Chapter 14: Generating Data
with Do Loops
OBJECTIVES
•
•
•
•
•
Understand iterative DO loops.
Construct a DO loop to perform repetitive calculations
Use DO loops to eliminate redundant code.
Use DO loop processing to conditionally execute code.
Generate multiple observations in one iteration of the
DATA step
• Construct nested DO loops
DO Loop Processing
• Statements within a DO loop execute for a
specific number of iterations or until a
specific condition stops the loop.
DATA statement
SAS statements
DO statement
iterated SAS statements
END statement
SAS statements
RUN statement
2
Repetitive Coding
• Compare the interest for yearly versus
quarterly compounding on a $50,000
investment made for one year at 7.5
percent interest.
• How much money will a person accrue in
each situation?
3
Repetitive Coding
data compound;
Amount=50000;
Rate=.075;
Yearly=Amount*Rate;
Quarterly+((Quarterly+Amount)*Rate/4);
Quarterly+((Quarterly+Amount)*Rate/4);
Quarterly+((Quarterly+Amount)*Rate/4);
Quarterly+((Quarterly+Amount)*Rate/4);
run;
4
Repetitive Coding
proc print data=compound noobs;
run;
PROC PRINT Output
Amount
Rate
Yearly
Quarterly
50000
0.075
3750
3856.79
What if you wanted to determine the quarterly compounded
interest after a period of 20 years
(80 quarters)?
5
DO Loop Processing
data compound(drop=Qtr);
Amount=50000;
Rate=.075;
Yearly=Amount*Rate;
do Qtr=1 to 4;
Quarterly+((Quarterly+Amount)*Rate/4);
end;
run;
6
The Iterative DO Statement
• The iterative DO statement executes statements
between DO and END statements repetitively,
based on the value of an index variable.
DO index-variable=specification-1 <,…specification-n>;
<additional SAS statements>
END;
• specification-1…specification-n can represent a
range of values or a list of specific values.
7
General Syntax of a DO Loop
DO index-variable = start TO stop BY increment;
SAS statements;
END;
• Index-variable stores the value of the current iteration of the
DO loop.
• Start, stop, increment values are set upon entry of the DO
loop.
• Can not be changed during the processing the DO loop
• Can be numbers, variables, or SAS expressions.
The Iterative DO Statement
• The BY clause is optional. If ignored, the increment is 1.
• Increment may be negative, which will process the DO
loop backwards. For this situation, Start should be larger
than Stop.
• When completing the DO loop, the value of the indexvariable is STOP_value+Increment, not the STOP_value.
Ex: Do month = 1 to 12;
pay = pay*1.05;
END;
Once the DO loop is completed, the value of month = 13, NOT
12.
9
The Iterative DO Statement
What are the values of each of the four index
variables?
Out of range
do i=1 to 12;
1 2 3 4 5 6 7 8 9 10 11 12 13
do j=2 to 10 by 2;
2 4 6 8 10 12
do k=14 to 2 by –2;
14 12 10 8 6 4 2 0
do m=3.6 to 3.8 by .05;
3.60 3.65 3.70 3.75 3.80 3.85
10
The Iterative DO Statement
DO index-variable=item-1 <,…item-n>;
• item-1 through item-n can be either all
numeric or all character constants, or they
can be variables.
• The DO loop is executed once for each
value in the list.
11
The Iterative DO Statement
How many times will each DO loop execute?
do Month='JAN','FEB','MAR';
3 times.
do Fib=1,2,3,5,8,13,21;
7 times.
do i=Var1,Var2,Var3;
3 times.
do j=BeginDate to Today() by 7;
The number of iterations depends
on the values of BeginDate and Today().
do k=Test1-Test50;
k is determined
by subtracting Test50 from Test1.
1 time. A single value of
12
Use ‘TO’ not ‘-’.
Performing Repetitive Calculations
On January 1 of each year, $5,000 is invested in
an account. Determine the value of the account
after three years based on a constant annual
interest rate of 7.5 percent.
data invest;
do Year=2001 to 2003;
Capital+5000;
Capital+(Capital*.075);
end;
run;
14
Performing Repetitive Calculations
proc print data=invest noobs;
run;
PROC PRINT Output
Year
Capital
2004
17364.61
NOTE: The Year is 2004, NOT 2003. The Capital
is not for 2003.
15
Performing Repetitive Calculations
Generate a separate observation for each year.
data invest;
do Year=2001 to 2003;
Capital+5000;
Capital+(Capital*.075);
output;
end;
Explicit output
run;
proc print data=invest noobs;
run;
16
Performing Repetitive Calculations
PROC PRINT Output
Year
Capital
2001
2002
2003
5375.00
11153.13
17364.61
Why is the value of Year not equal to 2004 in the last observation?
NOTE: The use of OUTPUT, the explicit output, controls the output
from 2001 to 2003.
17
Reducing Redundant Code
• The example that forecast the growth of each
division of an airline.
• Partial Listing of the data set: mylib.growth
18
Division
Num
Emps
Increase
APTOPS
FINACE
FLTOPS
205
198
187
0.075
0.040
0.080
Reducing Redundant Code
A Forecasting of Airline Growth
data forecast;
set mylib.growth(rename=(NumEmps=NewTotal));
Year=1;
NewTotal=NewTotal*(1+Increase);
output;
Year=2;
NewTotal=NewTotal*(1+Increase);
output;
Year=3;
NewTotal=NewTotal*(1+Increase);
output;
run;
What if you want to forecast growth over the next 30 years?
19
Reducing Redundant Code
Use a DO loop to eliminate the redundant
code in the previous example.
data forecast;
set mylib.growth(rename=(NumEmps=NewTotal));
do Year=1 to 3;
NewTotal=NewTotal*(1+Increase);
output;
end;
run;
20
Reducing Redundant Code
proc print data=forecast noobs;
run;
Partial PROC PRINT Output
Division
APTOPS
APTOPS
APTOPS
FINACE
New
Total
Increase
Year
220.38
236.90
254.67
205.92
0.075
0.075
0.075
0.040
1
2
3
1
What if you want to forecast the number of years it would take for
the size of the Airport Operations Division to exceed 300 people?
21
Conditional Iterative Processing
• You can use DO WHILE and DO UNTIL
statements to stop the loop when a
condition is met rather than when the
index variable exceeds a specific value.
• To avoid infinite loops, be sure that the
condition specified will be met.
22
The DO WHILE Statement
The DO WHILE statement executes statements in
a DO loop while a condition is true.
DO WHILE (expression);
<additional SAS statements>
END;
• expression is evaluated at the top of the loop.
• The statements in the loop never execute if the
expression is initially false.
23
The DO UNTIL Statement
The DO UNTIL statement executes statements in a DO
loop until a condition is true.
DO UNTIL (expression);
<additional SAS statements>
END;
• expression is evaluated at the bottom of the loop.
• The statements in the loop are executed at least
once.
24
Conditional Iterative Processing
Task:
Determine the number of years it would take
for an account to exceed $1,000,000 if
$5,000 is invested annually at 7.5 percent.
25
Conditional Iterative Processing
data invest;
do until(Capital>1000000);
Year+1;
Capital+5000;
Capital+(Capital*.075);
end;
run;
proc print data=invest noobs;
run;
26
Conditional Iterative Processing
PROC PRINT Output
Capital
1047355.91
Year
38
How could you generate the same result
with a DO WHILE statement?
27
The Iterative DO Statement with a
Conditional Clause
You can combine DO WHILE and DO UNTIL
statements with the iterative DO statement.
DO index-variable=start TO stop <BY increment>
WHILE | UNTIL (expression);
<additional SAS statements>
END;
This is one method of avoiding an infinite loop in DO
WHILE or DO UNTIL statements.
28
The Iterative DO Statement with a
Conditional Clause
Task:
• Determine the return of the account again.
• Stop the loop if 25 years is reached or more
than $250,000 is accumulated.
29
The Iterative DO Statement with a
Conditional Clause
data invest;
do Year=1 to 25 until(Capital>250000);
Capital+5000;
Capital+(Capital*.075);
end;
run;
proc print data=invest noobs;
run;
30
The Iterative DO Statement with a
Conditional Clause
PROC PRINT Output
31
Year
Capital
21
255594.86
Nested DO Loops
Nested DO loops are loops within loops.
When you nest DO loops,
– use different index variables for each loop
– be certain that each DO statement has a
corresponding END statement.
32
Nested DO Loops
Task:
• Create one observation per year for five
years and show the earnings if you invest
$5,000 per year with 7.5 percent annual
interest compounded quarterly.
33
Nested DO Loops
5x
data invest(drop=Quarter);
do Year=1 to 5;
Capital+5000;
do Quarter=1 to 4;
4x
Capital+(Capital*(.075/4));
end;
output;
end;
run;
proc print data=invest noobs;
run;
34
Nested DO Loops
PROC PRINT Output
Year
1
2
3
4
5
Capital
5385.68
11186.79
17435.37
24165.94
31415.68
How could you generate one observation for each quarterly
amount?
35
Nested DO Loops
Task:
• Compare the final results of investing $5,000
a year for five years in three different banks
that compound quarterly. Assume each bank
has a fixed interest rate.
mylib.Banks
36
Name
Rate
Calhoun Bank and Trust
State Savings Bank
National Savings and Trust
0.0718
0.0721
0.0728
Nested DO Loops
3x
37
data invest(drop=Quarter Year);
set mylib.banks;
Capital=0;
do Year=1 to 5;
Capital+5000;
do Quarter=1 to 4;
5x
4x
Capital+(Capital*(Rate/4));
end;
end;
run;
Nested DO Loops
proc print data=invest noobs;
run;
PROC PRINT Output
Name
Calhoun Bank and Trust
State Savings Bank
National Savings and Trust
38
Rate
Capital
0.0718 31106.73
0.0721 31135.55
0.0728 31202.91