敢请高手解惑一个数据问题

9# baoaibaobao
呵呵，非常感谢指正，学习了~

10# soporaeternus photo相当给力啊~

10# soporaeternus
呵呵，前辈的code很漂亮，不过小弟不是很理解啊，能大概解释一下，datediff是怎么得到的？on...的作用么？非常感谢！

1# 456852

I modify hssnow's program a bit. Hope this is what you want.
data a;
    input customer $ product $ date date9. category $;
datalines;
1               001 12Feb2010          A
1               002 13Feb2010          B
1               003 19Feb2010          B
1               004 01Mar2010          A
2               005 12Mar2010          B
2               006 15Apr2010          A
2               007 18Apr2010          B
;
run;


data b ;*(drop=iddate);
    set a;
    by customer;
    retain iddate;
    if first.customer then do;
       count=0;
       iddate=.;
    end;
    count+(    category='A');
    if count=1 and iddate=. then iddate=date;
    if iddate ne . then datediff=date-iddate;
    format     date iddate yymmdd10.;
run;

proc print;run;

try this;
proc sort data=aa;
by customer product date;
run;
data new(drop=datett);
datett=0;
do _n_=1 by 1 until(last.customer);
set aa;
by customer;
if _n_=1 then call missing(datediff);
  else  if datett=0 then call missing(datediff);
         else datediff=date-datett;
if category='A' then datett=date;
output;
end;
run;
需要注意的是楼主给出的数据集和给出的说明中两类日期不同，一个是3月1号 一个是2月22号...

shenliang_111 发表于 2011-8-24 10:46
try this;
proc sort data=aa;
by customer product date;

1. 再次简化：
   
2. data result2;
   
3. time=.;
   
4. do _n_=1 by 1 until(last.customer);
   
5.    set  zachary;
   
6.    by customer;
   
7.    datediff=date-time;
   
8.    time=ifn(category='A',date,time);
   
9.    output;
   
10. end;
    
11. run;

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shenliang_111 发表于 2011-12-21 14:19

1. sql方法（该方法的思路都是向soporaeternus学习的）
   
2. proc sql;
   
3. select a.customer,a.product,a.date,a.category,(a.date-b.date) as datediff
   
4. from (select *,monotonic() as n from zachary) a
   
5. left join (select *,monotonic() as n from zachary where category='A') b
   
6. on  a.customer=b.customer and a.n>b.n
   
7. group by 1,2
   
8. having calculated datediff=min(calculated datediff)
   
9. order by a.customer,a.product;
   
10. quit;

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