对列的计算实在是不好掌握啊！该怎样求？

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id	visit	date	value
1	1	1998-2-3	1.1
1	2	1998-2-13	1.2
1	3	1998-2-23	1.2
2	1	1998-2-5	1.2
2	2	1998-2-10	1.2
2	3	1998-2-20	1.2
3	1	1998-5-1	1.2
3	2	1998-5-2	1.3
3	3	1998-5-3	1.2

找出value的最大值，并且标注flag=‘yes’；
对于同一个id，valve相同的，找出最早的访问时间，并且标注flag=‘yes’；
就是变成这样

id	visit	date	value	flag
1	1	1998-2-3	1.1
1	2	1998-2-13	1.2	yes
1	3	1998-2-23	1.2
2	1	1998-2-5	1.2	yes
2	2	1998-2-10	1.2
2	3	1998-2-20	1.2
3	1	1998-5-1	1.2	yes
3	2	1998-5-2	1.3	yes
3	3	1998-5-3	1.2

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关键词：value Visit Flag alue eval visit

proc sort data=test;
by id visit date;
run;

proc sql;
create table maxvalue1 as
select distinct * from test
group by id
having value=max(value)
order by id, visit, date;
quit;

data maxvalue;
set maxvalue1;
by id visit date;
if first.id;
run;


proc sql;
create table dupvalue1 as
select * from test
group by value
having count(value)>1
order by id, visit,date;
quit;

data dupvalue;
set dupvalue1;
by id visit date;
if first.id;
run;

data wanted;
merge test maxvalue(in=a drop=value) dupvalue(in=b drop=value);
by id visit date;
if a or b then flag="Yes";
run;