topic Re: Reading DOB data in Statistical Procedures

Reading DOB data

GreggB — Wed, 04 Nov 2015 20:00:16 GMT

I have DOB values, some of which look like this: 9/23/1996 12:00:00 AM

Others are like this: 3/1/1998 12:00:00 AM

Using the length statement dob $10 works for the first one but not the 2nd. Instead, I get 3/1/1998 1

What I want is just the date:

9/23/1996

3/1/1998

Re: Reading DOB data

ballardw — Wed, 04 Nov 2015 20:07:06 GMT

You don't say if you want the time as well but to read the date I would start with an informat of ANYDTDTE. and then assign the mmddyy10. format for use. This way you will have an actual date that can be manipulated with many functions instead of a not very nice character value.

Re: Reading DOB data

GreggB — Wed, 04 Nov 2015 20:08:34 GMT

i want just the date.

Re: Reading DOB data

Reeza — Wed, 04 Nov 2015 20:17:11 GMT

Try anydtDTM for a date time variable. Then use the Datepart function to get only the date specific portion.

Re: Reading DOB data

ArtC — Wed, 04 Nov 2015 20:23:09 GMT

data have;
infile datalines truncover;
input cdate $30.;
datalines;
9/23/1996 12:00:00 AM
3/1/1998 12:00:00 AM
run;

data want;
set have;
date = input(scan(cdate,1,' '),anydtdte.);
format date date9.;
run;

Re: Reading DOB data

GreggB — Wed, 04 Nov 2015 20:23:39 GMT

I think I may have to forget the data step I'm using. The files I'm reading have changed from last year. Right now I have:

data student;

length dob $10 .................;

infile '.........';

input dob $;

new_dob=input(dob,yymmdd10.);

run;

Re: Reading DOB data

ballardw — Wed, 04 Nov 2015 20:30:56 GMT

Try:

data student;
infile '.........';
informat dob AnyDtDTE. ;
input dob ;
Format dob mmddyy10. ;
run;

and see if that work. I have a number of files where the data source appends a spurious "time" that I ignore by reading just the date part using similar code.

Re: Reading DOB data

Astounding — Wed, 04 Nov 2015 20:32:09 GMT

The program that you have is a good match for the data you have shown. You would run into a problem if your program were different. for example:

input dob $10.;

That would read in the extra character that you are seeing. But this would be fine:

length dob $ 10;

input dob $;

Another place you could go wrong is by adding an informat:

informat dob $10.;

But what you have posted is a good match and should be working.