Here is the program to identify no sale in any consecutive three month from a given product. I left how to remove them for you. It should be much easier.


*create testing data;
data t1;
   length yymm $4.;
   prodid=1980;
   do year=8 to 9;
      do month = 1 to 12;
            yymm=catt(put(year,z2.), put(month,z2.));
                sales=ceil(ranuni(123)*50);
                if yymm not in ('0801', '0802', '0812', '0903','0904','0905') then output;
                *output;
          end;
        end;
   prodid=1981;
   do year=8 to 9;
      do month = 1 to 12;
            yymm=catt(put(year,z2.), put(month,z2.));
                sales=ceil(ranuni(123)*50);
                *if yymm not in ('0802', '0812', '0903') then output;
                output;
          end;
        end;
        keep prodid yymm sales;
run;

***build a template with no missing month sales;
proc sql;
create table sales_info as
select prodid, input(catt(min(yymm),'01'), yymmdd6.) as LYYMM,
                input(catt(max(yymm),'01'), yymmdd6.) as HYYMM
  
  from t1
  group by prodid
  order by prodid  
  ;
  quit;

data prod_yymm;
   length yymm $4.;
   set sales_info;
   by prodid;
   date=LYYMM;
   do until (date>HYYMM);
      yymm=put(date,yymmdd6.);
      output;
          date= intnx('month', date,1);          
   end;
   keep prodid yymm;
run;

data check;
  merge prod_yymm t1;
  by prodid yymm;
  sales1=lag1(sales);
  sales2=lag2(sales);
  if first.prodid then cnt=0;
  cnt+1;
  if cnt in (1) then do;
    sales1=.; sales2=.;
  end;
  if cnt=2 then sales2=.;
  missing_3_month=(sum(sales, sales1, sales2)=.);
run;

proc print;run;

谢谢牛人！我看到论坛里好多sas的技术贴都是你回复的，膜拜！另外昨天我发完贴后又在继续想，下面是写的一个程序，请牛人帮忙看一下：
/******将0807改为200807，方便下面转换成日期形式**********/
data market.beer2;
set market.beer1;
sdate=month+200000;
drop month;
run;
/********将200807转换为日期2008-07形式**********/
data market.beer3;
set market.beer2;
time=mdy(mod(sdate,100),1,int(sdate/100));
format time yymmd.;
drop sdate;
run;
proc summary data=market.beer3 nway;
var time;
class procode;
output out=market.cal min=mintime max=maxtime;
run;

data market.beer4;
merge market.beer3 market.cal;
by procode;
keep procode time mintime _FREQ_;
run;
/*******计算时间差，这里算出的是天数********/
data Tmp1.beer7;
set Tmp1.beer4;
y=time-mintime;
run;
/******需要把算出的天数转换成月份，这里还没想到更好的方法，intck函数应该可以，不知如何实现*********/
data Tmp1.beer8;
set Tmp1.beer7;
interval=int((y+2)/30);
run;
/*******计算同一变量两两观测之间的差值，lag（）dif（）函数*******/
data Tmp1.beer9 replace;
set Tmp1.beer8;
z=lag(interval);
d=dif(interval);
run;
/******筛选出两两观测销售记录缺失之间不超过三个月的产品*******/
data Tmp1.new1 Tmp1.new2 replace;
set Tmp1.beer9;
if d >= 4 then output Tmp1.new1;
else output Tmp1.new2;
run;
data Tmp1.result replace;
merge Tmp1.new2 (in=a) Tmp1.new1 (in=b);
by procode;
if a=1 and b=0;
run;

想请问牛人的是，intck函数在这里应该怎么使？因为我想计算2008-09和2008-12之间月份的数，我写了几个都说语法错误，最后无奈，用了int函数。。。算是变相生成了结果吧。