About FriedEgg

FriedEgg · ‎10-14-2011

I would consider tables of this size to be small. I frequently interact with tables in SAS that are multiple terrabytes in size. Tens or billions of rows and thousands of columns wide. Your question is too broad to answer more plainly than: Proc SQL should have no problem working with your data, whether it is stored in sas, teradata or any number of other sources.

FriedEgg · ‎10-14-2011

I know someone who used this utility: https://github.com/BioStatMatt/sas7bdat to convert sas tables to other formats through R (http://www.r-project.org/) AppDev Studio is mainly for developing Applications that use SAS/IT and SCL to run processes in SAS in their background. If you wanted to go through the substanstial dev path to do such a thing you could accomplish your goals, but it is a pretty lofty scheme to do something so simple to accomplish with SAS itself. Eslink is a sort of homebrew version of the SAS Enhanced Viewer (which is build on the eclipse platform also, btw...) My real recommendation would be to follow Cynthia's plan. Create a batch program, it will take a file and convert it through sas to a given format, then open the converted file. Then you can change the filetype association for sas7bdat from SAS to your batch program.

FriedEgg · ‎10-14-2011

You need to change your encoding option on the import. Since these are OS specific here is a link where you can find the one you need. http://support.sas.com/documentation/cdl/en/nlsref/61893/HTML/default/viewer.htm#a002628552.htm then do something like this: filename in 'C:\Documents and Settings\Desktop\ExpMeet.tsv' encoding="uft-8"; proc import datafile=in out=base_data dbms=dlm replace; delimiter='09'x; datarow=2; run; Here is an additionaly resource which covers all of SAS's Language capabilities: SAS(R) 9.2 National Language Support (NLS): Reference Guide http://support.sas.com/documentation/cdl/en/nlsref/61893/HTML/default/viewer.htm#nlsrefwhatsnew902.htm

FriedEgg · ‎10-13-2011

Can you please provide code sample of what you are trying to do? You are trying to use the _NOLOG_ prefix and _PASSWORD suffix as part of a stored procedure, correct? If so, you should be aware that they only work when running on a stored process server. If you are trying to write other code and want to mask stored information look at this post: http://communities.sas.com/message/106240#106240

FriedEgg · ‎10-13-2011

Since these are large character variables you are storing it would be best to use compress=char.

FriedEgg · ‎10-13-2011

%let start=O; %let destination=117.153.4.238; data route_map; infile '/nas/sasbox/users/mkastin/gth2008.txt' dsd dlm='~' truncover; input @; _infile_=prxchange('s/ | => /~/',-1,_infile_); input @1 n :$1 (ip dest1 goto1 dest2 goto2 dest3 goto3 goto4) (:$40.); if ip="&destination" then call symput('dest',n); call symput(n,put(_n_,best.)); call symput('ip' || compress(strip(ip),'.'),put(_n_,best.)); run; data _null_; length route $40; retain route ' '; obsnum=symget("&start")*1; do until(obsnum=symget('ip' || compress(strip("&destination"),'.'))); set route_map point=obsnum; route=cats(of route n); array dest[3]; array goto[4]; j=0; i=1; do until(j>0); if i>3 then j=4; else if dest="&destination" then j=i; else if index(dest,'/') then if prxmatch('/' || strip(substr(prxchange('s/\./\\./',3,dest),1,length(strip(dest))-4)) || '/',"&destination") then j=i; i+1; end; obsnum=symget('ip' || compress(strip(goto ),'.')); end; route=cats(route,"&dest"); put 'SOLUTION FOUND: ' route; stop; run; SOLUTION FOUND: ONBMLDGFIJK

FriedEgg · ‎10-13-2011

Unfortuneatly I am not incredibly familiar with XMing, Whenever I interact with unix through windows I would choose to use a proper VNC rather than the X11 forwarding methods XMing seems to employ?

FriedEgg · ‎10-13-2011

This sounds like if anything, probably a configuration issue with your XMing installation. Can you use the keys in other programs through the same process?

FriedEgg · ‎10-13-2011

data foo; infile cards truncover; retain mlen_this 0 mlen_that 0; input wingband this :$100. that $100.; mlen_this=max(mlen_this,length(this)); mlen_that=max(mlen_that,length(that)); call symput('mlen_this','$' || put(mlen_this,best.)); call symput('mlen_that','$' || put(mlen_that,best.)); drop mlen:; cards; 1 abcdef ab 2 abc abcd 3 ab abc 4 abcd abcde 5 a a ; run; data bar; length this &mlen_this that &mlen_that; set foo; run;

FriedEgg · ‎10-13-2011

This is a fairly simple logic puzzle. You can easily solve by hand, once you understand the /24 piece. Basically is similar to saying 117.153.4.* (where star is a catch all from 0-255) O => 117.153.4.238(j) ============================================================================================================================================================ O 244.102.114.233 224.216.69.120 => 5.153.148.97 84.104.233.40 => 85.113.64.65 117.153.4.0/24 => 22.230.113.114 190.64.129.17 o-22.230.113.114(n) N 22.230.113.114 84.104.233.40 => 134.24.6.169 244.102.114.233 => 224.216.69.120 85.113.64.0/24 => 244.102.114.233 89.88.225.242 n-89.88.225.242(b) B 89.88.225.242 89.88.225.242 => 5.153.148.97 190.64.129.17 => 224.216.69.120 85.113.64.0/24 => 22.230.113.114 134.24.6.169 b-134.24.6.169(m) M 134.24.6.169 22.230.113.114 => 89.88.225.242 244.102.114.233 => 224.216.69.120 116.134.182.0/24 => 22.230.113.114 85.113.64.65 m-85.113.64.65(l) L 85.113.64.65 84.104.233.40 => 134.24.6.169 134.24.6.169 => 5.153.148.97 89.88.225.0/24 => 117.153.4.238 89.86.234.208 l-89.86.234.208(d) D 89.86.234.208 85.250.166.178 => 116.134.182.91 89.88.225.242 => 84.104.233.40 117.153.4.0/24 => 45.87.155.12 85.113.64.65 d-45.87.155.12(g) G 45.87.155.12 89.88.225.242 => 116.134.182.91 117.153.4.238 => 85.250.166.178 116.134.182.0/24 => 117.153.4.238 89.86.234.208 g-85.250.166.178(f) F 85.250.166.178 9.140.95.24 => 9.140.95.24 117.153.4.238 => 7.86.78.45 134.24.6.0/24 => 84.104.233.40 45.87.155.12 f-7.86.78.45(i) I 7.86.78.45 117.153.4.238 => 9.140.95.24 190.64.129.17 => 84.104.233.40 224.216.69.0/24 => 85.250.166.178 116.134.182.91 i-9.140.95.24(j) J 9.140.95.24 134.24.6.169 => 85.250.166.178 22.230.113.114 => 7.86.78.45 117.153.4.0/24 => 117.153.4.238 84.104.233.40 j-117.153.4.238(k)... done! ONBMLDGFIJK

FriedEgg · ‎10-13-2011

These are valid. Look into route summarization. It is the combination of IP Address and subnet mask. For example 117.153.4.0/24 is the ip address 117.153.4.0 with the subnet mask 255.255.255.0

FriedEgg · ‎10-12-2011

dkricond and dkrocond come to mind, however I am unable to remember the specifics circumstances they aid in off hand. proc options group=sasfiles; run; proc options group=errorhandling; run;

FriedEgg · ‎10-12-2011

Hi, Here is a new challenge from the google treasure hunt 2008. http://treasurehunt.appspot.com I found the first question rather entertaining if not a little troublesome. http://communities.sas.com/thread/31422?tstart=0 Question Below is a diagram of a computer network. The nodes are hosts on the network, and the lines between them are links. A packet is sent out from host O with a destination of 117.153.4.238. Which nodes does the packet pass through on its way to the destination? (include start and final node in your answer) Here is a network routing table you'll need to determine the path taken: Node Ip address Routing table entry Routing table entry Routing table entry Default route A 224.216.69.120 134.24.6.169 => 134.24.6.169 22.230.113.114 => 89.88.225.242 7.86.78.0/24 => 22.230.113.114 190.64.129.17 B 89.88.225.242 89.88.225.242 => 5.153.148.97 190.64.129.17 => 224.216.69.120 85.113.64.0/24 => 22.230.113.114 134.24.6.169 C 5.153.148.97 190.64.129.17 => 89.86.234.208 116.134.182.91 => 190.64.129.17 84.104.233.0/24 => 117.153.4.238 89.88.225.242 D 89.86.234.208 85.250.166.178 => 116.134.182.91 89.88.225.242 => 84.104.233.40 117.153.4.0/24 => 45.87.155.12 85.113.64.65 E 84.104.233.40 7.86.78.45 => 7.86.78.45 116.134.182.91 => 85.250.166.178 134.24.6.0/24 => 89.86.234.208 9.140.95.24 F 85.250.166.178 9.140.95.24 => 9.140.95.24 117.153.4.238 => 7.86.78.45 134.24.6.0/24 => 84.104.233.40 45.87.155.12 G 45.87.155.12 89.88.225.242 => 116.134.182.91 117.153.4.238 => 85.250.166.178 116.134.182.0/24 => 117.153.4.238 89.86.234.208 H 116.134.182.91 117.153.4.238 => 7.86.78.45 89.86.234.208 => 45.87.155.12 244.102.114.0/24 => 117.153.4.238 89.86.234.208 I 7.86.78.45 117.153.4.238 => 9.140.95.24 190.64.129.17 => 84.104.233.40 224.216.69.0/24 => 85.250.166.178 116.134.182.91 J 9.140.95.24 134.24.6.169 => 85.250.166.178 22.230.113.114 => 7.86.78.45 117.153.4.0/24 => 117.153.4.238 84.104.233.40 K 117.153.4.238 190.64.129.17 => 5.153.148.97 22.230.113.114 => 89.86.234.208 117.153.4.0/24 => 85.113.64.65 45.87.155.12 L 85.113.64.65 84.104.233.40 => 134.24.6.169 134.24.6.169 => 5.153.148.97 89.88.225.0/24 => 117.153.4.238 89.86.234.208 M 134.24.6.169 22.230.113.114 => 89.88.225.242 244.102.114.233 => 224.216.69.120 116.134.182.0/24 => 22.230.113.114 85.113.64.65 N 22.230.113.114 84.104.233.40 => 134.24.6.169 244.102.114.233 => 224.216.69.120 85.113.64.0/24 => 244.102.114.233 89.88.225.242 O 244.102.114.233 224.216.69.120 => 5.153.148.97 84.104.233.40 => 85.113.64.65 117.153.4.0/24 => 22.230.113.114 190.64.129.17 P 190.64.129.17 9.140.95.24 => 244.102.114.233 22.230.113.114 => 224.216.69.120 117.153.4.0/24 => 5.153.148.97 85.113.64.65 Enter the nodes the packet passes through below (Note: Answer must start with O, end with the destination node name, and contain only node names.)

FriedEgg · ‎10-12-2011

Fixed the approach with sas generated values. It finished in about one and half minutes total, so I am happy enough with it at this point: options fullstimer macrogen mlogic mprint spool; proc fcmp outlib=work.func.cipher; function isprime(num); n=0; j=3; do while(j<=int(sqrt(num)) and n=0); if mod(num,j)=0 then n+1; j+2; end; if n=0 then return(1); else return(0); endsub; run; %let cmplib=%sysfunc(getoption(cmplib)); options cmplib=(work.func &cmplib); data primes; pnum=1; prime=2; output; do prime=3 to 10**7 by 2; if isprime(prime) then do; output; pnum+1; call symput('max',prime); end; end; call symput('nobs',pnum); keep prime; run; %macro sumprimes(seq1,seq2,seq3,seq4,max,nobs); sasfile work.primes load; %do i=1 %to 4; data _s&&seq&i(sortedby=ptot); obsnum=1; cnt=1; iter=1; ptot=0; do until(obsnum+&&seq&i>&nobs or ptot>&max); set primes point=obsnum; ptot+prime; if cnt=&&seq&i then do; output; iter+1; cnt=1; obsnum=iter; ptot=0; end; else do; obsnum+1; cnt+1; end; end; stop; keep ptot; run; %end; sasfile work.primes close; data _null_; merge %do i=1 %to 4; _s&&seq&i(in=in&i) %end;; by ptot; if in1 and in2 and in3 and in4 and isprime(ptot) then put 'SOLUTION FOUND: ' ptot comma20.; run; %mend; %sumprimes(19,21,405,781,&max,&nobs) options cmplib=(&cmplib);

FriedEgg · ‎10-12-2011

Here is a method to solve without having sas generate the prime numbers but rather to download them from the internet... * The following was only tested under linux, it would require adjustment for other OS; x cd /home/friedegg; x 'for i in {1..50}; do wget "http://primes.utm.edu/lists/small/millions/primes$i.zip" && unzip "primes$i.zip" && rm -f "primes$i.zip"; done'; data primes; infile '/home/friedegg/primes*.txt' firstobs=3 termstr=crlf end=eof; input prime :??best. @@; if prime=. then delete; if eof then do; call symput('nobs',put(_n_,best.)); call symput('max',put(prime,best.)); end; run; %macro sumprimes(seq1,seq2,seq3,seq4,max,nobs); sasfile work.primes load; %do i=1 %to 4; data _s&&seq&i(sortedby=ptot); obsnum=1; cnt=1; iter=1; ptot=0; do until(obsnum+&&seq&i>&nobs or ptot>&max); set primes point=obsnum; ptot+prime; if cnt=&&seq&i then do; output; iter+1; cnt=1; obsnum=iter; ptot=0; end; else do; obsnum+1; cnt+1; end; end; stop; keep ptot; run; %end; data _null_; length ptots $32000; retain ptots ' '; merge %do i=1 %to 4; _s&&seq&i(in=in&i) %end;; by ptot; if in1 and in2 and in3 and in4 then do; ptots=catx(',', of ptots ptot); call symput('ptots',ptots); end; run; proc sql; select 'SOLUTION FOUND: ' || put(prime,best.) from primes where prime in (&ptots); quit; sasfile work.primes close; %mend; %sumprimes(19,21,405,781,&max,&nobs) SOLUTION FOUND: 9387361

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