Each order in the dataset has an entry and exit time from the sample
> > > >> >and thus, the the used of "first.time" and "last.time" in the logical
> > > >> >expression. What is required is a comparison of its price to the best
> > > >> >available prices that existed during this interval (betwen the entry
> > > >> >and exit times) and then record the highest price value and the
> > > >> >associated time of the event.
> > > >> >
> > > >> >For instance, order DGAE709 is of type 'B' and has an entry and exit
> > > >> >time of 08:07:48 and 08:08:00, respectively. We need to compare its
> > > >> >price ("lo_price") 122.50 with prices "B1pr" during its time spent in
> > > >> >the sample that is, between 08:07:48 and 08:08:00 and choose the
> > > >> >highest price value from this comparison and record the associated
> > > >> >event time ("event_time") the highest price value occurred. In this
> > > >> >case, this occurred at 08:07:52 when b1pr was 123.50.
> > > >> >
> > > >> >For order of teh type 'S', we choose the lowest price value by
> > > >> >comparing the price ("lo_price") of the order to "S1pr".
> > > >> >
> > > >> >I hope this clear up things a bit.
> > > >> >
> > > >> >Once again thanks for your interest shown.
> > > >> >
> > > >> >Take care.
> > > >> >
> > > >> >Dorian
> > > >> >
> > > >> >
> > > >> >
> > > >> >
> > > >> >
> > > >> >
> > > >> >Gerhard Hellriegel wrote:
> > > >> >> I don't understand the rules for the "best_price"!
> > > >> >> What should the
> > > >> >>
> > > >> >> First.time LE max(lo_price, B1pr) LE last.time
> > > >> >>
> > > >> >> tell?
> > > >> >>
> > > >> >> first.time is a logical value which means "first in group of
> > same-times"?
> > > >> >> 1. time is a bad var for grouping
> > > >> >> 2. you compare a logical value (0 or 1) with a price?? What should
> > that
> > > >> tell us?
> > > >> >>
> > > >> >> So that must be a missunderstanding, but what is the MEANING? What
> > > should be
> > > >> >> the logic?
> > > >> >>
> > > >> >> Regards,
> > > >> >> Gerhard
> > > >> >>
> > > >> >>
> > > >> >>
> > > >> >> On Thu, 28 Dec 2006 02:50:54 -0800, Dorian
> > <d.noel@ICMACENTRE.RDG.AC.UK>
> > > >> wrote:
> > > >> >>
> > > >> >> >Good morning,
> > > >> >> >
> > > >> >> >
> > > >> >> >I'm in need of some urgent advice on a SAS coding problem. I am
> > trying
> > > >> >> >to locate the highest/lowest price (and associated event time) that
> > > >> >> >existed during the lifespan of an order. This is determined by
> > > >> >> >comparing the price of the order to those that occurred during the
> > life
> > > >> >> >of the order and is done separately for two types of orders.
> > > >> >> >
> > > >> >> >Consider the following event dataset:
> > > >> >> >
> > > >> >> >Ohecode Time ohebs lo_price B1pr S1pr
> > > >> >> >
> > > >> >> >DGAE709 08:07:48 B 122.50 122.50 123.00
> > > >> >> >TANA05T 08:07:50 B 121.50 122.50 123.00
> > > >> >> >SGTIA05T 08:07:52 S 122.75 123.50 123.75
> > > >> >> >GJCXEUT 08:07:55 B 122.00 121.00 121.50
> > > >> >> >DGAE709 08:08:00 B 122.50 122.75 123.00
> > > >> >> >SGTIA05T 08:08:10 S 122.75 121.75 122.00
> > > >> >> >GUOLS5T 08:08:15 S 123.00 122.75 123.00
> > > >> >> >TANA05T 08:09:50 B 121.50 122.75 123.00
> > > >> >> >
> > > >> >> >I seek the following solution:
> > > >> >> >
> > > >> >> >Ohecode Time Best_price Event_time
> > > >> >> >
> > > >> >> >GJCXEUT 08:07:55 122.00 08:07:55
> > > >> >> >DGAE709 08:08:00 123.50 08:07:52
> > > >> >> >SGTIA05T 08:08:10 121.50 08:07:55
> > > >> >> >GUOLS5T 08:08:15 123.00 08:08:15
> > > >> >> >TANA05T 08:09:50 123.50 08:07:52
> > > >> >> >
> > > >> >> >Please note that the "best_price" for each order is derived as
> > > >> >> >follows:
> > > >> >> >
> > > >> >> >If ohebs = 'B' then best_price (and associated event_time) for the
> > > >> >> >particular order is:
> > > >> >> >
> > > >> >> > First.time LE max(lo_price, B1pr) LE last.time.
> > > >> >> >
> > > >> >> >Else if ohebs = 'S' then best_price (and associated event_time) for
> > > >> >> >the particular order is:
> > > >> >> >
> > > >> >> > First.time LE min(lo_price, S1pr) LE last.time.
> > > >> >> >
> > > >> >> >Thanks for the assistance.
> > > >> >> >
> > > >> >> >Take care.
> > > >> >> >
> > > >> >> >Dorian
> >
>
| 
