The biggest problem is that these programs replace the original version of TEMP.  If you need it (such as if you kater discover a logic error), you no longer have the original input available. Depending on the size of the data sets, you can be running up the bill.  Each step needs to read the entire data set and output all the observations.  It's easy to get into bad habits, such as using more steps that you actually need.  Typically, input and output are much more costly than data manipulation statements. ... View more

A DATA step can do all the calculations, if you keep track of what information is where.  I would use arrays for this: data want; array lebron {6} _temporary_; array stats {6} pts oreb dreb ast stl blk; array similar {6} similar_pts similar_oreb simil_dreb similar_ast similar_stl similar_blk; if _n_=1 then do; set player_traditional_usage_merged (where=(player="LeBron James")); lebron{1} = pts; lebron{2} = oreb; lebron{3} = dreb; lebron{4} = ast; lebron{5} = stl; lebron{6} = blk; end; set player_traditional_usage_merged; do _n_=1 to 6; similar{_n_} = stats{_n_} - lebron{_n_}; end; run; The temporary array cleans up after itself.  None of the variables fields get stored permanently once the DATA step ends.  You can pick any names you would like in the SIMILAR array.   It's untested code, but you have the data so easy enough for you to try it. ... View more

A minor tweak: since the LENGTH statement defines VISITCD, get rid of SUBSTR. visitcd = _name_; There's room to store only 3 characters, which is exactly the right outcome. ... View more

What combination of fields (if any) form a unique identifier for your records? ... View more

First, get rid of the "3" in the INPUT statement: input id $3 Just a dollar sign would be sufficient.  The "3" tells SAS to take the contents of column 3, which it always blank. Then sort your data: proc sort data=have; by id pnr pid time cn var1 var2 var3 var4 var5; run; Then pick out the observations that need to be investigated: data want; set have; by id pnr pid time cn var1 var2 var3 var4 var5; if first.var5 and last.var5 then delete; run; It's untested code, but easy enough to test since you have the data.  This just picks out the observations, but doesn't tell you where the differences lie. Follow-up:  right idea but wrong logic  will post an update later.   ... View more

It appears you have a sorted order to your data:  by ID month.  Let's pull the observations in a slightly different order: data want; set have (where=(case=1) in=have_case) have (where=(case=0) in=have_control); by id; retain any_case 'Y'; *condition 1; if first.id then do; if have_case then any_case='Y'; else any_case='N'; end; if have_control and any_case='Y' then delete; *condition 2; if have_control and not first.id then delete; run; The code is untested, since you have the data to work with.  But it looks right (even if it's mildly cumbersome). If there are multiple case records for an ID, this selects them all.  I'm not sure if that is your intention or not.  I'm interpreting the title vs. the description of the conditions as saying opposite things about that question. ... View more

Just clarifying .... Please confirm that ID 2 Month 8 should be removed. Please confirm that there is no YEAR in the data.  Just looking at MONTH is sufficient to determine which observation is earlier. ... View more

This is not difficult for SAS: data dsn; length phone $ 50; input phone phone phone; cards; S001 110064 9873628374 S002 110065 8447752736 S003 110023 9811537264 ; But why would you want to prohibit the most useful tools.  To illustrate, here's a task for you.  Try to eat a bowl of cereal and milk without using a spoon. ... View more

Since you're not sure what the problem is, here's my guess.  The intent is to output the original observation, then look at ID and output it that many additional times.  So the program could be: data want; set ds; output; do _n_ = 1 to id; output; end; run; ... View more

Believe it or not, there is some question as to what you are asking for here.  I'm going with the title:  zeros that follow nonzeros.   Here's one way: data want; set have; array counts {4} Mon_01 - Mon_04; zeros = 0; do n=2 to 4; if counts{n} = 0 and counts{n-1} ne 0 then zeros + 1; end; run; ... View more

I would verify each of your assumptions.   Confirm that work.consoli actually contains A, B, and C: proc contents data=work.consoli; run; Next, confirm that A, B, and C contain data: proc print data=work.consoli (obs=1); var a b c; run; That will probably clarify what needs to change, but try it and see. ... View more