Chapter 13 Processing Data Iteratively. Section 13.1 DO-Loop Processing

Chapter 13

Processing Data Iteratively

Section 13.1

DO-Loop Processing

3 3

Objectives Describe iterative DO loops. Use DO loops to generate data. Use DO loops to eliminate redundant code. Use DO-loop processing to conditionally execute

code.

4 4

DO-Loop ProcessingStatements within a DO loop execute for a specific number of iterations or until a specific condition stops the loop.

DATA statement<additional SAS statements>

DO statementiterated SAS statements

END statement<additional SAS statements>

RUN statement

DATA statement<additional SAS statements>

DO statementiterated SAS statements

END statement<additional SAS statements>

RUN statement

5 5

DO-Loop ProcessingYou can use DO loops to perform repetitive calculations generate data eliminate redundant code execute SAS code conditionally.

6 6

Repetitive CodingCompare 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?

7 7

Repetitive Codingdata 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;

8 8

Repetitive Codingproc print data=compound noobs;run;

Amount Rate Yearly Quarterly

50000 0.075 3750 3856.79

What if you want to determine the quarterly compounded interest after a period of 20 years (80 quarters)?

PROC PRINT Output

9 9

DO Loop Processingdata compound(drop=i); Amount=50000; Rate=.075; Yearly=Amount*Rate; do i=1 to 4; Quarterly+((Quarterly+Amount)*Rate/4); end;run;

1010

The Iterative DO StatementThe 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;

DO index-variable=specification-1 <,…specification-n>; <additional SAS statements>END;

index-variable names a variable whose value governs execution of the DO loop.• Index-variable is required.• Unless dropped, it is

included in the data setthat is being created.

specification-1…specification-n represents a range of values or a list of specific values.

specification denotes an expressionor a series of expressions. The iterative DO statement requires at least one specification argument

1111

The Iterative DO Statement

DO index-variable=start TO stop <BY increment>;DO index-variable=start TO stop <BY increment>;

start specifies the initial value of the index variable.

stop specifies the ending value of the index variable.

increment optionally specifies a positive or negative number to control the incrementing of index-variable. If increment is not specified, the index variable is increased by 1.

1212

The values of start, stop, and increment must be numbers or expressions that yield numbers are established before executing the loop.

Any changes to the values of stop or increment made within the DO loop do not affect the number of iterations.

Avoid changing the value of the index variable within the DO loop to prevent an infinite loop.


DO index-variable=start TO stop <BY increment>;DO index-variable=start TO stop <BY increment>;

1313

The Iterative DO StatementWhat 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;

1 2 3 4 5 6 7 8 9 10 11 12 13

2 4 6 8 10 12

14 12 10 8 6 4 2 0

3.60 3.65 3.70 3.75 3.80 3.85

Out of range

Out of range

Out of range

Out of range

...

1414


item-1 through item-n can be either all numeric or all character constants, or they can be variables.

Enclose character constants in quotation marks.

The DO loop is executed once for each value in the list.

DO index-variable=item-1 <,…item-n>;DO index-variable=item-1 <,…item-n>;

1515

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 StatementHow many times will each DO loop execute?

3 times

7 times

3 times

Unknown. The number of iterations dependson the values of BeginDate and Today().

One time. A single value of k is determined

by subtracting Test50 from Test1.

...

1616

DO Loop LogicDO loops iterate within the normal looping process of the DATA step.

Initialize PDV.

Initialize PDV.

Execute READ statement.


Execute program statements.

Execute program statements.

EOFmarker?

NO

...

1717

Define start,stop, and increment

values. Set INDEX = start.

Is INDEXout of

range?

Execute statements in the loop.Execute statements in the loop.

INDEX = INDEX + incrementINDEX = INDEX + increment

NO

Execute additional program statements.

Execute additional program statements.

Output observation to the SAS data set.

Output observation to the SAS data set.

YES

do index=start to stop by increment; SAS statementsend;

do index=start to stop by increment; SAS statementsend;

...

1818

DO Loop LogicDO loops iterate within the normal looping process of the DATA step.

StopDATAstep

YES

Initialize PDV.

Initialize PDV.



EOFmarker?

...

1919

Performing Repetitive CalculationsOn 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;

2020


Year Capital

0

_N_D

PDV

Repetitive Calculations: Compilation

R

...

2121


Year

.

Capital

0

_N_

1

DPDV

Repetitive Calculations: Execution

Initialize PDV to missing

R

...

2222


Year

2001

Capital

0

_N_

1

DPDV

Is Yearout ofrange?


R

...

2323

Year

2001

Capital

5000

_N_

1

DPDV

Repetitive Calculations: Executiondata invest; do Year=2001 to 2003; Capital+5000; Capital+(Capital*.075); end;run;

0 + 5000

R

...

2424


Year

2001

Capital

5375

_N_

1

DPDV

5000 + (5000 * .075)


R

...

2525


Year

2002

Capital

5375

_N_

1

DPDV

Year + 1


R

...

2626

Year

2002

Capital

5375

_N_

1

DPDV


R


Is Yearout ofrange?

...

2727

Year

2002

Capital

10375

_N_

1

DPDV

5375 + 5000



R

...

2828

Year

2002

Capital

11153.13

_N_

1

DPDV

10375 + (10375 * .075)



R

...

2929

Year

2003

Capital

11153.13

_N_

1

DPDV


Year + 1


R

...

3030

Year

2003

Capital

11153.13

_N_

1

DPDV



R

Is Yearout ofrange?

...

3131

Year

2003

Capital

16153.13

_N_

1

DPDV


11153.13+5000


R

...

3232

Year

2003

Capital

17364.61

_N_

1

DPDV


16153.13 + (16153.13 * .075)


R

...

3333

Year

2004

Capital

17364.61

_N_

1

DPDV


Year + 1


R

...

3434

Year

2004

Capital

17364.61

_N_

1

DPDV



R

Is Yearout ofrange?

...

3535

Year

2004

Capital

17364.61

_N_

1

DPDV



R

...

3636

Write observation to invest.


Automatic output

Year

2004

Capital

17364.61

_N_

1

DPDV


R

3737

Performing Repetitive Calculations

Year Capital

2004 17364.61

proc print data=invest noobs;run;

PROC PRINT Output

Why is Year = 2004?

3838


Year Capital

2004 17364.61


PROC PRINT Output

Why is Year = 2004?

Year was incremented to 2004, which was then out of range. SAS reached automatic output at the end of the DATA step and output 2004.

3939

This exercise reinforces the concepts discussed previously.

Exercise

4040

Exercises

Leonardo Fibonacci, who was born in the 12th century, studied asequence of numbers with a rule for determining the next number

in a sequence. 1, 1, 2, 3, 5, 8, 13, 21, 34 …

The Rule: Take a number and add it to the number in front of it to get the next number.

0+1=1, 1+1=2, 1+2=3, 2+3=5 …

You want to know the 31st Fibonacci value.

Hint: 1 is both the first and second Fibonacci number. Create three variables: Number – the actual Fibonacci numberPriorNumber – the previous number in the sequence (initially set to 0) Fib – the index value of your DO LOOP.

4141

Exercises

Output

The SAS System

Number

1346269

4242

Exercises

data fibonacci; drop Fib priornumber; Number = 1; PriorNumber = 0; do Fib = 1 to 30; number = number + priornumber;

priornumber = number - priornumber; end;run;proc print data=fibonacci noobs;run;

4343

Performing Repetitive CalculationsWhat if you want to see one row for each year?

data invest; do Year=2001 to 2003; Capital+5000; Capital+(Capital*.075); end;run;proc print data=invest noobs;run;

Year Capital

2004 17364.61

4444


data invest; do Year=2001 to 2003; Capital+5000; Capital+(Capital*.075); output; end;run;


Generate a separate observation for each year.

The OUTPUT statement within the DO loop writes one observation for each of the three iterations of the DO loop.

4545

Performing Repetitive CalculationsPROC PRINT Output

Year Capital

2001 5375.00 2002 11153.13 2003 17364.61

Why is the value of Year not equal to 2004 in the last observation?

4646


data invest; do Year=2001 to 2003; Capital+5000; Capital+(Capital*.075); output; end;run;

Generate a separate observation for each year.

The OUTPUT statement within the DO loop writes one observation for each of the three iterations of the DO loop.

Because there is an OUTPUT statement, SAS will not execute an automatic output at the RUN statement.

4747

Reducing Redundant CodeRecall the example that forecasts the growth of each division of an airline.

Partial Listing of prog2.growth

NumDivision Emps Increase

APTOPS 205 0.075 FINACE 198 0.040 FLTOPS 187 0.080

4848

A Forecasting Application (Review)data forecast; set prog2.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?

4949

Reducing Redundant CodeUse a DO loop to eliminate the redundant code in the previous example.

data forecast; set prog2.growth(rename=(NumEmps=NewTotal)); do Year=1 to 3; NewTotal=NewTotal*(1+Increase); output; end;run;

5050

Reducing Redundant Codeproc print data=forecast noobs;run;

Partial PROC PRINT Output

NewDivision Total Increase Year

APTOPS 220.38 0.075 1APTOPS 236.90 0.075 2APTOPS 254.67 0.075 3FINACE 205.92 0.040 1

5151


Exercise

5252

Exercises

Modify the program from the previous Fibonacci exercise.

You want to include one row for each Fibonacci Number.

Obs Number

1 1 2 1 3 2 4 3 5 5 6 8 7 13 8 21 9 34 . . . 27 196418 28 317811 29 514229 30 832040 31 1346269

5353

Exercises

data fibonacci; drop Fib PriorNumber; Number = 1; PriorNumber = 0; output; do Fib = 1 to 30; number = number + priornumber; priornumber = number-priornumber; output; end;run;proc print data=fibonacci;run;

5454

Conditional Iterative ProcessingWhat if you want to forecast the number of years that it would take for the size of the Airport Operations Division to exceed 300 people?

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.

5555

The DO WHILE StatementThe DO WHILE statement executes statements in a DO loop while a condition is true.

The expression is evaluated at the top of the loop. The statements in the loop never execute if the

expression is initially false.

DO WHILE (expression); <additional SAS statements>END;

DO WHILE (expression); <additional SAS statements>END;

The parentheses around the expression are required.

5656

The DO UNTIL StatementThe DO UNTIL statement executes statements in a DO loop until a condition is true.

The expression is evaluated at the bottom of the loop. The statements in the loop are executed at least once.

DO UNTIL (expression); <additional SAS statements>END;

DO UNTIL (expression); <additional SAS statements>END;

The parentheses around the expression are required.

5757

Conditional Iterative ProcessingDetermine the number of years that it would take for an account to exceed $1,000,000 if $5,000 were invested annually at 7.5 percent.

5858

Conditional Iterative Processing

data invest; do until(Capital>1000000); Year+1; Capital+5000; Capital+(Capital*.075); end;run;


5959

Conditional Iterative ProcessingPROC PRINT Output

Capital Year

1047355.91 38

How can you generate the same result with a DO WHILE statement?

6060

Conditional Iterative Processingdata invest; do while(Capital<=1000000); Year+1; Capital+5000; Capital+(Capital*.075); end;run;


6161

Conditional Iterative ProcessingPROC PRINT Output

Capital Year

1047355.91 38

6262


Exercise

6363

Exercises


You want to have all the Fibonacci numbers up to 100,000,000.

Obs Number

1 1 2 1 3 2 4 3 5 5 6 8 7 13 8 21 9 34 . . . 36 14930352 37 24157817 38 39088169 39 63245986 40 102334155

6464

Exercises

data fibonacci; drop PriorNumber; Number = 1; PriorNumber = 0; output; do until (number > 100000000); number = number + priornumber; priornumber = number - priornumber; output; end;run;proc print data = fibonacci;run;

6565

Exercises

data fibonacci; drop PriorNumber; Number = 1; PriorNumber = 0; output; do until (number > 100000000); number = number + priornumber; priornumber = numberpriornumber; if number < 100000000 then output; end;run;proc print data =fibonacci;run;

If you do not want to output the row that exceeds 100,000,000, modify the program.

6666


data invest; do until(Capital=1000000); Year+1; Capital+5000; Capital+(Capital*.075); end;run;proc print data=invest noobs;run;

What is the result of this DO LOOP?

6767


data invest; do until(Capital=1000000); Year+1; Capital+5000; Capital+(Capital*.075); end;run;proc print data=invest noobs;run;

The result is an infinite loop because Capital can never equal 1,000,000. Use >=.

What is the result of this DO LOOP?

6868

The Iterative DO Statement with aConditional ClauseYou 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 statementsEND;

DO index-variable=start TO stop <BY increment> WHILE | UNTIL (expression); additional SAS statementsEND;

This is one method of avoiding an infinite loopin DO WHILE or DO UNTIL statements.

6969

The Iterative DO Statement with a Conditional ClauseIn a DO WHILE statement, the conditional clause is checked after the index variable is incremented.

In a DO UNTIL statement, the conditional clause is checked before the index variable is incremented.

7070

The Iterative DO Statement with a Conditional ClauseDetermine the return of the account again.

Stop the loop if 25 years is reached or more than $250,000 is accumulated.

7171

The Iterative DO Statement with a Conditional Clausedata invest; do Year=1 to 25 until(Capital>250000); Capital+5000; Capital+(Capital*.075); end;run;


The loop will stop when Year=26 or

when Capital exceeds 250,000.Remember

7272

The Iterative DO Statement with a Conditional ClausePROC PRINT Output

Year Capital

21 255594.86

What stopped the DO loop?

7373

The Iterative DO Statement with a Conditional Clause

The expression (Capital>250000)

PROC PRINT Output

Year Capital

21 255594.86

What stopped the DO loop?

7474


Exercise

7575

Exercises


You want to have all the Fibonacci numbers up to 1,000,000,000 or to stop after 44 numbers, whichever comes first.

Obs Number

1 1 2 1 3 2 4 3 5 5 6 8 7 13 8 21 9 34 . . .

41 165580141 42 267914296 43 433494437 44 701408733

7676

Exercises

data fibonacci; drop Fib PriorNumber; Number = 1; PriorNumber = 0; output; do Fib=1 to 43 while (number <= 1000000000); number = number + priornumber; priornumber = numberpriornumber; output; end;run;proc print data=fibonacci;run;

7777

Nested DO LoopsNested 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.

7878

Nested DO LoopsCreate 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.

7979

data invest(drop=Quarter); do Year=1 to 5; Capital+5000; do Quarter=1 to 4; Capital+(Capital*(.075/4)); end; output; end;run;


Nested DO Loops

5x 4x

8080

Nested DO LoopsPROC PRINT Output

Year Capital

1 5385.68 2 11186.79 3 17435.37 4 24165.94 5 31415.68

How can you generate one observationfor each quarterly amount?

8181

Nested DO Loops

Move the OUTPUT statement inside the Quarter= loop.

PROC PRINT Output

Year Capital

1 5385.68 2 11186.79 3 17435.37 4 24165.94 5 31415.68

How can you generate one observationfor each quarterly amount?

8282

Nested DO LoopsCompare the final results of investing $5,000 a year for five years in three different banks that compound quarterly. Assume that each bank has a fixed interest rate.

prog2.Banks

Name Rate

Calhoun Bank and Trust 0.0718State Savings Bank 0.0721National Savings and Trust 0.0728

8383

Nested DO Loops

data invest(drop=Quarter Year); set prog2.banks; Capital=0; do Year=1 to 5; Capital+5000; do Quarter=1 to 4; Capital+(Capital*(Rate/4)); end; end;run;

4x5x

3x

This program is similar to the previous program. The changes are noted.

...

8484

Nested DO Loops

NAME

Calhoun Bank and Trust

RATE

0.0718

_N_

1

Partial PDV

D


(0.0718/4));

...

8585

Nested DO Loops

NAME

State Savings Bank

RATE

0.0721

_N_

2

Partial PDV

D


(0.0721/4));

...

8686

Nested DO Loops

NAME

National Savings and Trust

RATE

0.0728

_N_

3

Partial PDV

D


(0.0728/4));

8787

Nested DO Loops

PROC PRINT Output

Name Rate Capital

Calhoun Bank and Trust 0.0718 31106.73State Savings Bank 0.0721 31135.55National Savings and Trust 0.0728 31202.91


8888


Exercise

8989

Exercises

Several students are participating in one of three organized trips with their classmates. The Savings data set contains the student names, the trip they want to take, and the deposit they made for the trip. Students have three more months to make payments before the final payment is due. Each payment is 20 percent of the total cost of the trip.

Costs: The Ski trip $700 Beach trip $600 Washington D.C. $900

Create a data set called SavingsPlan. The data set will contain four rows for each student, one row for each month.

Create the following variables:

TotalCost – the cost of the trip, depending on the student choice

Month – numeric index variable for the DO loop

TotalPaid – initially the value of the deposit, but each month is increased by 20% of the total cost. (Notice that TotalPaid should not be more than TotalCost.)

FinalPayment – the final payment due the fourth month

Create a listing report by student ID.

9090

Exercises – Partial Output--------------------------- StudentID=1005 ---------------------------

Trip Total Total Final Name Choice Deposit Cost Paid Month Payment

Chaz Richardson Beach $235 $600 $355 1 . Chaz Richardson Beach $235 $600 $475 2 . Chaz Richardson Beach $235 $600 $595 3 . Chaz Richardson Beach $235 $600 $600 4 $5

--------------------------- StudentID=1154 ---------------------------

Total Total Final Name TripChoice Deposit Cost Paid Month Payment

Barbara Muir Washington D.C. $35 $900 $215 1 . Barbara Muir Washington D.C. $35 $900 $395 2 . Barbara Muir Washington D.C. $35 $900 $575 3 . Barbara Muir Washington D.C. $35 $900 $900 4 $325

--------------------------- StudentID=1155 ---------------------------

Total Total Final Name TripChoice Deposit Cost Paid Month Payment

Angel Reisman Washington D.C. $95 $900 $275 1 .Angel Reisman Washington D.C. $95 $900 $455 2 .Angel Reisman Washington D.C. $95 $900 $635 3 .Angel Reisman Washington D.C. $95 $900 $900 4 $265

9191

Exercises

data savingsplan; set prog2.savings; if tripchoice = 'Beach' then TotalCost = 600; else if tripchoice = 'Washington D.C.' then TotalCost = 900; else if tripchoice = 'Skiing' then TotalCost = 700; TotalPaid = Deposit; do Month = 1 to 3 while (TotalPaid < TotalCost); TotalPaid = TotalPaid + (TotalCost*.2); if TotalPaid > TotalCost then TotalPaid=TotalCost; output; end; FinalPayment = TotalCost- TotalPaid; TotalPaid = TotalCost; output; format Deposit TotalPaid TotalCost FinalPayment dollar5.;run;proc print data=savingsplan noobs; by studentid;run;

9292


Exercise – Section 13.1

Section 13.2

SAS Array Processing

9494

Objectives Describe the concepts of SAS arrays. Use SAS arrays to perform repetitive calculations.

9595

Performing Repetitive CalculationsEmployees contribute an amount to charity every quarter. The SAS data set prog2.donate contains contribution data for each employee.

The employer supplements each contribution by 25 percent.

Calculate each employee's quarterly contribution including the company supplement.

Partial Listing of prog2.donate

ID Qtr1 Qtr2 Qtr3 Qtr4

E00224 12 33 22 .E00367 35 48 40 30

9696

Performing Repetitive Calculationsdata charity; set prog2.donate; Qtr1=Qtr1*1.25; Qtr2=Qtr2*1.25; Qtr3=Qtr3*1.25; Qtr4=Qtr4*1.25;run;

proc print data=charity noobs;run;

9797

Performing Repetitive CalculationsPartial PROC PRINT Output


E00224 15.00 41.25 27.50 .E00367 43.75 60.00 50.00 37.50E00441 . 78.75 111.25 112.50E00587 20.00 23.75 37.50 36.25E00598 5.00 10.00 7.50 1.25

9898

Performing Repetitive CalculationsWhat if you want to similarly modify 52 weeks of data stored in Week1 through Week52?

data charity; set prog2.donate; Qtr1=Qtr1*1.25; Qtr2=Qtr2*1.25; Qtr3=Qtr3*1.25; Qtr4=Qtr4*1.25;run;


A DO loop should work, but you are using a different variable for each calculation.

9999


data charity; set prog2.donate;

do i=1 to 52; <Var> = <Var>*1.25;

end;run;

proc print data=charity;run;

You can substitute the variable name using a SAS array and write the DO loop.

What if you want to similarly modify 52 weeks of data stored in Week1 through Week52?

100100

Array ProcessingYou can use arrays to simplify programs that perform repetitive calculations create many variables with the same attributes read data compare variables perform a table lookup rotate SAS data sets by making variables into

observations or observations into variables.

101101

What Is a SAS Array?A SAS array is a group of variables whose order is important.

It is not a data structure, but a name given to a group of variables.

A SAS array is a temporary grouping of SAS variables that are

arranged in a particular order is identified by an array name exists only for the duration of the current DATA step is not a variable.

102102

What Is a SAS Array?SAS arrays are different from arrays in many other programming languages.

In SAS, an array is not a data structure. It is simply a convenient way of temporarily identifying a group of variables under one name.

103103

What Is a SAS Array?Each value in an array is called an element identified by a subscript that represents the position of

the element in the array.

When you use an array reference, the corresponding value is substituted for the reference.

104104

What Is a SAS Array?

ID QTR4QTR2 QTR3QTR1

CONTRIBCONTRIB

Firstelement

Secondelement

Thirdelement

Fourthelement

Array references

CONTRIB{1} CONTRIB{2} CONTRIB{3} CONTRIB{4}

Array name

...

105105

The ARRAY StatementThe ARRAY statement defines the elements in an array. These elements will be processed as a group. You refer to elements of the array by the array name

and subscript.

ARRAY array-name {subscript} <$> <length> <array-elements> <(initial-value-list)>;

ARRAY array-name {subscript} <$> <length> <array-elements> <(initial-value-list)>;

106106

The ARRAY Statementarray-name specifies the name of the array.

{subscript} describes the number and arrangement of elements in the array by using an asterisk, a number, or a range of numbers. Subscript is enclosed in braces ({}), brackets ([ ]), or parentheses ( () ). Subscript often has the form {dimension-size(s)}. {dimension-size(s)} is used to indicate a numeric representation of either the number of elements in a one-dimensional array or the number of elements in each dimension of a multidimensional array.

107107

The ARRAY Statement$ indicates that the elements in the array are

character elements. The dollar sign is not necessary if the elements in the array were previously defined as character elements.

length specifies the length of elements in the array that were not previously assigned a length.

array-elements names the elements that make up the array. Array elements can be listed in any order.

(initial-value-list) gives initial values for the corresponding elements in the array. The values for elements can be numbers or character strings. Must enclose all character strings in quotation marks.

108108

The ARRAY StatementThe ARRAY statement must contain all numeric or all character elements. It is

a group of variables of one type. must be used to define an array before the array name

can be referenced. creates variables if they do not already exist in the

PDV. is a compile-time statement.

109109

Defining an ArrayWrite an ARRAY statement that defines the four quarterly contribution variables as elements of an array.

array Contrib{4} Qtr1 Qtr2 Qtr3 Qtr4;

Firstelement

Secondelement

Thirdelement

Fourthelement


CONTRIBCONTRIB

...

110110

Defining an ArrayWrite an ARRAY statement that defines the four quarterly contribution variables as elements of an array.

array Contrib{4} Qtr1 - Qtr4;

Firstelement

Secondelement

Thirdelement

Fourthelement


CONTRIBCONTRIB

...

111111

Defining an ArrayVariables that are elements of an array need not have similar, related, or numbered names.

array Contrib2{4} Q1 Qrtr2 ThrdQ Qtr4;

QTR4QRTR2 THRDQQ1

CONTRIB2CONTRIB2

Firstelement

Secondelement

Thirdelement

Fourthelement

ID

...

112112

Defining an ArrayVariables that are elements of an array need not have similar, related, or numbered names.

array Contrib2{4} Q1 -- Qtr4;

QTR4QRTR2 THRDQQ1

CONTRIB2CONTRIB2

Firstelement

Secondelement

Thirdelement

Fourthelement

ID

113113

Processing an ArrayArray processing often occurs within DO loops. An iterative DO loop that processes an array has the following form:

To execute the loop as many times as there are elements in the array, specify that the values of index-variable range from 1 to number-of-elements-in-array.

DO index-variable=1 TO number-of-elements-in-array; additional SAS statements using array-name{index-variable}…END;

DO index-variable=1 TO number-of-elements-in-array; additional SAS statements using array-name{index-variable}…END;

114114

Processing an Array You must tell SAS which variable in the array to use in

each iteration of the loop. You can write programming statements so that the

index variable of the DO loop is the subscript of the array reference.For example, array-name{index-variable}).

When the value of the index variable changes, the subscript of the array reference (and therefore the variable that is referenced) also changes.

115115

Processing an Array To process particular elements of an array, specify

those elements as the range of the iterative DO statement.

By default, SAS includes index-variable in the output data set.

Use a DROP statement or the DROP= data set option to prevent the index-variable from being written to your output data set.

116116

CONTRIB{QTR}CONTRIB{QTR}

4

CONTRIB{4}

3

CONTRIB{3}

2

CONTRIB{2}

Processing an Arrayarray Contrib{4} Qtr1 Qtr2 Qtr3 Qtr4;do Qtr=1 to 4; Contrib{Qtr}=Contrib{Qtr}*1.25;end;

QTR4

QTR2

QTR3

QTR1

1

Value of index

variable Qtr

CONTRIB{1}

array reference

Firstelement

Secondelement

Thirdelement

Fourthelement

117117



118118

Performing Repetitive Calculationsdata charity(drop=Qtr); set prog2.donate; array Contrib{4} Qtr1 Qtr2 Qtr3 Qtr4; do Qtr=1 to 4; Contrib{Qtr}=Contrib{Qtr}*1.25; end; run;

...

119119


When Qtr=1

Qtr1=Qtr1*1.25;

data charity(drop=Qtr); set prog2.donate; array Contrib{4} Qtr1 Qtr2 Qtr3 Qtr4; do Qtr=1 to 4; Contrib{Qtr}=Contrib{Qtr}*1.25; end; run;

Contrib{1}=Contrib{1}*1.25;

...

120120


When Qtr=2

Qtr2=Qtr2*1.25;



...

121121


When Qtr=3

Qtr3=Qtr3*1.25;



...

122122


When Qtr=4

Qtr4=Qtr4*1.25;



123123




E00224 15.00 41.25 27.50 .E00367 43.75 60.00 50.00 37.50E00441 . 78.75 111.25 112.50E00587 20.00 23.75 37.50 36.25E00598 5.00 10.00 7.50 1.25


124124


Hint: Use variable lists when creating your array.

Exercise – Section 13.2

Section 13.3

Using SAS Arrays

126126

Objectives Use SAS arrays to create new variables. Use SAS arrays to perform a table lookup. Use SAS arrays to rotate a SAS data set.

127127



128128

Creating Variables with ArraysCalculate the percentage that each quarter's contribution represents of the employee's total annual contribution.

Base the percentage only on the employee's actual contribution and ignore the company contributions.



E00224 12 33 22 .E00367 35 48 40 30

129129

Creating Variables with Arraysdata percent(drop=Qtr); set prog2.donate; Total=sum(of Qtr1-Qtr4); array Contrib{4} Qtr1-Qtr4; array Percent{4}; do Qtr=1 to 4; Percent{Qtr}=Contrib{Qtr}/Total; end; run;

The second ARRAY statement creates four numeric variables: Percent1, Percent2, Percent3, and

Percent4.

130130

Creating Variables with Arrays

ID Percent1 Percent2 Percent3 Percent4

E00224 18% 49% 33% .E00367 23% 31% 26% 20%E00441 . 26% 37% 37%E00587 17% 20% 32% 31%E00598 21% 42% 32% 5%

proc print data=percent noobs; var ID Percent1-Percent4; format Percent1-Percent4 percent6.;run;


131131

Creating Variables with ArraysThe PERCENTw.d format multiplies values by 100 formats them with the BESTw.d format adds a percent sign (%) to the end of the

formatted value encloses negative values in parentheses allows room for a percent sign and

parentheses, even if the value is not negative.

132132

Creating Variables with ArraysCalculate the difference in each employee's actual contribution from one quarter to the next.


E00224 12 33 22 .E00367 35 48 40 30

Firstdifference

Seconddifference

Thirddifference


133133

Creating Variables with Arraysdata change(drop=i); set prog2.donate; array Contrib{4} Qtr1-Qtr4; array Diff{3}; do i=1 to 3; Diff{i}=Contrib{i+1}-Contrib{i}; end; run;

...

134134


When i=1

Diff1=Qtr2-Qtr1;

data change(drop=i); set prog2.donate; array Contrib{4} Qtr1-Qtr4; array Diff{3}; do i=1 to 3; Diff{i}=Contrib{i+1}-Contrib{i}; end; run;

Diff{1}=Contrib{2}-Contrib{1};

...

135135


When i=2

Diff2=Qtr3-Qtr2;



...

136136


When i=3

Diff3=Qtr4-Qtr3;



137137


ID Diff1 Diff2 Diff3

E00224 21 -11 .E00367 13 -8 -10E00441 . 26 1E00587 3 11 -1E00598 4 -2 -5

proc print data=change noobs; var ID Diff1-Diff3; run;


138138

Assigning Initial ValuesDetermine the difference between employee contributions and last year’s average quarterly goals of $10, $15, $5, and $10 per employee.

data compare(drop=i Goal1-Goal4); set prog2.donate; array Contrib{4} Qtr1-Qtr4; array Diff{4}; array Goal{4} Goal1-Goal4 (10,15,5,10); do i=1 to 4; Diff{i}=Contrib{i}-Goal{i}; end;run;

139139

Assigning Initial Values

ID Diff1 Diff2 Diff3 Diff4

E00224 2 18 17 .E00367 25 33 35 20E00441 . 48 84 80E00587 6 4 25 19E00598 -6 -7 1 -9

proc print data=compare noobs; var ID Diff1 Diff2 Diff3 Diff4;run;


140140

Assigning Initial ValuesElements and values are matched by position.

If there are more array elements than initial values, the remaining array elements are assigned missing values and SAS issues a warning.

You can separate the values in the initial value list with either a comma or a blank space.

Initial values are retained until a new value is assigned to the array element.

141141

data compare(drop=Qtr Goal1-Goal4); set prog2.donate; array Contrib{4} Qtr1-Qtr4; array Diff{4}; array Goal{4} Goal1-Goal4 (10,15,5,10); do Qtr=1 to 4; Diff{Qtr}=Contrib{Qtr}- Goal{Qtr}; end;run;


E00224 12 33 22 .E00367 35 48 40 30


PDV

Compile

...

142142


ID QTR3QTR1 QTR2 QTR4PDV


E00224 12 33 22 .E00367 35 48 40 30


...

143143


ID QTR3QTR1 QTR2 DIFF1 DIFF2QTR4PDV

DIFF3 DIFF4


E00224 12 33 22 .E00367 35 48 40 30


...

144144



DIFF3 DIFF4 GOAL2GOAL1 GOAL4GOAL3


E00224 12 33 22 .E00367 35 48 40 30


...

145145



DIFF3 GOAL2DIFF4 GOAL1 GOAL4 QTRGOAL3


E00224 12 33 22 .E00367 35 48 40 30


...

146146



DIFF3 GOAL2DIFF4 GOAL1 GOAL4 QTRGOAL3D D DD D


E00224 12 33 22 .E00367 35 48 40 30


The elements in the Goal array, Goal1, Goal2, Goal3, and Goal4, are created in the PDV and are used to calculate the values of Diff1, Diff2, Diff3, and Diff4. The values are subsequently excluded from the output data set compare using the DROP= data set option.

147147



148148

Performing a Table LookupYou can use the keyword _TEMPORARY_ instead of specifying variable names when you create an array to define temporary array elements.

data compare(drop=Qtr); set prog2.donate; array Contrib{4} Qtr1-Qtr4; array Diff{4}; array Goal{4} _temporary_ (10,15,5,10); do Qtr=1 to 4; Diff{Qtr}=Contrib{Qtr}-Goal{Qtr}; end;run;

149149

Performing a Table LookupArrays of temporary elements are useful when the only purpose for creating an array is to perform a calculation.

To preserve the result of the calculation, assign it to a variable. Temporary data elements do not appear in the output

data set Temporary data element values are always

automatically retained.

150150

Performing a Table Lookup

ID Diff1 Diff2 Diff3 Diff4

E00224 2 18 17 .E00367 25 33 35 20E00441 . 48 84 80E00587 6 4 25 19E00598 -6 -7 1 -9

proc print data=compare noobs; var ID Diff1 Diff2 Diff3 Diff4;run;


151151



152152

Rotating a SAS Data SetYou can use array processing to rotate, or transpose, a SAS data set. When a data set is rotated, the values of an observation in the input data set become values of a variable in the output data set.

The TRANSPOSE procedure is also used to create an output data set by restructuring the values in a SAS data set, transposing selected variables into observations.


E00224 12 33 22 .E00367 35 48 40 30

153153

ID Qtr Amount

Partial Listing of rotate


E00224 12 33 22 .E00367 35 48 40 30


E00224 1 12E00224 2 33E00224 3 22E00224 4 .E00367 1 35E00367 2 48E00367 3 40E00367 4 30

Rotating a SAS Data Set

...

154154

Rotating a SAS Data Setdata rotate(drop=Qtr1-Qtr4); set prog2.donate; array Contrib{4} Qtr1-Qtr4; do Qtr=1 to 4; Amount=Contrib{Qtr}; output; end; run;

155155

ID QTR3QTR1 QTR2 QTR AMOUNTQTR4D D D D


E00224 12 33 22 .E00367 35 48 40 30


ID Qtr Amount


data rotate(drop=Qtr1-Qtr4); set prog2.donate; array Contrib{4} Qtr1-Qtr4; do Qtr=1 to 4; Amount=Contrib{Qtr}; output; end; run;

Execute

PDV

...

156156


......


Initialize PDV to missing.

PDV


E00224 12 33 22 .E00367 35 48 40 30


ID Qtr Amount


...

157157


...223312E00224


PDV


E00224 12 33 22 .E00367 35 48 40 30


ID Qtr Amount


...

158158


...223312E00224



E00224 12 33 22 .E00367 35 48 40 30


PDV

ID Qtr Amount


...

159159


.1.223312E00224


PDV


E00224 12 33 22 .E00367 35 48 40 30


ID Qtr Amount


...

160160


.1.223312E00224

ID QTR3QTR1 QTR2 QTR AMOUNTQTR4D D D

Amount=Contrib{1};

D

12

PDV


E00224 12 33 22 .E00367 35 48 40 30


ID Qtr Amount


...

161161 Write out observation to rotate.


121.223312E00224

ID QTR3QTR1 QTR2 QTR AMOUNTQTR4D D DD

PDV


E00224 12 33 22 .E00367 35 48 40 30


ID Qtr Amount

E00224 1 12


...

162162


122.223312E00224

ID QTR3QTR1 QTR2 QTR AMOUNTQTR4D D

Amount=Contrib{2};

D D

33

PDV


E00224 12 33 22 .E00367 35 48 40 30


ID Qtr Amount

E00224 1 12


...



332.223312E00224


PDV


E00224 12 33 22 .E00367 35 48 40 30


ID Qtr Amount

E00224 1 12 E00224 2 33


...

164164


333.223312E00224

ID QTR3QTR1 QTR2 QTR AMOUNTQTR4D

Amount=Contrib{3};

D D D

22


E00224 12 33 22 .E00367 35 48 40 30


ID Qtr Amount

E00224 1 12 E00224 2 33


PDV

...



223.223312E00224


PDV


E00224 12 33 22 .E00367 35 48 40 30


ID Qtr Amount

E00224 1 12 E00224 2 33 E00224 3 22


...

166166


224.223312E00224

ID QTR3QTR1 QTR2 QTR AMOUNTQTR4

Amount=Contrib{4};

D D D D

.

PDV


E00224 12 33 22 .E00367 35 48 40 30


ID Qtr Amount

E00224 1 12 E00224 2 33 E00224 3 22


...



.4.223312E00224


PDV


E00224 12 33 22 .E00367 35 48 40 30


ID Qtr Amount

E00224 1 12 E00224 2 33 E00224 3 22 E00224 4 .


...

168168

PDV

.5.223312E00224

ID QTR3QTR1 QTR2 QTR AMOUNTQTR4DD D D


Implicit return. Continue processing observations from prog2.donate.


E00224 12 33 22 .E00367 35 48 40 30


ID Qtr Amount

E00224 1 12 E00224 2 33 E00224 3 22 E00224 4 .


169169

Rotating a SAS Data Set


proc print data=rotate noobs;run;

ID Qtr Amount

E00224 1 12 E00224 2 33 E00224 3 22 E00224 4 . E00367 1 35 E00367 2 48 E00367 3 40 E00367 4 30

Documents

Chapter 13 Processing Data Iteratively. Section 13.1 DO-Loop Processing