The simplest answer is a combination of 2 answers:
data have;
input Subject $ Value1 value2 value3 value4;
datalines;
10001 3 4 5 8
10007 4 6 4 9
10009 4 9 10 3
run;data need;
set have;
value = max(of value1 - value4);
run;
the dataset NEED will look as follows:
10001 3 4 5 8 8
10007 4 6 4 9 9
10009 4 9 10 3 10
Message was edited by: Amin Bardai
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