topic Re: Number of days around the fixed date within 365 days across date pairs in SAS Programming

Select date pair that includes the fixed date

Cruise — Mon, 15 Apr 2019 10:39:03 GMT

Hi folks,

My data has

- Date pairs (elig_beg and elig_end) for varying length of time intervals to mark insurance coverage periods. Everybody in this dataset has insurance, but, on and off basis. For example: 01/05/2005(elig_beg1)-01/08/2005(elig_end1) and 01/01/2007(elig_beg2)-01/01/2010(elig_end2). This patient had two coverage intervals. First coverage lasted 3 months and the last coverage lasted 3 years.

- Date_diagnosis : date of diagnosis of a disease

- ID: unique patient identifier

- Eligibility data in separate columns (elig_beg1 thru 50 and elig_end1 thru 50) and unique for each patients.

My goal is to output the date pair that included date_diagnosis as shown in the image below for the first row of "have" demo data.

For instance: 10DEC2004 fell in the date pair: elig_beg/elig_end1, thus, desired output is two variables such as "cov_beg/cov_end". If I identify this date pair (cov_beg and cov-end) that included the date of diagnosis I will be able to calculate the length of the coverage period of insurance that included date_diagnosis in it.


DATA HAVE; 
INPUT ID DATE_DIAGNOSIS elig_beg1	elig_end1	elig_beg2	elig_end2;
CARDS;
1	16415	16376	17713	17806	18808
2	16419	16376	17836	.	.
3	16428	16376	17836	17867	20819
4	16429	16376	16617	.	.
5	16771	16741	20850	.	.
6	16792	16741	20819	.	.
7	17153	17106	17286	.	.
8	17157	17106	17256	.	.
9	18235	18202	18352	.	.
10	18234	18202	20819	21124	21184
11	18597	18567	18839	19663	20027
12	18624	18567	18992	.	.
13	18605	18567	18961	19175	19205
14	18974	18932	19144	.	.
15	18963	18932	19631	.	.
16	18991	18932	19052	.	.
17	19339	19298	21184	.	.
18	19353	19298	19389	.	.
19	19331	19298	19631	.	.
20	19716	19663	20819	.	.
21	19345	19298	19601	.	.
22	20075	20028	20819	20851	20909
23	20065	20028	20361	.	.
;

Re: Number of days around the fixed date within 365 days across date pairs

SASKiwi — Sun, 14 Apr 2019 21:04:30 GMT

What happens in leap years?

Re: Number of days around the fixed date within 365 days across date pairs

Cruise — Sun, 14 Apr 2019 22:00:01 GMT

They shouldn’t affect the accuracy that much. I’d accept approximate rounding then.

Re: Number of days around the fixed date within 365 days across date pairs

SASKiwi — Sun, 14 Apr 2019 23:09:05 GMT

"Approximate rounding" isn't a sufficient definition to code from. To be more precise you could ignore 29 Feb if it falls within any of your date pairs.

Re: Number of days around the fixed date within 365 days across date pairs

Cruise — Mon, 15 Apr 2019 11:41:42 GMT

I think, I solved the problem using Patrick's solution provided in the earlier forum. @Patrick

https://communities.sas.com/t5/SAS-Programming/Check-if-date-points-fall-in-the-time-intervals-for-eligibility/m-p/549064#M152310

data P.long(drop=_: elig_beg: elig_end:); /*NROW=19,931*/ set MY_ACTUAL_DATA; /*NROW=19,931, N_SUBJ=10,341*/
  array _elig_beg {*} elig_beg:;
  array _end {*} elig_end:;
  attrib insurance_elig_beg_dt insurance_end_dt format=date9.;
  do _i=1 to dim(_elig_beg);
    if missing(_elig_beg[_i]) then continue;
    insurance_elig_beg_dt=_elig_beg[_i];
    insurance_end_dt=_end[_i];
    output;
  end;
run;

DATA P.long1; SET P.long;  /*NROW=19,931, N_SUBJ=10,341*/
IF insurance_elig_beg_dt<=DATE_DIAGNOSIS<=insurance_end_dt THEN COV=1; ELSE COV=0; 
RUN;

Proc sort data=P.long1;
By DISPLAY_ID COV;
Run;
Data P.long2; Set P.long1;
By DISPLAY_ID;
If last.DISPLAY_ID;
Run;

Re: Number of days around the fixed date within 365 days across date pairs

ErikLund_Jensen — Mon, 15 Apr 2019 12:39:51 GMT

Hi @Cruise

I think it can be done in the first step with a slight modification of the loop, thus eliminating the intermediate data set and the 3 following steps. It seems that all diagnosis dates fall within the first interval in your example data, so I took the liberty to change a couple of diagnosis dates to test the code.

DATA have; 
	format id 8. DATE_DIAGNOSIS elig_beg1 elig_end1 elig_beg2 elig_end2 date9.;
	informat DATE_DIAGNOSIS elig_beg1 elig_end1 elig_beg2 elig_end2 date9.;
	input ID DATE_DIAGNOSIS elig_beg1 elig_end1 elig_beg2 elig_end2;
	cards;
 1 10DEC2004 01NOV2004 30JUN2008 01OCT2008 30JUN2011
 2 14DEC2004 01NOV2004 31OCT2008 . .
 3 01DEC2008 01NOV2004 31OCT2008 01DEC2008 31DEC2016
 4 24DEC2004 01NOV2004 30JUN2005 . .
 5 01DEC2004 01NOV2005 31JAN2017 . .
 6 22DEC2005 01NOV2005 31DEC2016 . .
 7 18DEC2006 01NOV2006 30APR2007 . .
 8 22DEC2006 01NOV2006 31MAR2007 . .
 9 04DEC2009 01NOV2009 31MAR2010 . .
10 03DEC2009 01NOV2009 31DEC2016 01NOV2017 31DEC2017
11 01DEC2010 01NOV2010 31JUL2011 01NOV2013 31OCT2014
12 28DEC2010 01NOV2010 31DEC2011 . .
13 09DEC2010 01NOV2010 30NOV2011 01JUL2012 31JUL2012
14 13DEC2011 01NOV2011 31MAY2012 . .
15 02DEC2011 01NOV2011 30SEP2013 . .
16 30DEC2011 01NOV2011 29FEB2012 . .
17 12DEC2012 01NOV2012 31DEC2017 . .
18 26DEC2012 01NOV2012 31JAN2013 . .
19 04DEC2012 01NOV2012 30SEP2013 . .
20 24DEC2013 01NOV2013 31DEC2016 . .
21 18DEC2012 01NOV2012 31AUG2013 . .
22 18DEC2014 01NOV2014 31DEC2016 01FEB2017 31MAR2017
23 08DEC2014 01NOV2014 30SEP2015 . .
	;
run;

data want (drop=i); set have; 
  array beg {*} elig_beg:;
  array end {*} elig_end:;
  format cov_beg cov_end date9.;
  do i = 1 to dim(beg);
    if missing(beg{i}) then leave;
	if beg{i} <= DATE_DIAGNOSIS <= end{i} then do;
	    cov_beg = beg{i};
    	cov_end = end{i};
		cov_days = cov_end - cov_beg + 1;
 		leave;
	end;
  end;
run;

Re: Number of days around the fixed date within 365 days across date pairs

Cruise — Mon, 15 Apr 2019 12:42:10 GMT

More efficient indeed! Thank you.