About pp2014

pp2014 · ‎03-03-2015

Thanks for help Reeza. It worked...

pp2014 · ‎03-03-2015

Sorry below is the corrected code. But still not getting correct output.. Data Have; input Cust_id prd$ eff_dt anydtdte10.; format eff_dt date9.; cards; 1 SP 20140721 1 GL 20140822 1 GL 20140903 1 GL 20141001 1 SP 20141021 1 SP 20141105 1 SP 20141208 1 TA 20150103 1 TA 20150209 2 SP 20150110 2 TA 20150215 data want; set have; to_prd= prd; from_prd = lag(prd); run; proc print data = want noobs; var cust_id eff_dt from_prd to_prd; run; Output I am getting is as follows which is wrong: Cust_id eff_dt from_prd to_prd 1 21JUL2014 SP 1 22AUG2014 SP GL 1 03SEP2014 GL GL 1 01OCT2014 GL GL 1 21OCT2014 GL SP 1 05NOV2014 SP SP 1 08DEC2014 SP SP 1 03JAN2015 SP TA 1 09FEB2015 TA TA 2 10JAN2015 TA SP 2 15FEB2015 SP TA

pp2014 · ‎03-03-2015

Hi Reeza, I did run the Jag's code as below but output I am getting is not correct: Data Have; input Cust_id prd$ eff_dt anydtdte10.; format eff_dt date9.; cards; 1 SP 20140721 1 GL 20140822 1 GL 20140903 1 GL 20141001 1 SP 20141021 1 SP 20141105 1 SP 20141208 1 TA 20150103 1 TA 20150209 2 SP 20150110 2 TA 20150215 data want; set have; to_prd=lag(prd); from_prd = lag(prd); run; proc print data = want noobs; var cust_id eff_dt from_prd to_prd; run; Output I am getting is as follows which is wrong: Cust_id eff_dt from_prd to_prd 1 21JUL2014 1 22AUG2014 SP SP 1 03SEP2014 GL GL 1 01OCT2014 GL GL 1 21OCT2014 GL GL 1 05NOV2014 SP SP 1 08DEC2014 SP SP 1 03JAN2015 SP SP 1 09FEB2015 TA TA 2 10JAN2015 TA TA 2 15FEB2015 SP SP

pp2014 · ‎03-03-2015

Thanks Jag and Reeza. I wanted output to show switches from_Prod and To_Prod as follows for the following data: Data Have; input Cust_id prd eff_dt anydtdte10.; format eff_dt date9.; cards; 1 SP 20140721 1 GL 20140822 1 GL 20140903 1 GL 20141001 1 SP 20141021 1 SP 20141105 1 SP 20141208 1 TA 20150103 1 TA 20150209 2 SP 20150103 2 GL 20150209 Output should be as follows: Cust_id prd eff_dt From_Prod To_Prod 1 SP 20140721 1 GL 20140822 SP GL 1 GL 20140903 GL GL 1 GL 20141001 GL GL 1 SP 20141021 GL SP 1 SP 20141105 SP SP 1 SP 20141208 SP SP 1 TA 20150103 SP TA 1 TA 20150209 TA TA 2 SP 20150110 2 GL 20150215 SP GL Thanks...

pp2014 · ‎03-02-2015

I have the data as follows: I want the flag when the products switches from SP to some other product (Fr_SP) or from some other product to SP (To_SP) Data Have; input Cust_id prd eff_dt anydtdte10.; format eff_dt date9.; cards; 1 SP 20140721 1 GL 20140822 1 GL 20140903 1 GL 20141001 1 SP 20141021 1 SP 20141105 1 SP 20141208 1 TA 20150103 1 TA 20150209 I want the output as follows: Cust_id prd eff_dt Flag 1 SP 20140721 1 GL 20140822 Fr_SP 1 GL 20140903 1 GL 20141001 1 SP 20141021 To_SP 1 SP 20141105 1 SP 20141208 1 TA 20150103 Fr_SP 1 TA 20150209 Thanks

pp2014 · ‎02-11-2015

I am trying to assign the file name to the variable. I am getting error saying that symbolic variable names must be 32 or fewer character long. Which macro function I should be using so that I do not have to worry about the length of the file? %let fname = my_file_name_is_xyz_20150202090056.txt; data abc; src_file_nm = symget("&fname."); run;

pp2014 · ‎02-09-2015

I have several Text files Zipped on Linux. I am trying to use Acess Zip Method. How do I read the each file from the zip using infile statement. Suppose I have 1)File1 20150202.txt 2)ABC 20150204.txt 3)XYZ 20150309.txt Sometimes there may be more than 3 files. I am trying to use Zip Access Method on UNIX. I am using SAS 9.4 . filename fromunix zip '/myfiles/mytest.zip'; data _null_; infile fromunix(File1 20150202.txt); informat x $20 y $25; format x $ y $; input x y; run;

pp2014 · ‎02-05-2015

I was able to figure it out. I had to use dlm='09'x in the infile statement and then it worked...

pp2014 · ‎02-05-2015

Hi RW9, Thanks for the solution. I tried it but it does not seem to work. Below is my code without macro which runs on UNIX. The zipped file has 2015.02.04 Complete XYZ Data.txt file in it. When I run the below code I get the filenm variable with value = 2015.02.04 and not the full file name. I tried your code but somehow it gives me error. Can you please let me know how do I make it work on UNIX? Thanks a lot filename xx pipe "zipinfo -1 /PROD/in/2015.02.04_Complete_XYZ_Data.zip"; data check; infile xx ; length filenm $45; input filenm $; run;

pp2014 · ‎02-05-2015

I am trying to read the text file on UNIX using infile. Below is the part of the code: %if &&file&j.. = %str(2015.02.04 Complete XYZ Data.txt) %then %do; data rawdata; infile fromunix(%sysfunc(cats(&&file&j..))) dlm="|" dsd truncover lrecl=32707 firstobs=2; %end; But somehow macro does not work mainly because there are periods and spaces for the file name. If there is no period or space then it works. Is there a macro function I can use to make the above thing work with spaces and periods in the name of the file?

pp2014 · ‎02-04-2015

Thanks..

pp2014 · ‎02-04-2015

I have date string = "20150109103704" How do I convert it to 09JAN2015:10:37:04

pp2014 · ‎02-02-2015

Any Help???

pp2014 · ‎02-02-2015

I am trying to read the files from the folder on Linux using PIPE. There are currently 2 files (ABC_*.zip) in the folder. But when I tried to use the following code but it does not work. What may be wrong in the code? %let extl_dir=/PROD/PGM/; filename zl pipe "ls &extl_dir.ABC_*.zip"; data ziplist_a; infile pl end=last; length path $200 zipnm $50 filedt $15; input path $; zipnm=scan(path,-1,"/"); filedt=scan(scan(path,-1,"_"),1,"."); call symput('zip'||left(_n_), zipnm); call symput('path'||left(_n_), path); call symput('filedt'||left(_n_),filedt); if last then call symput('num_zip',_n_); run; NOTE: Numeric values have been converted to character values at the places given by: (Line):(Column). 97:25 98:26 99:28 100:36 NOTE: The infile ZL is: Unnamed Pipe Access Device, PROCESS=ls /PROD/PGM/ABC_*.zip, RECFM=V,LRECL=256 Stderr output: 'ls' is not recognized as an internal or external command, operable program or batch file. NOTE: 0 records were read from the infile PL. NOTE: The data set WORK.ZIPLIST_A has 0 observations and 3 variables. NOTE: DATA statement used (Total process time): real time 0.07 seconds cpu time 0.03 seconds

pp2014 · ‎10-22-2014

I have .csv file which I want to save as .xlsx file. I want to keep the original .csv file as it is but kind of copy .csv into .xlsx. I am using the following. It copies .csv file into .xlsx but I cannot open that .xlsx file. options xsync noxwait; x %sysfunc(quote(copy "c:\temp\test123.csv" "c:\temp\test123.xlsx")) ; Is there a way I can copy .csv into .xlsx file using sas? Thanks

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