Calculate Date Difference for Non-Consecutive Months

Question

I am trying to find gaps in enrollment and have a table set up like this:

ID	Enrollment _Month	Consecutive_Months
1	202403	1
1	202404	2
1	202405	3
1	202409	1
1	202410	2
1	202411	3
2	202401	1
2	202402	2
2	202407	1
2	202408	2

Using ID 1 as an example, the goal is to find the difference between enrollment months between 202405 (max consecutive months number in sequence) and 202409, where the "Consecutive Months" count restarts back to 1 for each ID. Is there a way to do this?

Is there another way to calculate a gap like this without using the consecutive months column?

Thank you!!

Convert the "monthdate" to a real date, for example first day of the month and then take datediff between that and LEAD(monthdate)? — siggemannen
– siggemannen, Commented Jan 15 at 22:55

ValNik · Accepted Answer · 2025-01-15 23:52:11Z

As already suggested in the comments, it would be better to make a Enrollment_Month column of the date type. Then you can use the functions of working with the date (DATEDIFF(month,...).

However, let's say you need it is an integer.
First, we use lag() to get the previous value.
Then calculate the difference in months between the two dates (YYYYMM). The rest of query is the typical gaps and islands task.

If currMonth (as Enrollment_Month) and prevMonth (as lag(Enrollment_Month)) are integers in format YYYYMM, we calculate difference as

(currMonth/100-prevMonth/100)*12+(currMonth%100-prevMonth%100)  dif

If dif>1, there is Gap.

See example

with A as (
  select ID,Enrollment_month currMonth 
    ,lag(Enrollment_Month,1,Enrollment_Month)over(partition by ID order by Enrollment_Month) prevMonth
  from test
)
,B as (
select *
  ,(currMonth/100-prevMonth/100)*12+(currMonth%100-prevMonth%100)  dif
  ,sum(case when (currMonth/100-prevMonth/100)*12+(currMonth%100-prevMonth%100)>1 then 1 else 0 end)
       over(partition by ID order by currMonth) seqN
from A
)
select *
  ,row_number()over(partition by ID,seqN order by currMonth) ConsecutiveMonths
from B

ID	currMonth	prevMonth	dif	seqN	ConsecutiveMonths
1	202403	202403	0	0	1
1	202404	202403	1	0	2
1	202405	202404	1	0	3
1	202407	202405	2	1	1
1	202409	202407	2	2	1
1	202410	202409	1	2	2
1	202411	202410	1	2	3
1	202503	202411	4	3	1
1	202504	202503	1	3	2
2	202401	202401	0	0	1
2	202402	202401	1	0	2
2	202407	202402	5	1	1
2	202408	202407	1	1	2
7	202410	202410	0	0	1
7	202411	202410	1	0	2
7	202412	202411	1	0	3
7	202501	202412	1	0	4
8	202410	202410	0	0	1
8	202501	202410	3	1	1
8	202502	202501	1	1	2
9	202412	202412	0	0	1
9	202501	202412	1	0	2

Shortly, the same

select *
  ,row_number()over(partition by ID,seqN order by currMonth) ConsecutiveMonths
from (
  select *
    ,(currMonth/100-prevMonth/100)*12+(currMonth%100-prevMonth%100)  dif
    ,sum(case when (currMonth/100-prevMonth/100)*12+(currMonth%100-prevMonth%100)>1 then 1 else 0 end)
       over(partition by ID order by currMonth) seqN
  from (
  select ID,Enrollment_month currMonth 
      ,lag(Enrollment_Month,1,Enrollment_Month)
            over(partition by ID order by Enrollment_Month) prevMonth
    from test
  )A
)B

We do not use column Consecutive_Months in this queries.

fiddle

THANK YOU SO MUCH, especially for the part about working with YYYYMM format :)

Roland Kwee · Accepted Answer · 2025-01-15 23:47:42Z

-1

Date differences are very easy to compute using Julian Date, which is an integer representation of a date. Oracle has conversion functions to and from Julian Date.

answered Jan 15 at 23:47

Roland Kwee

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