Date:         Tue, 9 Mar 2004 04:34:14 -0600
Reply-To:     Kevin Myers <KevinMyers@AUSTIN.RR.COM>
Sender:       "SAS(r) Discussion" <SAS-L@LISTSERV.UGA.EDU>
From:         Kevin Myers <KevinMyers@AUSTIN.RR.COM>
Organization: Posted via Supernews, http://www.supernews.com
Subject:      Re: SORTEDBY after MERGE

"Robert Burbidge" <RBurbidge@PHD.CO.UK> wrote in message
news:F83CD98B6E0ABB4B8B56E880DF92F7550EAA87@phd-exch1.phd.int...
> Hi all,
>
> Could someone explain why after merging two sorted tables, the result is
not sorted?

As explained further below, the result *DOES* remain sorted.  The resulting
output data set simply doesn't have the sorted by attribute set.  That does
*NOT* mean that the data set is not sorted.  It simply means that SAS no
longer knows whether it is sorted or not.


>
> <sas>
> data test1;
>   infile datalines;
>   input ID;
>   datalines;
>   2
>   1
>   4
>   5
>   7
>   3
> ;
> data test2;
>   infile datalines;
>   input ID;
>   datalines;
>   3
>   0
>   -2
>   0.1
>   7
>   .
> ;
> proc sort data = test1; by ID;
> proc sort data = test2; by ID;
> data test3; merge test1 test2; by ID;
> proc contents data = test3; run;
> </sas>
>
> There's an interesting (and divergent) thread pertaining to this in the
archives:
> http://listserv.uga.edu/cgi-bin/wa?A2=ind0110B&L=sas-l&D=0&P=15646
>
> It seems that any DATA step will copy the table but not the SORTEDBY, even
if the sort key is not changed (the docs suggest that the SORTEDBY is only
overwritten if the sort key is manipulated in some way.)  E.g.:
>
> <sas>
> data test4; set test1;  /* test1 is sorted */
> proc contents data = test4; run; /* test4 isn't! */
> </sas>

Incorrect.  Data set test4 *IS* sorted.  However, it simply doesn't have the
sorted by attribute set.  Why not?  Because 1) SAS doesn't try to figure out
all the kinds of possible manipulations in a data step that could have
affected the order of the output observations relative to the values of it s
variables, and 2) because data set test4 is a completely *different* data
set from test1 as far as SAS is concerned.  SAS does not "pass" the value of
the sorted by attribute from one data set to another, unless perhaps you are
doing something like a simple proc copy.

>
> What are the pitfalls of using the following?
>
> <sas>
> data test3 (sortedby = ID); merge test1 test2; by ID;
> </sas>

There are no pitfalls of doing this as long as you can be *certain* that you
haven't done anything to get the ID values out of order, which you haven't
in this case.


>
> [Release 8.02, Windows 5.1 (XP)]
>
> Thanks+rgds,
>
> Robert
>
>
>
>
>
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