Nope it's a question from sas base practice exam ... View more

@mmohotsi    Please be kind enough to un-mark your comment as a solution (since it is not a solution in any meaningful sense) and mark the link I provided as the solution. ... View more

Also I am sorry the problem does ask for the booksales. I did not share the problem statement. But I do get the idea of what you are trying to convey. TY Paige! ... View more

So the basic ideas are: ( 1) create an array that has columns of offer amount according to offer dates, and set them to 0, (2) loop over the dates (i.e., dates as index of the array) and fill in the offer amount of correspondence dates, 3) calculate the dates interval, which is 7 days, and find out the offer amount occurred in the time interval, and keep comparing all of the offer amount until the max amount is found, and then fill in the correspondence date. These are very advanced logic and technique, thanks a lot for reply, Kurt_Bremser! It's good learning experience for me and helps me understand how to improve my code and thinking! ... View more

SAS Energy Forecasting ... View more

I've spoken with my colleague about his absurd data formatting. He said he'll try to fix it. So, for now, this task is moot.   Thanks to all who took the time to respond. ... View more

Assign $HEX formats of suitable length (18 and 40) to the policy variables in both datasets and look what you find. I suspect leading blanks (hex 20) or unprintable characters (0D0A, for instance). ... View more

Hello, thank you for your answer. ... View more

Sorry, I'm not good enough to clearly state the problem and/or solution. That data step performed other functions and when I moved the age determination outside of the problematic data step it worked just fine and that's the most important thing. The data step performed other functions so I'm assuming one conflicted or one of the variables in my function was dependent on that and I was attempting to this age calculation. Moving it outside of the data step eliminated that conflict and it runs just fine. The key concept from Cynthia and Paige was that perhaps my function wasn't to blame (since I did not change it, I just moved it). So it wasn't the particular function at fault, which is what I had focused on. A good reminder to me to check other possibilities.  ... View more

@TimurShangareev wrote: Is there any method, where I can extract dataset name from expressions like: <work.ds(...where=(...))> <ds(... where=(...))> or somehow check that this expression is valid in macro calls (wthout PROC or DATA statements)? It would help if you explained in more detail what you actually mean. Perhaps some example use cases would help?   You mentioned macro calls.  So in macro code here are a couple of simple things. %let dsn=sashelp.class(where=(age>11)); %let valid=FALSE; %let dsid=%sysfunc(open(&dsn)); %if &dsid %then %do; %let valid=TRUE; %let dsid=%sysfunc(close(&dsid)); %end; %let name_only=%scan(&dsn,1,()); %let options=%substr(&dsn%str( ),%length(&name_only)+1); %let memname=%scan(&name_only,-1,.); %let libref=%scan(WORK.&name_only,-2,.); %put &=dsn &=valid ; %put &=name_only &=options; %put &=libref &=memname ; ... View more

Array that has been suggested is a good alternative. The reason why your macro is not working is because you are using DATA step variables on a %IF condition which is not correct. If you change your code to use an IF statement, you should get the expected results. %macro test; data test; merge v (in=a) trans (in=b); by site_number; if a; %let i = 7; %do %while (&i > 0); if (dt_&i ne . and start_date ge dt_&i) then do; applied_tier = tier_&i; end; %let i = %eval(&i - 1); %end; run; %mend test; options mprint mlogic symbolgen; %test; ... View more

This is perfect, thank you so much for your help!! ... View more

Ive found itfrom the web. Thanks ... View more

And @SAS-questioner , according to your if/else condition, if any row has all values<=0, the flag of that row is missing, regardless there is -7, -8, and -9 in that row or not. I think this is also the question @PaigeMiller prompted. See the example: data have; input q1 q2 q3 q4 q5 q6; datalines; 20 0 0 4 5 0 20 20 0 -1 -2 0 20 0 -7 1 2 0 0 0 0 0 0 0 0 -8 0 -1 -3 0 -8 -7 -7 -1 -1 0 1 2 3 0 -1 -20 -9 20 1 2 3 1 -7 -1 0 0 -8 -9 -1 -2 -3 -4 -5 -6 0 -1 -2 -3 0 -1 ; run; proc print data=have;run; data want (drop=i); set have; array q[*] q:; do i=1 to dim(q); if q[i]>0 then flag=1; else if q[i] in (-7,-8,-9) then flag=.; else if max(of q:)=0 and min(of q:)=0 then flag=0; end; run; proc print data=want;run; ... View more

@LinusH wrote: Agree with @PaigeMiller sequences are easier to maintain/use.   It was not clear from your example how many times the indicator can change, but I guess the maximum with this technique  would be 26. Good point. That's a problem if you need more than 26 letters. Another problem is that if the variable GROUP_ID goes above 9, then the next value of NEW_ID is 10a. Since you also have 1a, these will not sort properly. If you sort on these NEW_ID values, 10a, 10b, ... will be followed by 1a. And if GROUP_ID goes above 9, you will also need a LENGTH statement to make this work at all.   Plus I am skeptical that there is a valid use for identifiers like 1a and 10a and so on. Who can remember what they represent when looking at the output? I also worry that the OP has oversimplified a real world problem and any answers we give will not fit the real world problem. ... View more