About varunnakra

varunnakra · ‎03-24-2013

Thanks Peter. Actually, I knew that whenever an explicit output statement is specified, the default output action of a data step is over ruled, however, the part that i was missing here was the do end loop. The if then construct needs a do end loop to make the output statement a part of itself. Otherwise, if the output statement is kept outside then it implies an output every time. Hence, the importance of do end loop. Yes, the point to be noted is that, during compilation itself SAS gets to know that there is an explicit output statement whether it is in the conditional code or otherwise and hence it suppresses the default output action of the data step. Resolved! Thanks

varunnakra · ‎03-23-2013

Thanks Peter, however, the problem is not that it doesn't work when i omit the do end loop but the problem is that it works differently. w/o the do end loop the output is course professor Name Id maths profA maths profA maria 100 maths profA smith 101 english profB english profB rick 101 w/ the do end loop the output is course professor Name Id maths profA maria 100 maths profA smith 101 english profB rick 101 The moot point is why are the two obs with the values of name and id missing displayed as an o/p without the do end loop but not with the do end loop. What makes the data step not to produce an output after the first iteration when we're using the do end loop or am i missing something here?

varunnakra · ‎03-23-2013

Thanks for the suggestions Arthur and Peter. I looked at the data step debugging commands and ran some of them but not much success. I also looked at the usual control flow of a data step. The real puzzle is with the do end loop that is there in the conditional logic of if type ='s'. If I omit the do end loop and simply specify an output statement (or choose to even omit that as well) the data step gives an output where the student information is blank which makes sense also as it never went into the branch of if type ='s' to encounter an output statement. It just executed all the statements and produced a default output. however, with the do end loop, i dont get the default output from the data step which i should have as even the loop is mentioned in the conditional logic of if type ='s'. hence, this is still confounding for me... Please advise more !

varunnakra · ‎03-23-2013

data want(drop=type); retain Course Professor; input type $1. @; if type='C' then input course $ professor $; else if type='S' then /*do;*/ input Name $10. Id; /*output schedule; end;*/ datalines; c maths profA s maria 100 s smith 101 c english profB s rick 101 ; When the above code is run there we get the first obs w/o the name of the student and his id or they are blank. The next observations retain the course name and the prof name and have the name of the student and his/her id. If we omit the comments around the do end loop and around the output statement; we don't get the first obs with the blank values of the student and his id. I have some questions around these 1. how is the iteration of the data step related to the processing of input statements? If we don't specify an output statement explicitly the data step, shouldn't the data step produce an output and move to the next iteration once it encounters no other input statement? 2. even if we specify an explicit output statement (as done by removing the comments in the above code), why is it that we get an output only when the student information is read from the input buffer. AND, that happens only when do end loop is present. Why is that so? Please advise!

varunnakra · ‎11-28-2012

Thanks a lot everyone for the insights

varunnakra · ‎11-27-2012

Hi, I have to concatenate the values of different macro variables and make a list of them. The following code gives the value of Node as null/blank. It's not able to concatenate the null value of Node with another text data test1; length Node $5.; Node = ' '; Node = Node ||"A"; run; Even if I use an initial value of Node as 'X'. It doesn't concatenate 'X' with 'A'. The following code gives the value of Node as X data test1; length Node $5.; Node = 'X '; Node = Node ||"A"; run; I wanted to make just one variable Node and use it recursively (referring to its previous value each time), but since the above code wasnt working properly, I made separate variables each time and wrote the following code options symbolgen mlogic; options compress = yes; %let C1 = A; %let C2 = B; %let C3 = C; %let C4 = D; %macro test; data test; length Node1 $5.; Node1 = "&C1"; %do i = 2 %to 4; /*%if &i = 1 %then %let j = &i;*/ /*%else*/ %let j = %eval(&i-1); Node&i = Node&j || " &&C&i"; %end; run; %mend; %test The above does give an output value of Node4 as A B C D but uses 3 spaces between A and B but 1 space (as used in the code) between B, C and D, etc. Why dont the previous codes work? Why does the above code give 3 spaces between A and B. Please advise!

varunnakra · ‎11-03-2012

Thanks Q for taking the pain. Yes, Symbolgen did behave chimerically on my laptop too, however, the results were consistent with our expectations. FYI, i was using SAS 9.1. I'll try the code on 9.2 too. Now, it gives me a sense of relief. The SUGI paper referred by Cynthia was enlightening

varunnakra · ‎11-03-2012

After reading the threads that Cynthia referred to and the SUGI pdf, I can now calculate the correct answer to such multiple ampersand references, however, the log using Symbolgen doesn't show the 'correct' resolution steps at times (publishes && resolves to & more number of times than it is required OR, publishes && resolves to & out of order).. This happens when multiple ampersands occur before a text and not a macro variable, however, the result from the resolution is according to the forward scan and rescan rules. 1. Rule1 : Every pair of ampersands gets resolved to a single ampersand and that single ampersand is 'left alone' until the scanning of the complete string takes place 2. Rule2 : Since, multiple ampersands have been used, a rescan occurs following Rule1.

varunnakra · ‎11-03-2012

Thanks for the reply Quentin, however, what I posted was the Symbolgen output of resolution of ampersands in the log. %put &&&&&varun&m2&m3 1. && resolves to & 2. && resolves to & (at this stage we have three &'s in front of varun) 3. && resolves to & (at this stage we have two &'s in front of varun) 4. && resolves to & (at this stage we have one & in front of varun) 5. m2 resolves to y 6. m3 resolves to z 7. varunyz not resolved Statements 1 to 7 are the ones written in the log by symbolgen which suggest that in the first pass amp#1, amp#2 get resolved to one amp; amp#3, amp#4 get resolved to one amp. Now according to your explanation, amp#5 should have tried resolving varun, however, it doesn't and no message is printed for it. Instead, it resolves the resulting ampersands before it could approach resolving m2 and m3. Many thanks Cynthia, I will read all the documentation and posts that you suggested. I have been using symbolgen, MPRINT and MLOGIC in my macro programming, however, was just intrigued by the logic behind the multiple scanning resolution. A common usage wont let me go beyond &&var&i, however, I was just chasing this from a learning point of view. Thanks all

varunnakra · ‎11-02-2012

Finally, I am able to reply.. The problem was with my browser. Posting the message here. @Art, yes, that is true..but I wanted to understand the programming logic of resolving so many ampersands by SAS from a learning point of view. I also have your book and will read it thoroughly very soon..Great to have the opportunity to 'speak' to you here @Quentin, thanks for recommending Art's book.. Taking your example forward - %put &&&&m1&m2&m3 1. && resolves to & 2. && resolves to & (at this stage we have two &'s in front of m1) 3. m2 resolves to y 4. m3 resolves to z 5. && resolves to & 6. M1YZ resolves to foo This suggests that SAS is resolving the macro variables from left to right. It tries to resolve two ampersands into one ampersand until it reaches two ampersands in front of a text after which it resolves all the rest of the macro variables and then comes back to resolve the two leading ampersands. %put &&&&&m1&m2&m3 1. && resolves to & 2. && resolves to & (at this stage we have three &'s in front of m1) 3. m1 resolves to x 3. m2 resolves to y 4. m3 resolves to z 5. && resolves to & 6. XYZ resolves to foo2 This suggests that SAS is resolving the macro variables from left to right. It tries to resolve two ampersands into one ampersand until it reaches three ampersands in front of a macro variable after which it resolves all the macro variables one by one and then comes back to resolve the two leading ampersands. The difference in the two resolutions above is - in the first case when SAS had two ampersands in front of m1, it didnt interpret m1 as a macro variable, however, in the second case, when SAS reached three ampersands in front of m1, it interpreted m1 as a macro variable. So I thought of playing with it further and using a text 'varun' and not a macro variable in place of m1. %put &&&&varun&m2&m3 1. && resolves to & 2. && resolves to & (at this stage we have two &'s in front of varun) 3. m2 resolves to y 4. m3 resolves to z 5. && resolves to & 6. varunYZ not resolved. This suggests that SAS is resolving the macro variables from left to right. It tries to resolve two ampersands into one ampersand until it reaches two ampersands in front of a text after which it resolves all the rest of the macro variables and then comes back to resolve the two leading ampersands. %put &&&&&varun&m2&m3 1. && resolves to & 2. && resolves to & (at this stage we have three &'s in front of varun) 3. && resolves to & (at this stage we have two &'s in front of varun) 4. && resolves to & (at this stage we have one & in front of varun) 5. m2 resolves to y 6. m3 resolves to z 7. varunyz not resolved Now I couldn't understand when SAS reached 3 ampersands in front of a text 'varun', it didn't generate a warning at this stage that apparent symbolic reference 'varun' couldnt be resolved. Instead, it continued resolutions of the ampersands further till it reached one ampersand in front of varun. I am confounded by the order of resolution of macro variables when multiple ampersands are used. What logic does SAS follow? I think multiple ampersands for multiple macro variables at the same time may create some more confusion. P.S. I am doing this from a learning point of view; trying to understand the logic used by SAS here. Message was edited by: Varun Nakra.. I am not able to reply to the post here, hence, editing original message.

varunnakra · ‎11-02-2012

Thanks for the reply Art, however, what I am looking at is resolution of more than 2 ampersands such as - &&&&&m1&m2&m3. Why does &&&&m1&m2&m3 and &&&&&m1&m2&m3 have the same result? Please explain the logic behind resolution of such multiple ampersands..

varunnakra · ‎11-01-2012

I haven't understood how are multiple ampersands resolved? I formulated a logic that whenever two ampersands occur together, they get resolved into one ampersand (moving from left to right) and one can think of a bracket enclosing the entire text following that ampersand then..For e.g. &&m1&m2 should logically mean &(m1&m2) &((&m1&m2&m3), however, it doesnt work like this and &&&&&m1&m2&m3 gives the same answer as &&&&m1&m2&m3 Please enlighten... Message was edited by: Varun Nakra.. I am not able to reply to the post here, hence, editing original message.

varunnakra · ‎11-01-2012

Conclusively, one must use %eval in logical or arithmetic operations just to be sure that there is not an error when SAS doesnt perform an automatic conversion..

varunnakra · ‎11-01-2012

yes, i am sorry to have missed on the %'s and the &'s.. I was writing this code here feeling a little dizzy..

varunnakra · ‎10-31-2012

Please look at the code below - %macro testeval; %let cnt = 0; %do until(&cnt =5); %let cnt = &cnt + 1; %put cnt; %end; %mend; has the following output: 0 + 1 0 + 1 + 1 0 + 1 + 1 + 1 0 + 1 + 1 + 1 + 1 0 + 1 + 1 + 1 + 1 + 1 The loop goes till 5 iterations which means the integral sum is being calculated and the value of cnt is 5 as understood by the SAS system, however, as an output the values of cnt are not 1,2,3,4,5. If we use %eval(&cnt + 1), we can have those values of cnt but still, the SAS system executes the loop 5 times..Then why doesn't it output those values..or does this mean that we can perform logical and arithmetic operations on macro variables (integral values) without the use of %eval but can get their true values? Please enlighten!

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