Hi,   First of all, I apologize to up this post but it is probably the closest one that relates to my understanding issues. I feel like I am about to grasp all these concepts but there are still blurry parts concerning definitions.   - What's the difference between the "true responders" and the "responders" you mentioned in the definitions? Or is it simply language abuse?   - For the % Response, I don't understand what the "proportion of true responders" is. Is it the proportion of responders in each decile calculated with respect to the total number of responders (on the whole data set), or the proportion of responders in each decile with respect to the size of the decile (number of participants in each decile)? To clear things up, say the whole data set contains data about 20,000 individuals, among which 5,000 gave a positive response (say, a donation, associated with a primary event). With a certain predictive model, I get the "best 5%" of my population (which gives me in total 1,000 individuals, i.e. the ones most likely to donate provided that the predictive model is decent), "next best 5%", and so on, which gives me all my deciles. Now imagine that I have a great model and, among these 1,000 individuals, it appears that 760 of them donated. Then, what will my % Response be? Will it be 760/5000 (=15.2%) or rather 760/1000 (=76%)? My guess goes towards the first choice (15.2%), but the definition I gave might not correspond to % Response but rather to % Captured Response.   - I don't understand SAS EM's notion of "cumulative". To me, cumulative should add up to 100%, however it is never the case here (except for Cumulative % Captured Response). Therefore, I don't understand the notion of Cumulative % Response.   - I don't understand the notion of % Captured Response and of its Cumulative counterpart. What's the difference between % Response and % Captured Response?   I think that, with these aspects cleared up, I'll get a better understanding of the notion of lift.     Thanks in advance if anybody answers this message, Yann. ... View more

As presented at the 2013 Midwest SAS Users Group meeting: http://billqualls.com/mba/index.html  ... View more

@FreelanceReinh's solution might look too complicated, but it is the only one that really works. PRXCHANGE would also work well, but compiling a new pattern for every observation would be very expensive. ... View more

You cam also write this as: proc sql; select AGE ,sum((SEX='F') * WEIGHT) as WEIGHT_F ,sum((SEX='M') * WEIGHT) as WEIGHT_M from SASHELP.CLASS group by 1; quit;   ... View more

@frupaul I believe you should get hold of Ron cody's Learning SAS by Example: A Programmer's Guide, Second Edition.   It is only safe to be aware  @FreelanceReinh's standards are among the  best in the world although his humility/modesty might not show that , but the reality for us folks (like you and I) is that  we might find the learning curve bit too steep to read Dorfman alike authors. Therefore, my sincere and humble suggestion is to start  learning with detailed examples portrayed in Ron cody's , then perhaps look into Art carpenter's book and beyond. HTH & Thank you!   ... View more

HI @frupaul  For some reason, when i tested the solutions with million + records the two step solution previous apparently is the fastest(sighs). I was intrigued to test as @FreelanceReinh diligently pointed out a similar two step solution in the previous thread of yours was much faster and of course he knows i try to pay attention to those finer details to learn more. On my college lab machine, that holds true and in other machines it could be otherwise. FWIW, that was fun testing.      ... View more

Sorry too late, was sleeping.  Assuming i understood your requirement:   data Hierarchy; infile datalines dlm='' truncover; length CustID $10 L1 $20 L2 $20 L3 $20 L4 $20 L5 $20 L6 $20 L7 $20 L8 $20; input CustID $ L1 $ L2 $ L3 $ L4 $ L5 $ L6 $ L7 $ L8 $; cards; 1 A20 A30 A40 A50 A60 A70 A80 2 A40 A70 A80 A90 3 A70 A80 A90 A100 4 A30 A60 A70 ; run; data want; do n=1 by 1 until(l); set Hierarchy end=l; array t(*) l8-l1; array j(100) $ _temporary_; j(n)=coalescec(of t(*)); end; l=0; do until(l); set Hierarchy end=l; do _n_=1 to dim(t); if not missing(t(_n_)) and t(_n_) in j then t(_n_)=cats('XXX',t(_n_)); end; output; end; keep custid l:; run;     ... View more

You are mixing up macro and datastep.  Macro is nothing more than a find replace system.  So lets replace the macro part with the actual values: data _null_; If CP ^= CLB then put/ "Failed: Second word of Filename is " CP ". Should be CLB"; else put /"File is fine"; run; Now you can clearly see why your code does not work.  There is no CP variable, nor CLB variable hence nothing=nothing, so the file is fine.  If you put quotes around each part then it should work; %let test_filename=JJJ_CP_LUKAKU; %let T_word2 = %scan(&test_filename,2,_); %let Agreed_Word2 = CLB; data _null_; If "&T_word2" ^= "&Agreed_Word2" then put/ "Failed: Second word of Filename is &T_word2. Should be &Agreed_Word2."; else put /"File is fine"; run; That being said, its really not a good idea in any sense to be doing data processing in macro, as shown above it is just a find/replace system, it does nothing itself just generates some text.  A better way of writing the same thing would be: %let test_filename=JJJ_CP_LUKAKU; %let agreed_word2=CLB; data _null_; if scan("&test_filename.",2,"_") ne "&agreed_word2." then put "Failed: Second word of Filename is &t_word2. Should be &agreed_word2."; else put "File is fine"; run; (and do note the dots at the end of each macro variable - always a good idea to do this). ... View more