SAS BASE 题库

1.The following SAS program is submitted:

data WORK.TOTAL;

set WORK.SALARY;

by Department Gender;

if First.<_insert_code_> then Payroll=0;

Payroll+Wagerate;

if Last.<_insert_code_>;

run;

The SAS data set WORK.SALARY is currently ordered by Gender within Department. Which inserted code will accumulate subtotals for each Gender within Department?

A. Gender

B. Department

C. Gender Department

D. Department Gender

Answer: A

本题考察的是：first.var和last.var这两个知识点。本题的意思，数据集SALAR已经按照department 和gender排序，

现在计算每一个department下每一个gender的某一个变量值的累积和。

根据本题的意思，写了一个模拟程序：

data crackman;

input department $ gender $ salary@;

datalines;

market f 6000

market m 5000

market f 5500

market m 8000

market f 6000

market m 7000

sales f 6000

seles m 4000

sales f 6000

seles m 4000

sales f 6000

seles m 4000

;

proc sort data=crackman;

by department gender;

run;

data result;

set crackman;/*1*/

by department gender;

if first.gender then subtotal=salary;/*2*/

else subtotal+salary;/*3*/

if last.gender;/*4*/

run;/*5*/

这个DATA RESULT 部分程序执行的过程是这样（不是编译过程）：/*我把执行的语句分为标记1 2 3 4 5*/

1.1–2–4–1（先读数据—判断first.gender是否为1，第一个肯定是1——所以直接跳过else到if last.gender，last.gender=0所以—set crackman，继续读数据)

2.1–2–3–4–1

3.1–2–3–4–5–1(因为是第三个了，所以last.gender=1 然后到run,输出到result.

后面都是一样。

这里第三条语句其实===subtotal=subtotal+salary，只是这里简写了，体现程序的简洁性，但是对于入门同学来说增加了可读性的难度。

1.

1.hello说道：

2013 年 3 月 6 日 15:07

可以先理解下FIRST.变量和LAST.变量

data result;

set crackman;

by department gender;

first_department=first.department;

last_department=last.department;

first_gender=first.department;

last_gender=last.gender;

/* if first.gender then subtotal=salary;*/

/* else subtotal+salary;*/

/* if last.gender;*/

run;

proc print data=result;

run;

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2.Jessica说道：

2014 年 2 月 9 日 12:22

Hi 我看了很少一部分online tutor然后决定直接刷题库楼主自己写的那个example 怎么能看出来是first.gender=1，last.gender=0呢？你没有写equation在if后, 是说这是默认的么？

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