如果满足条件,我必须摆脱一个主题。
DATA:
Name Value1
A 60
A 30
B 70
B 30
C 60
C 50
D 70
D 40
我想要的是如果值= 30则两条线都不应该在输出中。
期望的outpu
Name Value1
C 60
C 50
D 70
D 40
我在proc sql中编写了一段代码
proc sql;
create table ck1 as
select * from ip where name in
(select distinct name from ip where value = 30)
order by name, subject, folderseq;
quit;
答案 0 :(得分:3)
将您的SQL更改为:
proc sql;
create table ck1 as
select * from ip where name not in
(select distinct name from ip where value = 30)
order by name, subject, folderseq;
quit;
答案 1 :(得分:0)
数据步骤方法:
data have;
input Name $ Value1;
datalines;
A 60
A 30
B 70
B 30
C 60
C 50
D 70
D 40
;;;;
run;
data want;
do _n_ = 1 by 1 until (last.name);
set have;
by name;
if value1=30 then value1_30=1;
if value1_30=1 then leave;
end;
do _n_ = 1 by 1 until (last.name);
set have;
by name;
if value1_30 ne 1 then output;
end;
run;
在某些情况下,另一种稍微快一些的方法可以在value1_30为1时避免使用第二个set语句(如果大多数都有30个,则速度更快,因此您只保留少量记录)。
data want;
do _n_ = 1 by 1 until (last.name);
set have;
by name;
counter+1;
if first.name then firstcounter=counter;
else if last.name then lastcounter=counter;
if value1=30 then value1_30=1;
if value1_30=1 then leave;
end;
if value1_30 ne 1 then
do _n_ = firstcounter to lastcounter ;
set have point=_n_;
output;
end;
run;
答案 2 :(得分:0)
另一个SQL选项......
proc sql number;
select
a.name,
a.value1,
case
when value1 = 30 then 1
else 0
end as flag,
sum(calculated flag) as countflagpername
from have a
group by a.name
having countflagpername = 0
;quit;