In the following SAS program, the input data files are sorted by the NAMES variable:
libname temp 'SAS-data-library';
data temp.sales;
merge temp.sales work.receipt;
by names;
run;
Which one of the following results occurs when this program is submitted?
A. The program executes successfully and a temporary SAS data set is created.
B. The program executes successfully and a permanent SAS data set is created.
C. The program fails execution because the same SAS data set is referenced for both read and write operations.
D. The program fails execution because the SAS data sets on the MERGE statement are in two different libraries.
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21 comments:
Answer: B
Even if the same named datasets are used in read and write, SAS first builds the observations in temporary memory and then overwrites the dataset
No, it doesn't work that way. SAS throws up an error stating the table doesn't exist. And merge principle works by reading the descriptor portion of the datasets first and then starts building up the new data set
so C is the answer
It is not C. it does not matter if you use the same name in read and write. you can still merge two data sets. the problem is the by statement.
you can sort them once the two sets are merges by using proc sort data=temp.sales; by names;
So the only choice for this question is D.
Answer is B
Hello,
The post are really catchy for Base SAS certification preparation
Yes the answer is B.
Option A - It cannot be A bcoz you are defining a library ref.hence output dataset is permanent dataset.
Option C- You can mention the same data set for read and write dataset. The same dataset will be overwritten(I tried it).
Option D - Well it doesn't matter if libraries are different as long as SAS knows where to find it. One library ref "temp" is already been mentioned and another one is "work".So it cant be D.
Only answer for the questions can be "B".
Yes , the answer is B
ans is B. I tried using below code..you can also try
libname testing "D:\";
data testing.test1;
input id $ age;
cards;
01 24
02 34
03 45
04 34
;
data test2;
input id $ gender $;
cards;
01 M
02 F
03 F
04 F
;
data testing.test1;
merge testing.test1 test2;
by id;
run;
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