SAME yymmn6 format dates are not recognized as equivalent

[appleorange]

This is very bizarre and I can't figure out for the life of me why but I have some data with birth dates and eligibility dates.  The data is originally in a txt file and I am reading in the birth_date as mmddyy10.  The eligibility_month data is in numeric format and appears like this:

Birth_date     ID      Eligibility_Month

01/01/2016  0001   201511

01/01/2016  0001   201512

01/01/2016  0001   201601

01/01/2016  0001   201602

01/01/2016  0001   201603

01/01/2016  0001   201604

01/01/2016  0001   201605

I've tried reading in eligibility_month as yymmn6. and numeric, then converting it to yymmn6. using:

data mom_elig_2016_3;
set mom_eligibility_2016_2; yymm=input(put(eligible_month_yyyymm, 6.), yymmn6.);
format yymm yymmn6.;
informat yymm yymmn6.;
date_my=birth_date;
format date_my yymmn6.;
informat date_my yymmn6.;
run;

The idea is to check if the birth_date = an equivalent entry in the eligibility_month.  I reformatted birth_date into the same format.

For some reason, SAS recognizes SOME of the date_my and yymm as equivalent but not others.  

data aa1; set mom_elig_2016_3;
if date_my=yymm; run;

So for example, this is inaccurate because I see records where date_my = yymm clearly equals each other but isn't picked up.  I must be missing something to do with formatting but I can't figure out what it is.

[singhsahab]

data have;
length Eligibility_Month 6.;
	input Birth_date  ddmmyy10. ID   Eligibility_Month ;
format birth_date ddmmyy10.;
datalines;
01/01/2016 0001 201511
01/01/2016 0001 201512
01/01/2016 0001 201601
01/01/2016 0001 201602
01/01/2016 0001 201603
11/11/2019 0002 201911
;
run;

data want;
set te;
o_mon=month(Birth_date);
o_yr=year(Birth_date);
if length(strip(put(o_mon,best.))) eq 1 then
birht_date_n=strip(strip(put(o_yr,best.))||'0'||strip(put(o_mon,best.)));
else birht_date_n=strip(strip(put(o_yr,best.))||strip(put(o_mon,best.)));
if input(birht_date_n,6.) eq Eligibility_Month then output;
drop o_mon o_yr birht_date_n;
run;

[PaigeMiller]

Formatting doesn't change the underlying value. Two dates can be formatted to look identical, but the underlying values are not equal, and so

if date_my=yymm;

would fail.

So the idea of formatting things to look the same won't help here.

The act of turning eligible_month_yyyymm to character and then back to numeric also seems unnecessary. If eligible_month is numeric string, you could turn it into an actual sas day via the INPUT command. If it is numeric, something like this would work:

yymm = mdy(mod(eligibility_month,100),1,floor(eligibility_month/100));

[Reeza · Accepted solution]

You never actually lose the day component it's still there in the data, just not displayed. So if the day of the month/years don't match then it won't match. Convert both to character or use INTNX() to set the birthdates to the beginning or end of the month as needed.

[appleorange]

OK, that's what I was missing. It seems like resetting the birth dates to the beginning of the month would be the simplest solution here, thanks!