The following SAS program is submitted:
data work.empsalary;
set work.people (in = inemp) work.money (in = insal);
if insal and inemp;
run;
The SAS data set WORK.PEOPLE has 5 observations, and the data set WORK.MONEY has 7 observations.
How many observations will the data set WORK.EMPSALARY contain?
A. 0
B. 5
C. 7
D. 12
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Other brand and product names are trademarks of their respective companies.
This blog is not responsible for any kind of copyright violation. This blog just collects the links hosted or posted by other servers/people/search engines.The creator of this page or the ISP(s) hosting any content on this site take no responsibility for the way you use the information provided on this site.If anybody has any copyright claim on it and doesn’t wish the information provided to be shown on our site, please do respond to this email. We shall remove them off immediately. Any inconvenience is regretted. Please do mention your exact grievance/problems with respect to certain third party links. We assure you that appropriate action will be taken off. Thank you
10 comments:
Answer: A
Since there is no BY statement..this is an example of concatenation…there is no match between each of the datasets…therefore 0 records in the output dataset
Thanks for Answer SASGuru....
wait a minute! shouldn't there be 12? i thought concatenation would append one file over the other.
nevermind. the "if insal and inemp;" statement create no matches, therefore no records.
In concatenation...SAS reads 1 dataset first and then the second. So if you specify IN=variable for both datasets and wants to know that whether both the dataset contributed to the current observation seems to be a false idea bcoz 1 dataset get reads first and then the second.So no matches.
Hence answer is A.
YES Amit that's the right explanation of the answer
This is not just a SET statement. It has a IN=variable for both datasets. BUT there is no by statement. So the answer is 0 match. Answer is A
Not enough info given... if you run
data test;
input obs name $ level;
cards;
1 Frank 1
2 Joan 2
3 Sui 2
4 Jose 3
5 Burt 4
6 Kelly .
7 Juan 1
;
run;
data test2;
input obs name $ level;
cards;
1 Frank 1
2 Joan 2
3 Sui 2
4 Jose 3
;
run;
data mergetest;
merge test(in=a) test2(in=b);
if a and b;
run;
then there are 4 obs in the merged set.
We are not told in the question that the two sets do/do not have matching rows.
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hi unknown run your code again this time without the merge and replace with set !
This is not a merge question it is a concatenation
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