如何在PROC SQL中使用MACRO在SAS中创建连续变量

Posted 2021-04-06

我想通过使用MACRO简化PROC SQL中的SAS代码。通过使用MARCO替换YYMM，我已经在数据步骤中取得了成功。 YYMM连续多个月。示例数据和我的原始代码如下。

data work.allredeem;
input no_code BONUS_0907 BONUS_0908   BONUS_0909   BONUS_0910   BONUS_0911   BONUS_0912   BONUS_1001   BONUS_1002   BONUS_1003   BONUS_1004   BONUS_1005   BONUS_1006   BONUS_1007   BONUS_1008   BONUS_1009   BONUS_1010   BONUS_1011   BONUS_1012;
DATALINES;
1   9   7   6   9   9   5   2   1   4   4   9   5   4   1   9   3   3   7
1   3   1   4   1   5   7   8   6   1   1   8   1   1   3   9   7   1   7
2   8   3   10  8   5   6   1   9   5   2   4   8   4   2   3   8   6   1
2   5   1   8   8   6   2   2   6   6   3   4   4   5   4   8   8   2   4
3   7   4   1   3   2   7   9   5   6   8   10  3   2   5   7   10  10  6
4   7   3   6   5   8   6   9   9   3   6   1   3   4   6   5   3   4   9
5   2   1   8   2   4   3   8   8   1   9   4   9   6   10  7   5   6   8
5   10  10  4   10  4   3   7   4   8   7   1   5   1   1   9   4   6   10
5   8   10  6   9   6   5   2   8   7   6   4   1   5   9   5   6   7   6
6   6   3   1   2   6   6   10  7   9   9   3   5   2   6   6   8   6   5
7   9   5   1   9   5   9   10  3   4   10  3   7   1   6   3   10  3   6
7   4   5   8   3   2   3   8   10  2   10  3   9   4   2   2   7   1   9
7   7   4   2   4   5   1   3   2   2   1   5   4   9   2   1   9   3   3
7   10  6   5   10  5   7   9   2   2   3   8   9   6   10  3   2   10  4
7   7   9   1   6   6   3   8   8   7   10  10  9   7   4   1   1   2   2
;
RUN;

PROC SQL;
create table allredeem2 as
SELECT no_code,  
sum(BONUS_0907) AS redeemNum0907,
sum(BONUS_0908)AS redeemNum0908,
sum(BONUS_0909)AS redeemNum0909,
sum(BONUS_0910)AS redeemNum0910,
sum(BONUS_0911)AS redeemNum0911,
sum(BONUS_0912)AS redeemNum0912,
sum(BONUS_1001)AS redeemNum1001,
sum(BONUS_1002)AS redeemNum1002,
sum(BONUS_1003)AS redeemNum1003,
sum(BONUS_1004)AS redeemNum1004,
sum(BONUS_1005)AS redeemNum1005,
sum(BONUS_1006)AS redeemNum1006,
sum(BONUS_1007)AS redeemNum1007,
sum(BONUS_1008)AS redeemNum1008,
sum(BONUS_1009)AS redeemNum1009,
sum(BONUS_1010)AS redeemNum1010,
sum(BONUS_1011)AS redeemNum1011,
sum(BONUS_1012)AS redeemNum1012
FROM allredeem
GROUP BY no_code;
QUIT;

我尝试使用MACRO如下，但它不起作用。

%MACRO  sub(year);
PROC SQL;
create table allredeem2 as
SELECT no_code,  sum(BONUS_&year) AS redeemNum&year 
FROM allredeem
GROUP BY no_code;
QUIT;
%MEND sub;
%sub(0907)
%sub(0908)
%sub(0909)
%sub(0910)
%sub(0911)
%sub(0912)
%sub(1001)
%sub(1002)
%sub(1003)
%sub(1004)
%sub(1005)
%sub(1006)
%sub(1007)
%sub(1008)
%sub(1009)
%sub(1010)
%sub(1011)
%sub(1012)

提前致谢。

答案

宏不工作的原因是因为您在宏调用中创建了相同的表，即您在每个宏调用中都在写表。在决赛桌中，你将no_code和与1012相关的值。这一点很容易在proc手段中处理。

在SQL中处理的一种方法是使用dictionary.columns，如下所示。

proc sql noprint;
select 
'sum('|| trim(name)||') as redeemNum'||compress(name,,'kd') into :sum separated by 
 ','
from dictionary.columns
where libname ='WORK'
and upcase(memname) =  upcase('allredeem')
and upcase(type) ='NUM'
and upcase(name) ne upcase("no_code");
%put &count;

proc sql;
create table want as 
select no_code, &sum
 from allredeem
 group by no_code;

另一种简单的处理方法是在数据步骤中使用array和sum语句。

  proc sort data=allredeem;
  by no_code;
  run;


data want;
set allredeem;
by no_code;
array bonus(*) BONUS0907 - BONUS0912   BONUS1001 - BONUS1012;
array redeem(*) redeemNum0907 - redeemNum0912   redeemNum1001 - redeemNum1012;
do i = 1 to dim(bonus);
if first.no_code then redeem(i) = bonus(i);
else redeem(i) +bonus(i);
end;
drop bonus: i;
if last.no_code;
run;