Hi, Paul D. The recursive quicksort for the entire array fails for large N but for the sliced array like the one the original poster asked can work for smaller N.   With regards, KachiM ... View more

Hi, Paul D. I have had no problem with a large array. I have tested on an array sized to 1E7 with no fatal exception. Here is the log. I am giving spec. of my Desktop and SAS version used. 98 options fullstimer; 99 data _null_; 100 array k[10000000] _temporary_ ; 101 do i = 1 to 10000000; 102 k[i] = ceil(ranuni(123) * 1e8); 103 end; 104 call qsort(k, 2, 10); 105 do i = 1 to 20; 106 put k[i]=; 107 end; 108 run; k[1]=75039609 k[2]=12466523 k[3]=17838965 k[4]=18768585 k[5]=22111402 k[6]=32091203 k[7]=35711708 k[8]=39808191 k[9]=78643830 k[10]=90603339 k[11]=77618419 k[12]=43607415 k[13]=96750024 k[14]=26370295 k[15]=71393449 k[16]=55485760 k[17]=53125006 k[18]=86134461 k[19]=14207808 k[20]=86042479 NOTE: DATA statement used (Total process time): real time 0.25 seconds user cpu time 0.18 seconds system cpu time 0.01 seconds memory 79715.28k OS Memory 96800.00k Timestamp 02/14/2024 12:13:10 PM Step Count 10 Switch Count 0 My Desktop has the following specs: Processor: Intel core i5-10400 CPU @ 2.90 GHz Ram: 8.00 GB System Type: 64-bit OS, x64-based processor SAS 9.4 M6 was used. ... View more

Here is a FCMP based Quick Sort Routine just for your reference.  proc fcmp outlib = work.cmpar.lib; subroutine qsort(v[*], left, right); outargs v; if left >= right then return; call swap(v, left, ceil((left + right) / 2)); last = left; do i = left to right; if v[i] < v[left] then call swap(v, last, i); end; call swap(v, left, last); call qsort(v, left, last - 1); call qsort(v, last + 1, right); endsub; subroutine swap(v[*], i, j); outargs v; temp = v[i]; v[i] = v[j]; v[j] = temp; endsub; quit; options cmplib = work.cmpar; /* Sort values from 2 to end of the Array k[*] */ data _null_; array k[5] _temporary_ (100, 2, 70, 12, 50); call qsort(k, 2, dim(k)); SORTED = catq('d',' ', of k[*]); put SORTED =; /* SORTED=100 2 12 50 70 */ run; ... View more

You can use SAS array and examine the values of each column to match with 1, and count them. data have; input A B C D E; datalines; 1 2 1 1 2 2 2 1 2 2 2 3 3 2 2 1 2 3 1 1 ; run; data want; set have; array x[*] A B C D E; count = 0; do i = 1 to dim(x); if x[i] = 1 then count + 1; end; drop i; run; proc print data = want; run; ... View more

Without using Functions and Procs, it can be done by programming. Assuming you have an array of medium size you can use simple and fairly easy insertion sort to sort the elements in place. The duplicate items will be marked as missing values in the array. If you have larger array of size in millions, use Quicksort but you have to modify the code to catch the duplicate items. Hope this interests you. data _null_; array k[8] _temporary_ (2 3 5 2 7 6 7 1); do i = 1 to dim(k) - 1; do j = i+1 to dim(k); if k[i] = k[j] then do; k[j] =.; continue; end; if k[i] > k[j] then do; t = k[i]; k[i] = k[j]; k[j] = t; end; end; end; do i = 1 to dim(k); put k[i] =; end; run; k[1]=1 k[2]=. k[3]=2 k[4]=3 k[5]=5 k[6]=6 k[7]=. k[8]=7 ... View more

There are ways to run programs with limited resources. Consult Paul Dorfman's papers. If you explain your issue with a constructed SAS dataset and your requirements, someone can come out with an efficient way to tackle memory issues.  ... View more

In your original post you wanted roll < 2 and roll > 2.  If your choice is to get one output data set when roll > 2 OR if you want it for roll < 2 here is one way:   %let choice = >; data final; if _n_=1 then do; declare hash aa(); aa.definekey('roll'); aa.definedata('dat'); aa.definedone(); roll=2; dat="Match"; aa.add(); p_roll = roll; drop p_roll; end; call missing(dat); set one; if roll &choice. p_roll and aa.find() ne 0 then output; run; If you want for roll < 2, change the macro variable into "<"   If you want for roll < 2 and roll > 2 in one data step here is a way: data lower upper; if _n_=1 then do; declare hash aa(); aa.definekey('roll'); aa.definedata('dat'); aa.definedone(); roll=2; dat="Match"; aa.add(); p_roll = roll; drop p_roll; end; call missing(dat); set one; if roll > p_roll and aa.find() ne 0 then output upper; else if roll < p_roll and aa.find() ne 0 then output lower; run; ... View more

It can be done but it is an unusual requirement with hash objects. You set the variable 'dat' to missing and replace the statement   if aa.find() =0 then output; as  if aa.find() NE 0 then output;   ... View more

A Data Step approach. Sort the data set. Remove the pair of observations that starts with negative AMOUNT. data have; input Clientid Purchase_Date:mmddyy10. item_name $ amount:dollar4.; format Purchase_Date mmddyy10. amount dollar4.; datalines; 1 01/01/2020 book $20 1 01/01/2020 book $-20 2 03/01/2020 pen $2 3 04/01/2020 game $70 2 03/01/2020 pen $-2 1 06/01/2020 bike $200 ; run; proc sort data = have out = have_sorted; by Clientid Purchase_Date item_name amount; run; data want; do until(last.Clientid); set have_sorted; by Clientid; if amount < 0 then do; prev = 1; delete; end; else if amount >= 0 and prev = 1 then do; delete; prev = .; end; end; drop prev; run; ... View more

I understood that a  Retain statement defined by SAS Causes a variable that is created by an INPUT or assignment statement to retain its value from one iteration of the DATA step to the next. ... View more

As said before_temporary_ array values are by defualt retained. But variables as expressed in array are not retained. Array values are refreshed as data step iterates. Here is an example and see the LOG. data _null_; set sashelp.class; array k[*] age height weight; put k[1] = k[2] = k[3] =; run; ... View more

Need only one array to check and replace. Reeza has already said this. I give an illustration with 3 cells.   data have; Revenue1_sum = 1; Revenue2_sum = -0.1; Revenue3_sum = -0.2; run; data want; set have; array aaa[3] Revenue1_sum Revenue2_sum Revenue3_sum; do i = 1 to 3; if aaa[i] < 0 and aaa[i] ne . then aaa[i] = 0; end; drop i; run; proc print data = want; run; ... View more