About Kurt_Bremser

Kurt_Bremser · ‎03-13-2014

How about getting the number of lines first with wc -l (or simply divide the file size by record size if you have fixed length records), then split that into 8 parts and do 8 data steps in succession where you supply firstobs= and obs= on the infile statement? I would not recommend running 8 concurrent jobs against the same file and physical devices, it might cause delays.

Kurt_Bremser · ‎03-12-2014

First, your example has 2 seemingly small but significant errors: proc sort data=short_state_soc22 ; by state; run; %macro srt(); data wide_state; set short_state_soc22; by state; array soc22{22} occ_11 occ_13 occ_15 occ_17 occ_19 occ_21 occ_23 occ_25 occ_27 occ_29 occ_31 occ_33 occ_35 occ_37 occ_39 occ_41 occ_43 occ_45 occ_47 occ_49 occ_51 occ_53 ; %do i=11 %to 53 %by 2; retain soc22; if soc_22 ="&i" %then soc22[occ_&i]=count; *"%then" is wrong, should be "then"; %end; if last.state=1 %then output; *"%then" is wrong, should be "then"; run; %mend srt; %srt; Remember that the macro facility is just a code generator. What your macro does looks like this: data wide_state; set short_state_soc22; by state; array soc22{22} occ_11 occ_13 occ_15 occ_17 occ_19 occ_21 occ_23 occ_25 occ_27 occ_29 occ_31 occ_33 occ_35 occ_37 occ_39 occ_41 occ_43 occ_45 occ_47 occ_49 occ_51 occ_53 ; retain soc22; if soc_22 ="11" then soc22[occ_11]=count; retain soc22; if soc_22 ="13" then soc22[occ_13]=count; retain soc22; if soc_22 ="15" then soc22[occ_15]=count; retain soc22; if soc_22 ="17" then soc22[occ_17]=count; retain soc22; if soc_22 ="19" then soc22[occ_19]=count; retain soc22; if soc_22 ="21" then soc22[occ_21]=count; retain soc22; if soc_22 ="23" then soc22[occ_23]=count; retain soc22; if soc_22 ="25" then soc22[occ_25]=count; retain soc22; if soc_22 ="27" then soc22[occ_27]=count; retain soc22; if soc_22 ="29" then soc22[occ_29]=count; retain soc22; if soc_22 ="31" then soc22[occ_31]=count; retain soc22; if soc_22 ="33" then soc22[occ_33]=count; retain soc22; if soc_22 ="35" then soc22[occ_35]=count; retain soc22; if soc_22 ="37" then soc22[occ_37]=count; retain soc22; if soc_22 ="39" then soc22[occ_39]=count; retain soc22; if soc_22 ="41" then soc22[occ_41]=count; retain soc22; if soc_22 ="43" then soc22[occ_43]=count; retain soc22; if soc_22 ="45" then soc22[occ_45]=count; retain soc22; if soc_22 ="47" then soc22[occ_47]=count; retain soc22; if soc_22 ="49" then soc22[occ_49]=count; retain soc22; if soc_22 ="51" then soc22[occ_51]=count; retain soc22; if soc_22 ="53" then soc22[occ_53]=count; if last.state=1 then output; run; Decide for yourself which code is more efficient and readable! All those ifs eat a lot of CPU power, BTW. The macro facility lets you create dynamic code. If the code remains static, do it in the data step itself. Edit: replaced %then with then

Kurt_Bremser · ‎03-12-2014

What you want is this: proc sort data=short_state_soc22 ; by state; run; data wide_state (drop=soc soc_22 count i); set short_state_soc22; by state; array soc22{22} occ_11 occ_13 occ_15 occ_17 occ_19 occ_21 occ_23 occ_25 occ_27 occ_29 occ_31 occ_33 occ_35 occ_37 occ_39 occ_41 occ_43 occ_45 occ_47 occ_49 occ_51 occ_53 ; retain soc22; if first.state then do i = 1 to 22; soc22 = .; end; soc22[(soc_22-9)/2] = count; if last.state then output; run;

Kurt_Bremser · ‎03-12-2014

But why does it start to display the mantissa at row 61 instead of row 53?

Kurt_Bremser · ‎03-12-2014

%if soc_22 ="&i" %then soc22[occ_&i]=count; In this macro statement, soc_22 is not a variable, but a value! Therefore you are comparing (the string!) soc_22 to "11", "13" and so on, which is never true. The same goes for %if last.state last.state is just a string when the macro executes (long before the data step even is compiled and starts) and the whole construct has no meaning.

Kurt_Bremser · ‎03-12-2014

first. and last. are only generated for variables in the "by" list. Try this: data int; set old; int_day = datepart(Transaction_Date); run; proc sort data=int; by Acc_Nr Transaction_Date; run; data wanted (drop=int_day); set int; by Acc_Nr int_date; if first.int_date; run;

Kurt_Bremser · ‎03-11-2014

How about doing it the other way around, by counting workdays? IF DLVR_STRT_DTS = . THEN C= 'N/A'; ELSE DO; SELECT (REGION); WHEN ('Americas') DO; check = 0; do iter_date = DATEPART(ORD_ENTRY_DATE) to DATEPART(DLVR_STRT_DTS); if (WEEKDAY(iter_date) not IN (1,7) and INPUT(PUT(iter_date,AMERHOL.),9.) ne 1 then check = check + 1; * count the actual number of workdays; end; IF check le 3 THEN C = 'MADE'; ELSE C = 'MISS'; END;

Kurt_Bremser · ‎03-10-2014

Just one thing: you might consider to mark one of the answers as correct so that your initial question ist listed as "answered". Likewise for helpful posts.

Kurt_Bremser · ‎03-03-2014

Do this: data wang; set have; format time_numeric time8.; time_numeric = input(substr(time,1,2) !! ':' !! substr(time,3,2) !! ':' !! substr(time,5,2),time8.); run;

Kurt_Bremser · ‎03-03-2014

Do this: data want; set have; by id; retain product; if first.id then product = 1; product = product * return; if last.id then output; run;

Kurt_Bremser · ‎02-28-2014

Determine how large your data set is (physically) (sum of size of vars * number of records) Depending on that, you might consider exporting the data to a flat file and sort that externally (linux). In my own experience I have to say that SAS performance in z/OS is surprisingly bad, after migrating to a 2-CPU pSeries (with AIX) we noticed that it ran circles around the MF. Also keep in mind that SAS generates a utility file while sorting, and then writes the sorted data back. You should make sure that the source and target of the sort are not located where your WORK library is (or the place where the UTILLOC system option points to). From your description I thing that you are heavily I/O bound, that's why the trick with dividing the data set did not make a difference.

Kurt_Bremser · ‎02-27-2014

The first @ in the input statement positions the cursor at column 1; The second, trailing(!) @ tells SAS to keep the contents of the input buffer and not read a new line. The input statement without the trailing @ switches to the next line of the infile after reading

Kurt_Bremser · ‎02-27-2014

In CD34_NUM = input(CD34, comma21.2); replace the format with comma21. The format works like that: - read in all numeric digits as a natural number - if there is a number X after the dot in the format, divide by 10^X Your numbers were converted as 1.00, 2.50, 3.50 respectively instead of 100, 250, 350

Kurt_Bremser · ‎02-27-2014

If at first you just want to make sure that you read everything that's in a line, try the following: %let maxrl=500; * or any number that is appropriate; data infile1; infile"xxxx" lrecl=&maxrl truncover; input whole_line $&maxrl.; run;

Kurt_Bremser · ‎02-27-2014

If your SAS is installed on a UNIX, the error message will include the process number of the locking process. ps -fp processnumber will give you the user's ID.

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