求教sas数据汇总问题

是 否 +2 论坛币

k人 参与回答

经管之家送您一份

应届毕业生专属福利!

求职就业群

赵安豆老师微信：zhaoandou666

经管之家联合CDA

送您一个全额奖学金名额~ !

立即领取

感谢您参与论坛问题回答

经管之家送您两个论坛币！

+2 论坛币

有这么一组数据，

customer_no	card_no
1	00002
1	00003
1	00006
1	00008
2	00005
2	00012
3	00015
4	00020

我现在要分别汇总customer_no=1、2、3、4客户所拥有的卡数，请问该怎么做？
附：我用if first.customer_no then cardsum=0; cardsum=cardsum+1;的做法显示说我未初始化first.customer_no.
然后我就用proc sql来做，这是我的代码
proc sql;
create table count as
select count(distinct customer_no) as customer
  from a1
  ;
  quit;
但还是做不出正确答案。

扫码加我 拉你入群

请注明：姓名-公司-职位

以便审核进群资格，未注明则拒绝

分享0 收藏1 回帖

关键词：总问题 Customer proc sql Distinct custom 汇总 数据 求教

proc sql;
create table count as
select count(distinct customer_no) as customer
  from a1
group by customer_no
  ;
  quit;

data wkn;
input customer_no $ card_no$;
cards;
1 00002
1 00003
1 00006
1 00008
2 00005
2 00012
3 00015
4 00020
;
run;
proc sql;
create table wkn1 as
select customer_no,count(*) as cardsum
from wkn
group by customer_no;
quit;
试试看

2# pobel
哈哈。我猜，结果都是1吧。

jingju11 发表于 2010-7-22 12:10
2# pobel
哈哈。我猜，结果都是1吧。

马虎了

proc sql;
create table count as
select distinct customer_no ,   count(distinct card_no) as cardnum
  from a1
group by customer_no
  ;
  quit;

那我第一种方法可否做的出来？

data have;
input customer_no $ card_no$;
cards;
1 00002
1 00003
1 00006
1 00008
2 00005
2 00012
3 00015
4 00020
;
run;

data wanted;
    set have;
        by customer_no;
        retain cardnum;
        if first.customer_no then cardnum=0;
        cardnum+1;
        if last.customer_no;
        keep customer_no cardnum;
run;

6# reolin

你需要加上by ×××；那个x = x+1;可以直接写成x+1；即可。本质上还是把by statement给漏了

proc sql;
        select distinct customer_no,count(*)
        from a1
        group by customer_no;
quit;

proc sort data=a1;
        by customer_no;
run;

data a2;
        set a1;
        by customer_no;
        if first.customer_no then count=1;
        else count+1;
run;

data a2;
        retain count;
        length count;
        set a1;
        by customer_no;
        if first.customer_no then count=1;
        else count=count+1;
run;

谢谢各位高手，尤其是pobel和jingju11 ！