About gsk

Amir · ‎07-06-2021

Hi @anuajax, Ideally, it is best practice to create a new question with what you are now asking and you can supply a link back to this question for context. For each example you have given what result do you want to see and what did you get when you tried the existing solution? Also, when supplying data, it is a lot easier for everyone to help you if you supply data using a data step with datalines, as can be seen in the solution. Thanks & kind regards, Amir.

hhinohar · ‎09-23-2020

Here is a quick sample. data have; input id visit answer $ score; datalines; 1 1 Y 10 1 2 N 5 1 3 N 3 1 4 N 2 1 5 N 9 1 6 N 11 1 7 Y 2 2 1 Y 3 2 2 N 5 2 3 Y 1 2 4 N 2 2 5 N 4 3 1 N 3 3 2 N 5 3 3 Y 2 3 4 Y 1 3 5 N 2 3 6 N 3 3 7 N 2 3 8 Y 5 3 9 Y 4 3 10 N 1 3 11 Y 5 3 12 Y 2 ; run; data want; merge have have(drop=id visit rename=(answer=lead_answer score=lead_score) firstobs=2) end=eof; retain flag_y; lag_answer=lag(answer); lag_score=lag(score); if lag_answer="N" and lag_score>=5 and lead_answer="N" then do; flag_y="Y"; end; else do; flag_n="N"; end; if lag(id) ne id then flag_y=""; flag=ifc(flag_y="Y", "Y", ifc(flag_n="N", "N", "")); keep id visit answer score flag; run;

ballardw · ‎06-09-2020

It may also help to show the code where you intend to use the resulting variables.

gsk · ‎03-13-2020

What I tried so far: first, sort proc sort data=test out=test1; by subjid seq; run; data test2; set test1; by subjid seq; if length(date2) >= 10 and not last.subjid then do; *<-by the way, both dates can contain time components; date2_n= input(date2, yymmdd10.)-1; date2_c= strip(put(date2_n, yymmdd10.)); *<- I need a character version of the new date later; end; run; But this subtracts all not last dates even though the next date1 has time component

Ksharp · ‎03-07-2020

How about this one ? data have1; input ID question $ subcat $ response $; datalines; 1 Q1 a Y 1 Q1 b Y 1 Q1 c Y 1 Q2 a Y 1 Q2 b Y 1 Q2 c N 1 Q2 d N 1 Q2 e Y 2 Q1 a N 2 Q1 b Y 2 Q1 c Y 2 Q2 a N 2 Q2 b Y 2 Q2 c N 2 Q2 d N 2 Q2 e Y ; proc report data=have1 nowd ; columns ('ID' id) question,subcat,response; define id/group ' '; define question/across ' ' nozero; define subcat/across ' ' nozero; define response/group ' ' nozero ; run;

Reeza · ‎02-28-2020

Show your log with the issue. I'm guessing it terminated due to an error in your code and you're looking at partial output. Otherwise, please provide some code that replicates the issue.

gsk · ‎11-09-2019

Thank you everyone! Either input(dateA, yymmdd10.)=input(dateB,yymmdd10.) or dateA=:dateB works In the end, I did strip(dateA)=:strip(dateB). Thanks a lot!

Ksharp · ‎11-04-2019

%let list = 001 002 003 004; data have; input @1 subjects $10.; cards; SUBJ-001-M SUBJ-002-J SUBJ-005-K SUBJ-007-K SUBJ-004-N ; run; data want; set have; call scan(subjects,2,p,l,"-"); if findw("&list",scan(subjects,2,"-")) then subjects=cats(substr(subjects,1,p-1),substr(subjects,p+l+1)); run;

Kurt_Bremser · ‎10-31-2019

I guess you wanted to do this: coalesce(data1.variable,data2.var2*data2.var3,data2.var4) as new_var

novinosrin · ‎10-25-2019

Since you mentioned Missing as valid count,Basically I am resetting missing to a non missing junk value before I count distinct using IFC function.

Tom · ‎08-19-2019

If you just want to split the string into smaller byte lengths it is trivial. So assume you have existing dataset named HAVE which has a long variable named SOURCE. Here is code to generate 9 variables named SPLIT1 to SPLIT9 that are each 200 bytes long. data want ; set have ; array split [9] $200 ; do _n_=1 to 9; split[_n_]=substr(source,1+(_n_-1)*200,200); end; run; Note that if your string has UTF-8 (or other multi-byte encoding) then you might want to add more complexity to prevent splitting a multi-byte character between two string.

Rick_SAS · ‎07-10-2018

I don't fully understand what you are saying, but I guess I don't have to. Your question seems to be "how can I compute k variances in SAS/IML where each one is computed by leaving out one of the k groups." Attached is a program that I hope will demonstrate the programming technique, even if I am misinterpreting some of the details. For background, I suggest you read my article about the "UNIQUE-LOC" technique. data Have; call streaminit(1234); do Group = 1 to 10; StdDev = round(1 + rand("Uniform"), 0.05); /* Group-specific Std Dev */ do i = 1 to 50; x = rand("Normal", 0, StdDev); output; end; end; run; proc iml; use Have; read all var {Group x}; close; OverallVar = var(x); u = unique(Group); k = ncol(u); /* number of groups */ variance = j(1, k, .); do i = 1 to k; idx = loc( Group ^= u[i] ); /* omit the i_th group */ variance[i] = var( x[idx] ); /* compute statistic on remaining k-1 groups */ end; print variance; est = ssq( OverallVar - variance ) / (k/(k-1)); print est;

Astounding · ‎04-15-2018

The exterior macro has already identified the next name to process as &VAR. Here's one way to search: %if %sysfunc(findw(&numlist, &var)) %then %do; If for any reason you want to stick to purely macro language functions, it can be done but is clumsy: %if %index(%str( &numlist ), %str( &var )) %then %do; The leading and trailing blanks are now treated as significant characters, a requirement for the proper result. You don't want the search for VAR1 to locate a match if the software finds VAR11.

Astounding · ‎04-14-2018

SQL does this for you. You just have to capture the information before it disappears: proc sql noprint; select name into :numlist separated by ' ', / * HERE.. I don't know what to write. something like len(&numlist) or count(name) into :numvalcount is what I want to do. */ from dictionary.columns where libname='MET' and memname='DEMA_669' and type='num' AND FORMAT^='DATE9.' order by name; %let numvalcount = &sqlobs; select name into :datelist separated by ' ', / * HERE.. I don't know what to write. */ from dictionary.columns where libname='MET' and memname='DEMA_669' and type='num' AND FORMAT='DATE9.' order by name; %let datevalcount = &sqlobs; select name into :charlist separated by ' ', / * HERE.. I don't know what to write. */ from dictionary.columns where libname='MET' and memname='DEMA_669' and type='char' order by name; %let charvalcount = &sqlobs; quit;

art297 · ‎04-13-2018

%let numlist=a c e; %let datelist=b d; %let charlist=f z; data test; length lists $255; array list(7) $; lists=catx(' ',"&numlist.", "&datelist.","&charlist."); i=1; do while (scan(lists,i) ne ''); list(i)=scan(lists,i); i+1; end; call sortc(of list(*)); call symput('alllist',catx(' ',of list(*))); run; %put &alllist.; Art, CEO, AnalystFinder.com

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