Not sure if I got the question right, but this looks like a possibility: data want; set have; if response ne 'NE' then replacement_date = date; if date = . then date = replacement_date; drop replacement_date; retain replacement_date; run; ... View more

The easy way to count observations in a SAS data set is PROC FREQ: proc freq data=DB; tables resp1 * resp2 * resp3 / list missing; run; ... View more

If all you want to do is use one format for numerics and a different format for character variables, you don't need to do so much work.  You could use:   data want; set have; format _numeric_ numformat. _character_ $charformat.; run; Of course you have to supply the format names.  Perhaps there is more to the problem that I am failing to see. ... View more

As a non-statistician, here's the approach I took about 10 years ago.  There was clustering software available for purchase, but it was quite expensive.  So here is my workaround approach.   Sum up each 30 minutes:  total N and # failues.   Start at the beginning of the data and compute:  what would happen if we added in the second 30 minutes to the first?  Would the success rate be significantly different for the combination vs. the first?  If so, the second segment begins a  new cluster.  If not, the first two segments combined become cluster #1.  Keep on incorporating a segment until the cluster becomes different.   You get to fiddle with the segment size and the definition of "significantly different".   In the end, this approach produced very similar results to off-the-shelf software that cost hundreds of thousands of dollars.  On the other hand, if you start at the end of the data and work backwards in time, the results would change dramatically.  Also note, clusters are not defined "within the day".  More like a "bucket" than a "cluster", a cluster could end up spanning more than a day or less than a day, and could include portions of two (or more) days.  At any rate, perhaps this gives you some ideas you could apply.   ... View more

To get help,  you will need to understand the question being asked.  You might have access to the person asking the question, but we certainly don't.  These types of questions readily pop out: If a hospitalization lasts for 3 weeks does that count as one hospitalization or as three hospitalizations? If a hospitalization lasts forr 10 days, does that count as one, two, or one-and-a-half hospitalizations? Does any of the counting depend on the day of the week in which a hospitalization begins? Basically, what is it you are trying to count. The programming is relatively straightforward.  But the key to solving it correcly is understanding the question. ... View more

A small tweak for the sake of efficiency: set random_A(in=ina keep=target) random_B(in=inb keep=target); ... View more

@Tom  is right about the correct starting point.  Since you have steps that are obscure and difficult to explain. start by examining data values before the MERGE.  See if any truncated values occur before any merging takes place.   Eventually, you will need to correct the program for all the other variables that SAS mentions.  But truncation of PID is a separate issue. ... View more

You're right, the retain statement causes the problem. In a simple manner. Retain tells SAS to let the value sit there and don't reset it just you begin a new observation. IF THEN is irrelevant, although you capitalize on it by adding an ELSE statement: else close_dte_eff = " "; ... View more

The usual place to begin would be to examine the data.  In particular take a look at COHORT_FLAG and see if it contains unusual values.  They don't have to be unprintable characters.  They could be values of 2 or 3 instead of what you expect.  So just run a PROC FREQ to be sure.     Consider how you set up your logic:   IF cohort_flag = 1 AND HLT_ALZDEM = 1 THEN ADRD_group = 1; ELSE ADRD_group = 0; IF cohort_flag = 0 THEN ADRD_group = .; IF HLT_ALZDEM ^in (1,2) THEN ADRD_group = .; This logic could easily give you a value of 0 for ADRD_group when cohort_flag is 3.  Remember, the ELSE statement applies to either of these conditions: COHORT_FLAG is not 1 HLT_ALZDEM is not 1 That could be a logic problem resolved by checking the data.  In fact, a general approach to solving this would be to take some observations that you believe are coming out incorrectly, and examine the incoming data values for that observation. ... View more

I would actually welcome an "in-between" solution.  Don't hide the queue, just make it a separate queue.  A sort of "might be spam" queue that everyone can see, until further investigation takes place.  Users can still read and respond if they would like. ... View more

One key issue to resolve is this.  What is in your TIME variables?  Are they character strings, or are they numeric measurements of time (number of seconds)? ... View more

It looks like your program drops the original variabes.  Nothing wrong with that, but if you don't need them this simplification comes to mind:   data want; length city gender name $ 1; set have; drop Name Gender City; left_to_right=catx(' ',name,gender,city); right_to_left=catx(' ',city,gender,name); run; Just shorten the variables, and forget about SUBSTR. ... View more