BookmarkSubscribeRSS Feed
🔒 This topic is solved and locked. Need further help from the community? Please sign in and ask a new question.
ballardw
Super User

At the end of an iterated do loop such as

do I= 1 to 8;

you find the value of I to be one more than the maximum value (unless you have some code to skip out of the loop early).

 

So in your code (slightly reformatted for legibility)

32 do i=1 to 8;
33    mindt = min(of dat[*]);
34    maxdt = max(of dat[*]);
35    if res{i}^='0' then _res{i}=i;
36 end;
37
38 if _res{i}=. then do;

line 38 uses the value of the index variable I after ending the loop and will have a value of 9. So the array index is out of range as you define the array _res to have 8 elements. Line 44 will have the same issue.

 

Note that you will get the exact same value for

33 mindt = min(of dat[*]); 
34 maxdt = max(of dat[*]); 

So these really do not need to be in any sort of loop.

 

 

skom1
Calcite | Level 5

Got it. Thank You very much

SAS Innovate 2025: Register Now

Registration is now open for SAS Innovate 2025 , our biggest and most exciting global event of the year! Join us in Orlando, FL, May 6-9.
Sign up by Dec. 31 to get the 2024 rate of just $495.
Register now!

How to Concatenate Values

Learn how use the CAT functions in SAS to join values from multiple variables into a single value.

Find more tutorials on the SAS Users YouTube channel.

SAS Training: Just a Click Away

 Ready to level-up your skills? Choose your own adventure.

Browse our catalog!

Discussion stats
  • 16 replies
  • 1531 views
  • 1 like
  • 7 in conversation