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.
-> SQLfiddle demo.
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
Related
I have 3 tables, a user table, an admin table, and a cust table. Both admin and cust tables are foreign keyed to the user_account table. Basically, every user has a user record, and the type of user they are is determined by if they have a record in the admin or the cust table.
user admin cust
user_id user_id | admin_id user_id | cust_id
--------- ---------|---------- ---------|---------
1 1 | a 2 | dd
2 4 | b 3 | ff
3
4
Then I have a login_history table that records the user_id and login timestamp every time a user logs into the app
login_history
user_id | login_on
---------|---------------------
1 | 2022-01-01 13:22:43
1 | 2022-01-02 16:16:27
3 | 2022-01-05 21:17:52
2 | 2022-01-11 11:12:26
3 | 2022-01-12 03:34:47
I would like to create a view that would contain all dates for the first day of each week in the year starting from jan 1st, and a count column that contains the count of unique admin users that logged in that week and a count of unique cust users that logged in that week. So the resulting view should contain the following 53 records, one for each week.
login_counts_view
week_start_date | admin_count | cust_count
-----------------|-------------|------------
2022-01-01 | 1 | 1
2022-01-08 | 0 | 2
2022-01-15 | 0 | 0
.
.
.
2022-12-31 | 0 | 0
Note that the first week (2022-01-01) only has 1 count for admin_count even though the admin with user_id 1 logged in twice that week.
Below is the current query I have for the view. However, the tables are pretty large and it takes over 10 seconds to retrieve all records from the view, mainly because of the left joined date comparisons.
CREATE VIEW login_counts_view AS
SELECT
week_start_dates.week_start_date::text AS week_start_date,
count(distinct a.user_id) AS admin_count,
count(distinct c.user_id) AS cust_count
FROM (
SELECT
to_char(i::date, 'YYYY-MM-DD') AS week_start_date
FROM
generate_series(date_trunc('year', NOW()), to_char(NOW(), 'YYYY-12-31')::date, '1 week') i
) week_start_dates
LEFT JOIN login_history l ON l.login_on::date BETWEEN week_start_dates.week_start_date::date AND (week_start_dates.week_start_date::date + INTERVAL '6 day')::date
LEFT JOIN admin a ON a.user_id = l.user_id
LEFT JOIN cust c ON c.user_id = l.user_id
GROUP BY week_start_date;
Does anyone have any tips as to how to make this query execute more efficiently?
Idea
Compute the pseudo-week of each login date: partition the year into 7-day slices and number them consecutively. The pseudo-week of a given date would be the ordinal number of the slice it falls into.
Then operate the joins on integers representing the pseudo-weeks instead of date values and comparisons.
Implementation
A view to implement this follows:
CREATE VIEW login_counts_view_fast AS
WITH RECURSIVE Numbers(i) AS ( SELECT 0 UNION ALL SELECT i + 1 FROM Numbers WHERE i < 52 )
SELECT CAST ( date_trunc('year', NOW()) AS DATE) + 7 * n.i week_start_date
, count(distinct lw.admin_id) admin_count
, count(distinct lw.cust_id) cust_count
FROM (
SELECT i FROM Numbers
) n
LEFT JOIN (
SELECT admin_id
, cust_id
, base
, pit
, pit-base delta
, (pit-base) / (3600 * 24 * 7) week
FROM (
SELECT a.user_id admin_id
, c.user_id cust_id
, CAST ( EXTRACT ( EPOCH FROM l.login_on ) AS INTEGER ) pit
, CAST ( EXTRACT ( EPOCH FROM date_trunc('year', NOW()) ) AS INTEGER ) base
FROM login_history l
LEFT JOIN admin a ON a.user_id = l.user_id
LEFT JOIN cust c ON c.user_id = l.user_id
) le
) lw
ON lw.week = n.i
GROUP BY n.i
;
Some remarks:
The epoch values are the number of seconds elapsed since an absolute base datetime (specifically 1/1/1970 0h00).
CASTS are necessary to convert doubles to integers and timestamps to dates as mandated by the signatures of postgresql date functions and in order to enforce integer arithmetics.
The recursive subquery is a generator of consecutive integers. It could possibly be replaced by a generate_series call (untested)
Evaluation
See it in action in this db fiddle
The query plan indicates savings of 50-70% in execution time.
I need to write mysql query which will group results by difference between timestamps.
Is it possible?
I have table with locations and every row has created_at (timestamp) and I want to group results by difference > 1min.
Example:
id | lat | lng | created_at
1. | ... | ... | 2020-05-03 06:11:35
2. | ... | ... | 2020-05-03 06:11:37
3. | ... | ... | 2020-05-03 06:11:46
4. | ... | ... | 2020-05-03 06:12:48
5. | ... | ... | 2020-05-03 06:12:52
Result of this data should be 2 groups (1,2,3) and (4,5)
It depends on what you actually want. If youw want to group together records that belong to the same minute, regardless of the difference with the previous record, then simple aggregation is enough:
select
date_format(created_at, '%Y-%m-%d %H:%i:00') date_minute,
min(id) min_id,
max(id) max_id,
min(created_at) min_created_at,
max(created_at) max_created_at,
count(*) no_records
from mytable
group by date_minute
On the other hand, if you want to build groups of consecutive records that have less than 1 minute gap in between, this is a gaps and islands problem. Here is on way to solve it using window functions (available in MySQL 8.0):
select
min(id) min_id,
max(id) max_id,
min(created_at) min_created_at,
max(created_at) max_created_at,
count(*) no_records
from (
select
t.*,
sum(case when created_at < lag_created_at + interval 1 minute then 0 else 1 end)
over(order by created_at) grp
from (
select
t.*,
lag(created_at) over(order by created_at) lag_created_at
from mytable t
) t
) t
group by grp
I have a table called Position, in this table, I have the following, dates are inclusive (yyyy-mm-dd), below is a simplified view of the employment dates
id, person_id, start_date, end_date , title
1 , 1 , 2001-12-01, 2002-01-31, 'admin'
2 , 1 , 2002-02-11, 2002-03-31, 'admin'
3 , 1 , 2002-02-15, 2002-05-31, 'sales'
4 , 1 , 2002-06-15, 2002-12-31, 'ops'
I'd like to be able to calculate the gaps in employment, assuming some of the dates overlap to produce the following output for the person with id=1
person_id, start_date, end_date , last_position_id, gap_in_days
1 , 2002-02-01, 2002-02-10, 1 , 10
1 , 2002-06-01, 2002-06-14, 3 , 14
I have looked at numerous solutions, UNIONS, Materialized views, tables with generated calendar date ranges, etc. I really am not sure what is the best way to do this. Is there a single query where I can get this done?
step-by-step demo:db<>fiddle
You just need the lead() window function. With this you are able to get a value (start_date in this case) to the current row.
SELECT
person_id,
end_date + 1 AS start_date,
lead - 1 AS end_date,
id AS last_position_id,
lead - (end_date + 1) AS gap_in_days
FROM (
SELECT
*,
lead(start_date) OVER (PARTITION BY person_id ORDER BY start_date)
FROM
positions
) s
WHERE lead - (end_date + 1) > 0
After getting the next start_date you are able to compare it with the current end_date. If they differ, you have a gap. These positive values can be filtered within the WHERE clause.
(if 2 positions overlap, the diff is negative. So it can be ignored.)
first you need to find what dates overlaps Determine Whether Two Date Ranges Overlap
then merge those ranges as a single one and keep the last id
finally calculate the ranges of days between one end_date and the next start_date - 1
SQL DEMO
with find_overlap as (
SELECT t1."id" as t1_id, t1."person_id", t1."start_date", t1."end_date",
t2."id" as t2_id, t2."start_date" as t2_start_date, t2."end_date" as t2_end_date
FROM Table1 t1
LEFT JOIN Table1 t2
ON t1."person_id" = t2."person_id"
AND t1."start_date" <= t2."end_date"
AND t1."end_date" >= t2."start_date"
AND t1.id < t2.id
), merge_overlap as (
SELECT
person_id,
start_date,
COALESCE(t2_end_date, end_date) as end_date,
COALESCE(t2_id, t1_id) as last_position_id
FROM find_overlap
WHERE t1_id NOT IN (SELECT t2_id FROM find_overlap WHERE t2_ID IS NOT NULL)
), cte as (
SELECT *,
LEAD(start_date) OVER (partition by person_id order by start_date) next_start
FROM merge_overlap
)
SELECT *,
DATE_PART('day',
(next_start::timestamp - INTERVAL '1 DAY') - end_date::timestamp
) as days
FROM cte
WHERE next_start IS NOT NULL
OUTPUT
| person_id | start_date | end_date | last_position_id | next_start | days |
|-----------|------------|------------|------------------|------------|------|
| 1 | 2001-12-01 | 2002-01-31 | 1 | 2002-02-11 | 10 |
| 1 | 2002-02-11 | 2002-05-31 | 3 | 2002-06-15 | 14 |
Postgres version 9.4.18, PostGIS Version 2.2.
I removed some of the details about the tables from this question because I doubt it's needed to answer the question. I can add those details back if necessary.
Desired result:
I want a total count for each week of year and hour of day (0100 to 5223). I'm able to successfully generate a series of 0100 to 5223 (actually up to 5300), and I'm able to get a total count for each week of year and hour of day individually, but i'm unable to combine the queries so that weeks of year/hours of day with a zero county still show up. I want to combine the count result with the generate_series (and ideally divide that result by 30) to get something like below.
MM-DD | count_not_zero | count_not_zero_divided_by_30
-------+----------------+----------------------------
0100 | 10 | 33.3
0101 | 0 | 0
0102 | 0 | 0
...
0123 | 0 | 0
0200 | 3 | 10
0201 | 10 | 33.3
...
5223 | 20 | 66.6
Here are my individual queries that work...that I want to combine:
SELECT DISTINCT f_woyhh(d::timestamp) as woyhh
FROM generate_series(timestamp '2018-01-01', timestamp '2018-12-31', interval '1 hour') d
GROUP BY woyhh
ORDER by woyhh asc;
SELECT dt, count(*) FROM
(SELECT f_woyhh((time)::timestamp at time zone 'utc' at time zone 'america/chicago')
AS dt,
EXTRACT(YEAR FROM time) AS ctYear, count(*)
AS ct
FROM counties c
INNER JOIN ltg_data d ON ST_contains(c.the_geom, d.ltg_geom)
WHERE countyname = 'Milwaukee' AND state = 'WI' AND EXTRACT(YEAR from time) > '1987' GROUP BY dt, EXTRACT(YEAR from time))
AS count group by dt;
The result from the second query above is (and skips zero count dt, which I don't want):
dt | count
-------+-------
0100 | 10
0104 | 5
0108 | 4
...
Conclusion:
I'm trying to combine the above working individual queries into a single query that provides a three a three column result--woyhh, count, and count divided by 30. And I want to include woyhh that have zero in the county, so that I have a complete set of woyhh.
Thanks for any help!!
I found the answer. I'll be posting it tomorrow, but I wanted to put this on today so no one unnecessarily works on this question. I apologize for the formatting.
WITH CTE_Dates AS (SELECT DISTINCT f_woyhh(d::timestamp) as dt
FROM generate_series(timestamp '2018-01-01', timestamp '2018-12-31', interval '1 hour') d),
CTE_WeeklyHourlyCounts AS (SELECT dt, count(*) as ct
FROM (SELECT f_woyhh((time)::timestamp at time zone 'utc' at time zone 'america/chicago') as dt,
EXTRACT(YEAR FROM time) as ctYear, count(*) as ct
FROM counties c
INNER JOIN ltg_data d on ST_contains(c.the_geom, d.ltg_geom)
WHERE countyname = 'Milwaukee' AND state = 'WI' AND EXTRACT(YEAR from time) > '1987'
GROUP BY dt,
EXTRACT(YEAR from time)) as count group by dt),
CTE_FullSTats AS (SELECT CTE_Dates.dt as dt, CAST(CTE_WeeklyHourlyCounts.ct as decimal) as ct
FROM CTE_Dates LEFT JOIN CTE_WeeklyHourlyCounts ON CTE_WeeklyHourlyCounts.dt = CTE_Dates.dt
GROUP BY CTE_Dates.dt, CTE_WeeklyHourlyCounts.ct, CTE_WeeklyHourlyCounts.dt) SELECT dt, COALESCE(ct, 0)
AS count, round(((COALESCE(ct,0) * 100) / 30),0) as percent FROM CTE_FullStats
GROUP BY dt, ct ORDER BY dt;
Bit stuck on a problem. Trying to find the difference between two dates in postgreSQL.
I have a table emp with many employees in it:
emp_id, date
1, 31-10-2017
1, 08-08-2017
1, 02-06-2017
I want it to look like this:
emp_id, max_date, penultimate_date, difference
1, 31-10-2017, 08-08-2017, 84 days
Obviously you can use max(date) and group by the emp_id, however how do you retrieve the penultimate date. I have used a few functions like:
order by date desc limit 1 offset 1
I have also tried to put these in sub queries but that hasn,t worked as there are many employee numbers and I need one row for each employee.
Can anyone help???
Thanks,
pp84
as kindly suggested by #Haleemur Ali, order by date desc limit 1 offset 1 would not work with several emp_id:
t=# with d(emp_id, date)as (values(1, '31-10-2017'::date),(1, '08-08-2017'),(1, '02-06-2017' ),(2,'2016-01-01'),(2,'2016-02-02'),(2,'2016-03-03'))
select distinct emp_id
, max(date) over (partition by emp_id) max_date
, nth_value(date,2) over (partition by emp_id) penultimate_date
, max(date) over (partition by emp_id) - nth_value(date,2) over (partition by emp_id) diff
from d
;
emp_id | max_date | penultimate_date | diff
--------+------------+------------------+------
2 | 2016-03-03 | 2016-02-02 | 30
1 | 2017-10-31 | 2017-08-08 | 84
(2 rows)
Time: 0.756 ms
WITH emps (emp_id, date) AS (
VALUES (1, '2017-10-31'::DATE)
, (1, '2017-08-08'::DATE)
, (1, '2017-08-08'::DATE)
)
SELECT DISTINCT ON (emp_id)
emp_id
, "date" max_date
, LEAD("date") OVER w penultimate_date
, "date" - LEAD("date") OVER w difference
FROM emps
WINDOW w AS (PARTITION BY emp_id)
ORDER BY emp_id, date DESC
When ordered in descending order, the LEAD("date") w will give the value of the date value from the next row.
The DISTINCT ON limits the resultset to 1 row (the first row encountered) per emp_id.
With our ordering this first row must contain the greatest date, and the LEAD(...) over w therefore returns the penultimate date. This gives us the following result:
emp_id | max_date | penultimate_date | difference
--------+------------+------------------+------------
1 | 2017-10-31 | 2017-08-08 | 84
(1 row)