About antonbcristina

antonbcristina · ‎11-08-2023

The reason your merge results in more observations is because certain matches were not found. You'll want to identify which observations were not found by using a version of the code below. Create indicator variables (call them a and b) to identify observations coming from the two datasets. You can use an IF statement to filter observations that belong to one dataset (a=1) but not the other (b=0). data want; merge one(in=a) two(in=b); by id; if a=1 and b=0; /* Observations in dataset one that are not present in dataset two */ run;

antonbcristina · ‎11-08-2023

Idealy you'd want them to be the same length. However, it doesn't seem to be the case for your datasets otherwise you should see the following in the log: WARNING: Multiple lengths were specified for the BY variable <variable name> by input data sets. This might cause unexpected results.

antonbcristina · ‎11-08-2023

If merging two datasets results in more observations than the larger of the two individual datasets, than it means that certain matches were not found. It helps to see the log as well to identify the issue. Feel free to post it and we can investigate.

antonbcristina · ‎11-08-2023

You mentioned that one dataset's ID start at 1 but the other starts at 1001. If you know IDs 1 and 1001 are the same individual then their observations should match. You'll have to change one of your ID variables by either adding or subtracting 1000.

antonbcristina · ‎10-30-2023

You could use a PROC SQL subquery to find the (distinct?) IDs with records in 2023 and then pull all records matching those IDs. proc sql; select id, year from have where id in (select id from have where year=2023); quit;

antonbcristina · ‎10-23-2023

As suggested, do this in stages. First scan the character string, confirming you have the right number (as a character string), then convert it using the INPUT function. DATA CI; countyCI = "95% C.I.: 7.7 - 10.7"; countylowerCI_char = SCAN(countyCI, 2, ':-'); countylowerCI_num = input(countylowerCI_char,best10.); RUN;

antonbcristina · ‎10-13-2023

Unfortunately using the IN operator here won't work since SAS interprets this statement: if diagnoses in ('J45.10', 'J45.20', 'J45.30') then AI=1; as: if diagnoses ='J45.10' or diagnoses ='J45.20' or diagnoses ='J45.30' then AI=1; Based on the example you provided for the values of Diagnoses, none of these conditions would ever result to True since they are much longer strings. What you could use instead is the FIND function which would allow you to search for a substring of characters within the variable Diagnoses like so: if find(diagnoses,"J45.10",'i')>0 then AI=1; Since the FIND fuction will return the starting position of the substring within the string, you're looking for a value >0 to confirm whether it is present.

antonbcristina · ‎10-12-2023

That's right @Tom, thanks for clarifying!

antonbcristina · ‎10-12-2023

I like to think of the subsetting WHERE statement like a pre-processing of the data, and the subsetting IF statement as a post-processing of the data. You can only filter using WHERE on variables that already exist in the Program Data Vector (PDV) at the beginning of the DATA step. To filter on any variables created in the DATA step, you must use IF. The other important thing to note is that the IF statement can only be used inside of the DATA step, whereas the WHERE statement can be used in both DATA and PROC steps.

antonbcristina · ‎10-10-2023

Considering you might want to solve this in SAS, the following will work. You want to use an array to iterate through the available measures (var1-var4). You will assign first to be the first non-missing value and assign last to be the last non-missing value. data want; set have; array visits [*] var1-var4; do i = 1 to dim(visits); if first=. then first=visits[i]; if visits[i]^=. then last=visits[i]; end; run;

antonbcristina · ‎09-12-2023

Suppose you had the chosen values in a dataset have2 (completely made up based on the values you're displaying): ID Selected 101 656 501 999 502 999 Then a PROC SQL left join will do the trick: proc sql; select t1.id, t1.value, t2.selected from have as t1 left join have2 as t2 on t1.id = t2.id; quit;

antonbcristina · ‎09-12-2023

Congratulations Ben! So very proud of you 🙂

antonbcristina · ‎08-28-2023

We’re excited to kick-off our SAS Explore Certification Review today to help prepare you for the SAS® Certified Specialist: Base Programming Using SAS® 9.4 performance-based certification exam. Our virtual workshop will run on August 28 and August 31, 2023 from 1:00 - 4:30 pm ET. During these sessions, we will review the exam logistics, highlight important concepts, practice sample certification exam questions, and provide additional materials for you to utilize as you study. By now you should have received an automated email with details for activating the Extended Learning Page and joining the class. Please use this post to ask questions and collaborate with classmates.

antonbcristina · ‎06-30-2023

Hi @lichee, you're right PROC SURVEYSELECT will do the trick here with the GROUPS=n option. First you'll want to make sure the data is sorted by the variables that will define your strata. proc sort data=person_fl out=sorted; by gender age_group; run; Then use PROC SURVEYSELECT using the GROUPS=n option and the STRATA statement. proc surveyselect data=sorted groups=4 out=sampled; strata gender age_group; run; You'll want to be careful and choose a number for groups no bigger than the number of observations in the smallest stratum, otherwise this will throw an error. I tried running this with GROUPS=4 on your sample dataset and got an error. Here's the documention explaining the GROUPS option: https://documentation.sas.com/doc/en/pgmsascdc/9.4_3.4/statug/statug_surveyselect_syntax01.htm#statug.surveyselect.selectgroups

antonbcristina · ‎06-28-2023

Hi @PaigeMiller, you're right the original code provided returned no errors. The error only appears if you run the subquery alone: PROC SQL; select ID from has_personID_var; quit; I'm running this on the same version of SAS.

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