Hello   May I ask for an advise? I'm trying to read 2D data to proc optmilp and I'm getting error: "ERROR: The symbol 'student' has no value at line XXX column YYY"   I think, it is because I'm trying to load data where not all combinations of indices are present. student class payment 1 A 1 2 A 2 2 B 3 3 A 4 3 B 5     So here, you can see that there is not combination of studen 1 in class B. Could this be a reason for the error?   Code:   proc optmilp; set student; set <str> class; num income {student, class}; read data work.input into [student class] payment = income; ... quit;   I need to have data in the single table and filling missing records (1-B) would probably lead to numerically more difficult task. I tried to use only a single index (composed of student-class) and it worked, but for the problem, I need to access these two indexes independently. ... View more

Hi   PROC OPTMILP returns _value_ columns in the PRIMALOUT= output. The documentation only states: "specifies the value of the variable in the current solution". Would anyone know its mathematical meaning?   When running the optimization for a binary variable x (searching for "1" if a term corresponding to a combination of parameters should contribute to the cost function and "0" when the term should vanish), the _value_ is in interval [0, 1]. I guess, the best solutions, i.e., the ones I want, are given by _value_=1. Because of constrains, each (original) observation can have only one combination of parameters. However, the same combination of parameters is found multiple times in the output table; one once with _value_=1 but several times for _value_<1.   I'll assume that the final solution (for optimized combination of observations) is a data set "where _value_=1". It'd be mathematically reasonable. How can I interpret lower values? For example, what "kind of a solution" do I get if All observations except a (random) single one have value=1 and the one is, e.g., the second highest ones? For all original observations, the second largest _value_ is selected. Or the lowest one.   (Technical note: with original observation I mean a record ID in the input table. The output table has multiple rows for this ID, one corresponding to each distinct _value_.) ... View more

Hi   I need to use a string above 32767 characters. Logic is more complicated, but here is and example: A data step has two variables below the limit as inputs. The variables need to be concated. The previous step is done to scan (and similar functions such as tranwrd) for the whole string. It is necessary because, e.g., a relevant substring may be divided, having one part at the end of one variable and the second part at the beginning of the second source. I do NOT try to save the string above 32767, just to use it in a single non-CAS data step. You can imagine that the scan function in the example returns a string much shorter (the example should produce a length of "2+length(varB)" because " aabbbBB..." is in the middle). I'm aware that the current example could be handled in a different way, such "find the end of A" and add "B". However, this is just due to simplification and I really need the whole concated string.   Not working:   data test; length var1 var2 $32000; var1 = repeat('A', 31000); var1 = catt(var1," aa"); var2 = repeat('B', 31000); var2 = catt("bbb", var2); output; run; data test; set test; length v varchar(70000); v=catt(var1, var2); /* put v; */ l=length(v); put l; length v2 varchar(70000); v2=scan(v, 2 ,' '); /* put v2; */ l2=length(v2); put l2; run;     Working (the same with shorter variables):   data test; length var1 var2 $32000; var1 = repeat('A', 1000); var1 = catt(var1," aa"); var2 = repeat('B', 1000); var2 = catt("bbb", var2); output; run; data test; set test; length v varchar(70000); v=catt(var1, var2); /* put v; */ l=length(v); put l; length v2 varchar(70000); v2=scan(v, 2 ,' '); /* put v2; */ l2=length(v2); put l2; run;     For the not working code, not even lengths are printed correctly (both outputs results in 1). The warrning from the log: NOTE: One or more variables were converted because the data type is not supported by the V9 engine. For more details, run with options MSGLEVEL=I. WARNING: In a call to the CATT function, the buffer allocated for the result was not long enough to contain the concatenation of all the arguments. The correct result would contain 62008 characters, but the actual result might either be truncated to 32767 character(s) or be completely blank, depending on the calling environment. The following note indicates the left-most argument that caused truncation. NOTE: Argument 2 to function CATT('AAAAAAAAAAAA'[12 of 32000 characters shown],'bbbBBBBBBBBB'[12 of 32000 characters shown]) at line 85 column 4 is invalid. var1=AAAAAAAA ...... [The whole A and B variables are printed here.] .... BBBBBBB v= l=1 v2= l2=1 _ERROR_=1 _N_=1 Any suggestions how to overcome it, i.e., how to get a long varchar from two smaller variables and use it in the scan function?   ... View more

Hi   May I ask someone to help me with following code?   %let url=xxx; %let token=_very_long_token_; filename out temp; proc http url="&url" method="GET" out=out; headers "Authorization" = "Token token = &token" ; run; %put Received code: &SYS_PROCHTTP_STATUS_CODE.; %put Received phrase: &SYS_PROCHTTP_STATUS_PHRASE.;   It which gives result: %put Received code: &SYS_PROCHTTP_STATUS_CODE.; Received code: 401 %put Received phrase: &SYS_PROCHTTP_STATUS_PHRASE.; Received phrase: Unauthorized   When the same is done in bash, it works correctly: url=xxx; token=_very_long_token_; curl -k -H "Authorization: Token token=\"$token\"" "$url" Why the SAS code does not work identically to the bash? ... View more