Somewhat related...this is a BRAND NEW edition. Be prepared for stuff like that. I took the base certification in April and it was based on the 4th edition. When I was looking at the 4th edition, there were some similar issues with the chapter quizzes. Someone else I knew who was studying for the same exam had the book for the 3rd edition. There were multiple instances where the answer was A in the 3rd edition, but the same question had a different answer for A in the 4th edition and it was wrong. It definitely made me think about it more until I was convinced that I was right and the book was wrong.    Also, I just looked at the table of contents for the fifth edition. I heard they were going to do this, but it looks like they don't have anything on reading in raw data!!! No column input, list input, formatted input. I don't see any pointer controls, creating variables with INPUT statements, missover options...nothing. None of that fun stuff! I'm not sure if that's covered in the exam, but not in the new edition.   Anyway, good luck. I'm glad I learned all that reading in raw data, but it was definitely a challenge to keep all the different input options straight, so IMO, you're lucky to be taking it now, assuming that's not in the exam.  ... View more

It's not because it is before 1960. It is because it is before 1926, which is the default yearcutoff= when using date7. formats. Change the yearcutoff= option to 1910 and it works. ... View more

Ah YES! COALESCE! ... View more

I did it slightly different, since the other solutions didn't seem to resolve keeping the address. I still think a solution using a data step with a by statement is probably better.   proc sql; create table want as select distinct t1.name ,case when t2.name ne '' then t2.birthdate else t1.birthdate end as birthdate ,case when t2.name ne '' then t2.source else t1.source end as source ,t1.address from have as t1 left join (select * from have where source = 'NEW') as t2 on t1.name=t2.name where t1.address ne '' ; quit; ... View more

Hi Cynthia,  I'm with you. I was thinking the same thing when I approached the question and I only answered the question at hand. My only thought is they have a list of known emails addresses and will end up looping through comments to determine if it exists anywhere in the string and aren't trying to pull out an assortment of email addresses. Because yeah, that will be impossible if the comment is structured like that and they don't have anything to notate the start and end.  Thanks for posting the code you did. I'm going to grab that - I'm sure it will come in handy some day! ... View more

My pleasure! I tried to edit my last code block to add a method to do it all in proc sql - you won't need the sort this way. Here it is again so it is readable:   proc sql; create table want1 as select id ,medclass ,sum(txduration) as sum from have group by id, medclass; create table want2 as select medclass ,mean(sum) as mean from want1 group by medclass; quit; ... View more

data want (drop=comments); set temp; email=substr(comments,find(comments,'xyz@gmail.com'),length('xyz@gmail.com')); run; Is the email always going to be the same? If not, you could do something similar looking for the @ and then the first instance of .com (or .edu or whatever after that.) ... View more