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

2 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

3 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?

4 Repetitive Coding data compound; Amount=50000; Rate=.075; Yearly=Amount*Rate; Quarterly+((Quarterly+Amount)*Rate/4); run;

5 Repetitive Coding proc print data=compound noobs; run; Amount Rate Yearly Quarterly What if you wanted to determine the quarterly compounded interest after a period of 20 years (80 quarters)? PROC PRINT Output

6 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;

7 The iterative DO statement executes statements between DO and END statements repetitively, based on the value of an index variable. specification-1…specification-n can represent a range of values or a list of specific values. The Iterative DO Statement DO index-variable=specification-1 ; END;

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.

9 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 index- variable 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. The Iterative DO Statement

10 The Iterative DO Statement What are the values of each of the four index variables? do i=1 to 12; do j=2 to 10 by 2; do k=14 to 2 by –2; do m=3.6 to 3.8 by.05; Out of range

11 The Iterative DO Statement 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. DO index-variable=item-1 ;

12 do Month='JAN','FEB','MAR'; do Fib=1,2,3,5,8,13,21; do i=Var1,Var2,Var3; do j=BeginDate to Today() by 7; do k=Test1-Test50; The Iterative DO Statement How many times will each DO loop execute? 3 times. 7 times. 3 times. The number of iterations depends on the values of BeginDate and Today(). 1 time. A single value of k is determined by subtracting Test50 from Test1. Use ‘TO’ not ‘-’.

13 DO Loop Logic DO loops iterate within the normal looping process of the DATA step. Stop DATA step YES Initialize PDV Initialize PDV Execute "read" statement Execute "read" statement EOF marker?

14 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;

15 Performing Repetitive Calculations Year Capital proc print data=invest noobs; run; PROC PRINT Output NOTE: The Year is 2004, NOT The Capital is not for 2003.

16 Performing Repetitive Calculations data invest; do Year=2001 to 2003; Capital+5000; Capital+(Capital*.075); output; end; run; proc print data=invest noobs; run; Generate a separate observation for each year. Explicit output

17 Performing Repetitive Calculations Year Capital 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 PROC PRINT Output

18 Reducing Redundant Code The example that forecast the growth of each division of an airline. Partial Listing of the data set: mylib.growth Num Division Emps Increase APTOPS FINACE FLTOPS

19 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?

20 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;

21 Reducing Redundant Code proc print data=forecast noobs; run; Partial PROC PRINT Output New Division Total Increase Year APTOPS APTOPS APTOPS FINACE 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?

22 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.

23 The DO WHILE statement executes statements in a DO loop while a condition is true. expression is evaluated at the top of the loop. The statements in the loop never execute if the expression is initially false. The DO WHILE Statement DO WHILE (expression); END;

24 The DO UNTIL Statement The DO UNTIL statement executes statements in a DO loop until a condition is true. expression is evaluated at the bottom of the loop. The statements in the loop are executed at least once. DO UNTIL (expression); END;

25 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.

26 Conditional Iterative Processing data invest; do until(Capital> ); Year+1; Capital+5000; Capital+(Capital*.075); end; run; proc print data=invest noobs; run;

27 Conditional Iterative Processing PROC PRINT Output How could you generate the same result with a DO WHILE statement? Capital Year

28 The Iterative DO Statement with a Conditional Clause You can combine DO WHILE and DO UNTIL statements with the iterative DO statement. This is one method of avoiding an infinite loop in DO WHILE or DO UNTIL statements. DO index-variable=start TO stop WHILE | UNTIL (expression); END;

29 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.

30 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;

31 The Iterative DO Statement with a Conditional Clause PROC PRINT Output Year Capital

32 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.

33 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.

34 data invest(drop=Quarter); do Year=1 to 5; Capital+5000; do Quarter=1 to 4; Capital+(Capital*(.075/4)); end; output; end; run; proc print data=invest noobs; run; Nested DO Loops 5x5x4x4x

35 PROC PRINT Output How could you generate one observation for each quarterly amount? Nested DO Loops Year Capital

36 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 Name Rate Calhoun Bank and Trust State Savings Bank National Savings and Trust

37 Nested DO Loops data invest(drop=Quarter Year); set mylib.banks; Capital=0; do Year=1 to 5; Capital+5000; do Quarter=1 to 4; Capital+(Capital*(Rate/4)); end; run; 4x4x 5x5x 3x3x

38 Nested DO Loops PROC PRINT Output Name Rate Capital Calhoun Bank and Trust State Savings Bank National Savings and Trust proc print data=invest noobs; run;