Thank you, Data_Null, It work perfectly. ... View more

I correct that error. I have to sort data prior to processing. The calculate start at 3 as a rule I have to follow. For the macro, I can do it with 4, 5.... HHC ... View more

Thank you all for helping me with this problem. HHC ... View more

oh, what I want to do is to get the Max and Min value for the prior 4 record. That's why I give the price2(4). So it look like there is no way to do that moving Min/Max with array? HHC ... View more

Hi, I try to do the moving Highest and Lowest of price using this array approach. Somehow I got the error "Array subscript out of range ..." Could you help me to correct it? Thank you, HHC data have;   input object time price level;   datalines; 1 1     5     9 1 2     2     3 1 3     8     5 1 4     10    25 1 5     12    63 2 1     15     29 2 2     12     3 2 3     18     15 2 4     10    15 2 5     18    6 3 1     15     29 3 2     22     32 3 3     8     52 3 4     11    50 3 100     1200    6 ;run; data want2; array price2(4) _temporary_; kk=0; do until (last.object); set have; by object; kk+1; price2(kk)=price; end; max_record=kk; nn=0; do until (last.object); set have; by object; nn+1; highest=0; lowest=10000; do i=nn+1 to max_record;   if price2(i)>highest then highest=price2(i);   if price2(i)<lowest then lowest=price2(i); end; end; run; ... View more

Thank you all for your help. HHC ... View more

Hi Everyone, The discussion here really helps me a lot. I am new to Hash process so I will not have any question here. Based on the simulated data, I find the array approach is so efficient. I would like to make myself clear about it and below, I try to teach myself with explanation of each step. I have 3 questions and it would be very much helpful if you could help me to clarify them. Thank you so much and have a nice Holiday. HHC ********************************************************************************** *My 1st question is the how SAS process the code: 1. at the benning, SAS goes to section 2 and collect prc and time record from 1 to min(10,000 , last.object) 2. then SAS goes to Section 3, take the first record of have and run the checking with the array price(); *Once is done, SAS redo the 2 section with 2nd record of have I wonder why SAS doesn't run the section 1 from the beginning to the end before moving onto section 3. What line of code prevent SAS from doing so? *My 2nd question is after finishing 1 record of have, why don’t we reset nn=0;     end;     output; nn=0;   end; *My erd question is: what is the role of :;   call missing (of prc(*));   call missing (of tim(*)); ***************************************************************************************************; data want(keep=object time price level greater exit_prc); *Section 1: declare array specification;   array prc(10000) _temporary_;   array tim(10000) _temporary_;   kk=0;   call missing (of prc(*));   call missing (of tim(*)); *Section 2: gathering records in the array prc() and tim(). These array is Max 10,000 records;   do until (last.object);     set have;     by object;     kk+1;     prc(kk)=price;    *start from price(0+1)?;     tim(kk)=time;   end; *Section 3: the main code of checking first record with price above level;   ntot=kk;    *the ntot is the value of record in the prc()/tim() array, Max 10,000;   nn=0;    *Reset nn=0 for below do loop;   do until (last.object);     set have;    *open file have, start with the first record;     by object;     nn+1;        *start with nn=0;     greater=0;     do i=nn+1 to ntot; *first I value =1 to the max ntot record above, max 10,000;        if (prc(i)>level) then do;             greater=tim(i);             exit_prc=prc(i);             leave; *exit the checking loop to OUTPUT;             end;     end;     output;   end; run; ... View more

I think I will add a new variable for the last record of each object to control for the LOOP. data have; set have; by object; if last.object then last=1;run; data want (keep=object time price level greater);   set have nobs=totalobs;   i+1;   found=0;   do j=i+1 to totalobs until (found=1 or last=1) ;     set have(keep=object time price rename=(object =obj time=greater price=price2)) point=j;     if price2 > level and obj=object then found=1;   end;   if found=0 then greater=0; run; ... View more