I am trying to group dates within a 1 year interval given an identifier by labeling which is the earliest date and which is the latest date. If there are no dates within a 1 year interval from that date, then it will record it's own date as the first and last date. For example originally the data is:
id | date
____________
a | 1/1/2000
a | 1/2/2001
a | 1/6/2000
b | 1/3/2001
b | 1/3/2000
b | 1/3/1999
c | 1/1/2000
c | 1/1/2002
c | 1/1/2003
And the output I want is:
id | first_date | last_date
___________________________
a | 1/1/2000 | 1/2/2001
b | 1/3/1999 | 1/3/2001
c | 1/1/2000 | 1/1/2000
c | 1/1/2002 | 1/1/2003
I have been trying to figure this out the whole day and can't figure it out. I can do it for cases id's with only 2 duplicates, but can't for greater values. Any help would be great.
SELECT id
, min(min_date) AS min_date
, max(max_date) AS max_date
, sum(row_ct) AS row_ct
FROM (
SELECT id, year, min_date, max_date, row_ct
, year - row_number() OVER (PARTITION BY id ORDER BY year) AS grp
FROM (
SELECT id
, extract(year FROM the_date)::int AS year
, min(the_date) AS min_date
, max(the_date) AS max_date
, count(*) AS row_ct
FROM tbl
GROUP BY id, year
) sub1
) sub2
GROUP BY id, grp
ORDER BY id, grp;
1) Group all rows per ( id, year
), in subquery sub1
. Record min and max of the date. I added a count of rows ( row_ct
) for demonstration.
2) Subtract the row_number()
from the year in the second subquery sub2
. Thus, all rows in succession end up in the same group ( grp
). A gap in the years starts a new group.
3) In the final SELECT
, group a second time, this time by ( id, grp
) and record min, max and row count again. Voilá. Produces exactly the result you are looking for.
Related answers:
Return array of years as year ranges
Group by repeating attribute
select id, min ([date]) first_date, max([date]) last_date
from <yourTbl> group by id
Use this ( SQLFiddle Demo ):
SELECT id,
min(date) AS first_date,
max(date) AS last_date
FROM mytable
GROUP BY 1
ORDER BY 1
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