@Astounding   Left to my own devices, my approach would have been to restructure the data to look like   Data AlternateInput; input id mydate:mmddyy10. datetype; datalines; 1 1/1/2010 start 1 4/1/2010 end 1 4/24/2010 start 1 6/1/2010 end 1 9/1/2010 start 1 10/15/2010 end 2 3/1/2010 start 2 5/1/2010 end 2 6/1/2010 start 2 10/1/2010 end ; and then try to figure out how to walk the ID's to find the periods.  Probably best I'm not left to solve it myself. Ha. ... View more

Using the data from above:   data have2; set input; by id; prior_end = lag(end_date); if first.id or (start_date - prior_end) > 45 then grouping + 1; run; data want; set have2; by id grouping; if first.grouping then new_start = start_date; retain new_start; drop start_date; rename new_start = start_date; if last.grouping; run; proc print data = want noobs; var id start_date end_date; format start_date date9. end_date date9.; run; Seems to yield The SAS System start_ id date end_date 1 01JAN2010 01JUN2010 1 01SEP2010 15OCT2010 2 01MAR2010 01OCT2010     I wish I understood either of the above approaches.  Lol.  ... View more

@RW9 I'm not sure that quite works.   If I do this: Data Input; input id start_date:mmddyy10. end_date:mmddyy10.; datalines; 1 1/1/2010 4/1/2010 1 4/24/2010 6/1/2010 1 9/1/2010 10/15/2010 2 3/1/2010 5/1/2010 2 6/1/2010 10/1/2010 ; run; data output; set input; retain start_date_ end_date_; by id; if first.id or start_date_=. then start_date_=start_date; else do; if last.id or end_date-lag(end_date) > 45 then do; end_date_=end_date; output; start_date_=.; end_date_=.; end; end; run; proc print data = output noobs; var id start_date_ end_date_; format start_date_ date9. end_date_ date9.; run; I get this:   The SAS System start_ id date_ end_date_ 1 01JAN2010 15OCT2010 2 01MAR2010 01OCT2010 which I don't think is exactly what OP wants.  It seems to skip over the 6/1 to 9/1 break for ID 1 in which OP is interested.  ... View more

Honestly, I'd probably work around it by making an image that contained both the logo and the title text I wanted in the way I wanted it presented and use that.         ... View more

I'm not going to open zip files from an unknown source in this day and age and I doubt anyone else will either. Post some data in a data step and you will likely get a better response.     ... View more

How many files and how often?   I usually just pull them in with file > import data >sas import wizard.   You can have it save the steps and then maybe use that generated import statement as a base for a macro if you have a lot to do.  ... View more

I guess more sxplicitly- I have a list of MRNs that fit this critieria.  In my larger data set how do I create a variable to which I can assign a "y" or "n" to these MRNs based on whether or not they are on this new table?  Thanks!!!     I have no idea.  We haven't seen even any made-up data.  I have no idea what you want to do. You said you already had things in Y/N format.    You'll have to provide more information and explanation for further help.  ... View more

Is each record an admission with a patient identifier and admission characteristics, or is each record a patient with some nonnormalized data structure?   Let us start someplace. Assume data where each record is an admit with a patient and a diagnosis.   data admits; input patientid admitdate:mmddyy10. opiod:$1. psych:$1. else:$1.; datalines; 1 1/1/2017 n y n 1 2/1/2017 n n y 1 3/1/2017 y n n 2 1/1/2017 n n y 2 2/1/2017 n n y 2 3/1/2017 y n n 3 1/1/2017 n y n 3 2/1/2017 n n y 3 3/1/2017 n y n 4 4/1/2017 n y n 4 5/1/2017 n n y 4 6/1/2017 y n n 4 7/1/2017 y n n 4 8/1/2017 y n n 5 6/1/2017 y n n 5 7/1/2017 n n n 5 8/1/2017 n y n ; run;   Then let's select anybody who is a y on opiod and also a y on pysch with a date before their y on opiod. In this data 1 and 4.  5 has a psych after and 2 doesn't have a psych.   proc sql; create table opiod_with_psych as select distinct a.patientid from admits a inner join admits b on a.patientid = b.patientid where a.opiod = 'y' and b.psych = 'y' and b.admitdate < a.admitdate; quit; That gives us   The SAS System   patientid 1 4   Start with that and then let's modify. ... View more

I don't understand what you want or why you are doubling up your observations.   I get that readmit 1 of 2/20 is beyond 30 days of drug start date 1/17 and is a 1.   But then i would assume that readmit 2 of 5/3 is beyond drug start date of 3/20 by 30 days and is also a one. 7/10 is a 1. 9/20 is a 30 days more than 8/7 and is a one. 11/9 is a one.   12/6 is less than 30 days from 11/23 and is a zero.   Obviously, I don't see what you are trying to do.   More explanation?   ... View more

@Kurt_Bremser   Trying to use Kurt's solution posted above   data mydata; input id:$1. include:$1. exclude:$3.; datalines; a . abc a . xyz b j abf b . gtj ; run; data newdata; set mydata; retain myinclude concat ; by id; length myinclude $10.; length concat $10.; if first.id then myinclude = ""; if first.id then concat = ""; myinclude = catx('',myinclude,include); concat = catx(',',concat,exclude); if last.id then output; drop include exclude; run; I can make   The SAS System           id myinclude concat a   abc,xyz b j abf,gtj     ... View more

I would think that would be a valid approach.  I personally don't know how to do that, but as I understand things some sort of model, like a regression, would work.    I would think one thing you would want to be careful about is drawing income conclusions based on data with a 20 percent imputation rate. ... View more

If you don't want to drop those records, your best bet is probably to impute income data.   You could assign those with missing income the median or modal income. You could create a second model to estimate income based on the demographic factors you do have,     ... View more