i have a list with peoples id and date, the list say when a person Entered to website (his id and date).
how can i show for all the dates how many people enter the site two days in a row?
the data ( 30,000 like this in diffrent dates)
01/03/2019 4616
01/03/2019 17584
01/03/2019 7812
01/03/2019 34
01/03/2019 12177
01/03/2019 7129
01/03/2019 11660
01/03/2019 2428
01/03/2019 17514
01/03/2019 10781
01/03/2019 7629
01/03/2019 11119
I succeeded to show the amount of pepole enter the site on the same day but i didnt succeeded to add a column that show the pepole that enter 2 days in row.
date number_of_entrance
2019-03-01 7099
2019-03-02 7021
2019-03-03 7195
2019-03-04 7151
2019-03-05 7260
2019-03-06 7169
2019-03-07 7076
2019-03-08 7081
2019-03-09 6987
2019-03-10 7172
select date,count(*) as number_of_entrance
fROM [finalaa].[dbo].[Daily_Activity]
group by Date
order by date;
how can i show for all the dates how many people enter the site two days in a row?
I would just use lag():
select count(distinct person)
from (select t.*,
lag(date) over (partition by person order by date) as prev_date
from t
) t
where prev_date = dateadd(day, -1, date);
Your code suggests SQL Server, so I used the date functions in that database.
If you want this per date:
select date, count(distinct person)
from (select t.*,
lag(date) over (partition by person order by date) as prev_date
from t
) t
where prev_date = dateadd(day, -1, date)
group by date;
You can use a subquery which returns the number of common entrances in 2 days:
select
t.date,
count(*) as number_of_entrance,
(
SELECT COUNT(g.id) FROM (
SELECT id
FROM [Daily_Activity]
WHERE date IN (t.date, t.date - 1)
GROUP BY id
HAVING COUNT(DISTINCT date) = 2
) g
) number_of_entrance_2_days_in_a_row
FROM [Daily_Activity] t
group by t.date
order by t.date;
Replace id with the 2nd column's name in the table.
Related
I would like to know the number of active users by month, I use SQL Server 2017.
I have an AuditLog table like:
- UserID: int
- DateTime: datetime
- AuditType: int
UserID DateTime AuditType
------------------------------
1 2022-01-01 1
1 2022-01-15 4
1 2022-02-20 3
2 2022-01-10 8
2 2022-03-10 1
3 2022-03-20 1
If someone has at least one entry in a given month then he/she is treated as active.
I would like to have a result like:
Date Count
2022-01 2
2022-02 1
2022-03 2
I think you can combine the function Month(datetime) in the GROUP BY with the Count function SELECT COUNT(UserID)
SELECT (CAST(YEAR(C.DATE)AS CHAR(4))+'-'+CAST(MONTH(C.DATE)AS CHAR(2)))YEAR_MONTH,COUNT(C.USER_ID)CNTT
FROM AUDITLOG AS C
GROUP BY (CAST(YEAR(C.DATE)AS CHAR(4))+'-'+CAST(MONTH(C.DATE)AS CHAR(2)))
ORDER BY (CAST(YEAR(C.DATE)AS CHAR(4))+'-'+CAST(MONTH(C.DATE)AS CHAR(2)));
Here is solutions,
Select [Date],count(1) as Count From (
select Cast(cast(d.DateTime as date) as varchar(7)) as [Date],UserId
from AuditLog d
Group by Cast(cast(d.DateTime as date) as varchar(7)),UserId
) as q1 Group by [Date]
Order by 1
Hope, it will works.
GROUP DATE (Year and Month) either combine or separate and count distinct userId
SELECT CONVERT(VARCHAR(7), [DateTime], 126)[Date], COUNT(DISTINCT UserID)[Count]
FROM AuditLog
GROUP BY CONVERT(VARCHAR(7), [DateTime], 126)
How can I get all user_id values from the data below, for all rows containing the same user_id value over consecutive months from a given start date in the date column.
For example, given the below table....
date
user_id
2018-11-01
13
2018-11-01
13
2018-11-01
14
2018-11-01
15
2018-12-01
13
2019-01-01
13
2019-01-01
14
...supposing I want to get the user_id values for consecutive months prior to (but not including) 2019-01-01 then I'd have this as my output:
user_id
m_year
13
2018-11
13
2018-12
13
2019-01
probably can be applied windows function
If you want to aggregate on a user and the year-months
select
t.user_id,
to_char(date_trunc('month',t.date),'YYYY-MM') as m_year
from yourtable t
where t.date < '2019-02-01'::date
group by t.user_id, date_trunc('month',t.date)
order by t.user_id, m_year
But if you only want those with consecutive months, then a little extra is needed.
select
user_id,
to_char(ym,'YYYY-MM') as m_year
from
(
select t.user_id
, date_trunc('month',t.date) as ym
, lag(date_trunc('month',t.date))
over (partition by t.user_id order by date_trunc('month',t.date)) as prev_ym
, lead(date_trunc('month',t.date))
over (partition by t.user_id order by date_trunc('month',t.date)) as next_ym
from yourtable t
where t.date < '2019-02-01'::date
group by t.user_id, date_trunc('month',t.date)
) q
where (ym - prev_ym <= '31 days'::interval or
next_ym - ym <= '31 days'::interval)
order by user_id, ym
user_id | m_year
------: | :------
13 | 2018-11
13 | 2018-12
13 | 2019-01
db<>fiddle here
you don't need a window function in this specific query. Just try :
SELECT DISTINCT ON (user_id) user_id, date_trunc('month', date :: date) AS m_year
FROM your_table
In postgres, I want to output the persons who have the highest no. of "discussed" requests for each month, irrespective of the year i.e. there should be 12 outputs.
ID PERSON REQUEST DATE
4 datanoise opened 2010-09-02
5 marsuboss opened 2010-09-02
6 m3talsmith opened 2010-09-06
7 sferik opened 2010-09-08
8 sferik opened 2010-09-09
8 dtrasbo discussed 2010-09-09
8 brianmario discussed 2010-09-09
8 sferik discussed 2010-09-09
9 rsim opened 2011-09-09
.....more tuples to follow
*This is just a small part of the databse. also assume that the dataset is big enough that all months are represented in the date column.
Test data:
CREATE TEMPORARY TABLE foo( id SERIAL PRIMARY KEY, name INTEGER NOT NULL,
dt DATE NULL, request BOOL NOT NULL );
INSERT INTO foo (name,dt,request) SELECT random()*1000,
'2010-01-01'::DATE+('1 DAY'::INTERVAL)*(random()*3650), random()>0.5
FROM generate_series(1,100000) n;
SELECT * FROM foo LIMIT 10;
id | name | dt | request
----+------+------------+---------
1 | 110 | 2014-11-05 | f
2 | 747 | 2015-03-12 | t
3 | 604 | 2014-09-26 | f
4 | 211 | 2011-12-14 | t
5 | 588 | 2016-12-15 | f
6 | 96 | 2012-02-19 | f
7 | 17 | 2018-09-18 | t
8 | 591 | 2018-02-15 | t
9 | 370 | 2015-07-28 | t
10 | 844 | 2019-05-16 | f
Now you have to get the count per name and month, then get the max count, but that won't give you the name that has the maximum, which requires joining back with the previous result. In order to do the group by only once, it is done in a CTE:
WITH totals AS (
SELECT EXTRACT(month FROM dt) mon, name, count(*) cnt FROM foo
WHERE request=true GROUP BY name,mon
)
SELECT * FROM
(SELECT mon, max(cnt) cnt FROM totals GROUP BY mon) x
JOIN totals USING (mon,cnt);
If several names have the same maximum count, they will be returned both. To keep only one, you can use DISTICT ON:
WITH (same as above)
SELECT DISTINCT ON (mon) * FROM
(SELECT mon, max(cnt) cnt FROM totals GROUP BY mon) x
JOIN totals USING (mon,cnt) ORDER BY mon,name;
You can also use DISTINCT ON to keep only one row per month, specified by the ORDER clause, in this cas by count desc, so it keeps the highest count.
SELECT DISTINCT ON (mon) * FROM (
SELECT EXTRACT(month FROM dt) mon, name, count(*) cnt FROM foo
WHERE request=true GROUP BY name,mon
)x ORDER BY mon, cnt DESC;
...or you could hack an argmax() function by sticking the primary key into an array passed to max(), which means it will return the id of the row which has the maximum value:
SELECT mon, cntid[1] cnt, name FROM
(SELECT mon, max(ARRAY[cnt,id]) cntid FROM (
SELECT EXTRACT(month FROM dt) mon, name, count(*) cnt, min(id) id FROM foo
WHERE request=true GROUP BY name,mon
) x GROUP BY mon)y
JOIN foo ON (foo.id=cntid[2]);
Which one will be faster?...
given your table is named t01 and the colum date is date1 (and in string format):
create temp table t02 as
select extract(month from CAST(date1 as date)) as month, person, count(*) nb from t01 where request = 'discussed' group by 1, 2 ;
create temp table t03 as
select month, max(nb) max_nb from t02 group by 1 ;
the result is :
select month , person from t02 a natural join t03 b where a.nb = b.max_nb;
https://rextester.com/BYMM84335[ : run here]1
I would recommend distinct on. If you want to combine all the months into a single "uber-month":
select distinct on (extract(month from date)) person, extract(month from date), count(*) as num_discussed
from t
where request = 'discussed'
group by person, extract(month from date)
order by extract(month from date), num_discussed desc;
Distinct on is a very handy Postgres extension. It returns on row per "group", which is defined by the expressions in parentheses. The row is the "first" one determined by the order by clause.
If you want the highest month regardless of year:
select distinct on (extract(month from date)) person, date_trunc('month', date), count(*) as num_discussed
from t
where request = 'discussed'
group by person, date_trunc('month', date)
order by extract(month from date), num_discussed desc;
I have a table that shows when a user signs up for a subscription and when their membership will expire. A user can purchase a new subscription even if their current one is in force.
userid|purchasedate|expirydate
1 |2019-01-01 |2019-02-01
2 |2019-01-02 |2019-02-02
3 |2019-01-03 |2019-02-03
3 |2019-01-04 |2019-03-03
I need a SQL query that will GROUP BY the date and return the number of active subscriptions on that date. So it would return:
date |count
2019-01-01|1
2019-01-02|2
2019-01-03|3
2019-01-04|3
Below is for BigQuery Standard SQL
#standardSQL
SELECT day, COUNT(DISTINCT userid) active_subscriptions
FROM (SELECT AS STRUCT MIN(purchasedate) min_date, MAX(expirydate) max_date FROM `project.dataset.table`),
UNNEST(GENERATE_DATE_ARRAY(min_date, max_date)) day
JOIN `project.dataset.table`
ON day BETWEEN purchasedate AND expirydate
GROUP BY day
You can test, play with above using dummy data from your question as in below example
#standardSQL
WITH `project.dataset.table` AS (
SELECT 1 userid, DATE '2019-01-01' purchasedate, DATE '2019-02-01' expirydate UNION ALL
SELECT 2, '2019-01-02', '2019-02-02' UNION ALL
SELECT 3, '2019-01-03', '2019-02-03' UNION ALL
SELECT 3, '2019-01-04', '2019-03-03'
)
SELECT day, COUNT(DISTINCT userid) active_subscriptions
FROM (SELECT AS STRUCT MIN(purchasedate) min_date, MAX(expirydate) max_date FROM `project.dataset.table`),
UNNEST(GENERATE_DATE_ARRAY(min_date, max_date)) day
JOIN `project.dataset.table`
ON day BETWEEN purchasedate AND expirydate
GROUP BY day
with below output
Row day active_subscriptions
1 2019-01-01 1
2 2019-01-02 2
3 2019-01-03 3
4 2019-01-04 3
5 2019-01-05 3
6 2019-01-06 3
... ... ...
... ... ...
31 2019-01-31 3
32 2019-02-01 3
33 2019-02-02 2
34 2019-02-03 1
35 2019-02-04 1
... ... ...
... ... ...
61 2019-03-02 1
62 2019-03-03 1
You need a list of dates and count(distinct):
select d.dte, count(distinct t.userid) as num_users
from (select distinct purchase_date as dte from t) d left join
t
on d.dte >= t.dte and
d.dte <= t.expiry_date
group by d.dte
order by d.dte;
EDIT:
BigQuery can be fickle about inequalities in the on clause. Here is another approach:
select dte, count(distinct t.userid) as num_users
from t cross join
unnest(generate_date_array(t.purchase_date, t.expiry_date, interval 1 day)) dte
group by dte
order by dte;
You can use a where clause to filter down to particular dates.
I make the table name 'test_expirydate' and use your data
and this one work
select
tb1.expirydate,
count(*) as total
from test_expirydate as tb1
left join (
select
expirydate
from test_expirydate as tb2
group by userid
) as tb2
on tb1.expirydate >= tb2.expirydate
group by tb1.expirydate
I don't sure is it work in other case or not but it fine with current data
Oh, I interpret that the left column should be the expiration date.
I have a table with the following info
|date | user_id | week_beg | month_beg|
SQL to create table with test values:
CREATE TABLE uniques
(
date DATE,
user_id INT,
week_beg DATE,
month_beg DATE
)
INSERT INTO uniques VALUES ('2013-01-01', 1, '2012-12-30', '2013-01-01')
INSERT INTO uniques VALUES ('2013-01-03', 3, '2012-12-30', '2013-01-01')
INSERT INTO uniques VALUES ('2013-01-06', 4, '2013-01-06', '2013-01-01')
INSERT INTO uniques VALUES ('2013-01-07', 4, '2013-01-06', '2013-01-01')
INPUT TABLE:
| date | user_id | week_beg | month_beg |
| 2013-01-01 | 1 | 2012-12-30 | 2013-01-01 |
| 2013-01-03 | 3 | 2012-12-30 | 2013-01-01 |
| 2013-01-06 | 4 | 2013-01-06 | 2013-01-01 |
| 2013-01-07 | 4 | 2013-01-06 | 2013-01-01 |
OUTPUT TABLE:
| date | time_series | cnt |
| 2013-01-01 | D | 1 |
| 2013-01-01 | W | 1 |
| 2013-01-01 | M | 1 |
| 2013-01-03 | D | 1 |
| 2013-01-03 | W | 2 |
| 2013-01-03 | M | 2 |
| 2013-01-06 | D | 1 |
| 2013-01-06 | W | 1 |
| 2013-01-06 | M | 3 |
| 2013-01-07 | D | 1 |
| 2013-01-07 | W | 1 |
| 2013-01-07 | M | 3 |
I want to calculate the number of distinct user_id's for a date:
For that date
For that week up to that date (Week to date)
For the month up to that date (Month to date)
1 is easy to calculate.
For 2 and 3 I am trying to use such queries:
SELECT
date,
'W' AS "time_series",
(COUNT DISTINCT user_id) COUNT (user_id) OVER (PARTITION BY week_beg) AS "cnt"
FROM user_subtitles
SELECT
date,
'M' AS "time_series",
(COUNT DISTINCT user_id) COUNT (user_id) OVER (PARTITION BY month_beg) AS "cnt"
FROM user_subtitles
Postgres does not allow window functions for DISTINCT calculation, so this approach does not work.
I have also tried out a GROUP BY approach, but it does not work as it gives me numbers for whole week/months.
Whats the best way to approach this problem?
Count all rows
SELECT date, '1_D' AS time_series, count(DISTINCT user_id) AS cnt
FROM uniques
GROUP BY 1
UNION ALL
SELECT DISTINCT ON (1)
date, '2_W', count(*) OVER (PARTITION BY week_beg ORDER BY date)
FROM uniques
UNION ALL
SELECT DISTINCT ON (1)
date, '3_M', count(*) OVER (PARTITION BY month_beg ORDER BY date)
FROM uniques
ORDER BY 1, time_series
Your columns week_beg and month_beg are 100 % redundant and can easily be replaced by
date_trunc('week', date + 1) - 1 and date_trunc('month', date) respectively.
Your week seems to start on Sunday (off by one), therefore the + 1 .. - 1.
The default frame of a window function with ORDER BY in the OVER clause uses is RANGE BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW. That's exactly what you need.
Use UNION ALL, not UNION.
Your unfortunate choice for time_series (D, W, M) does not sort well, I renamed to make the final ORDER BY easier.
This query can deal with multiple rows per day. Counts include all peers for a day.
More about DISTINCT ON:
Select first row in each GROUP BY group?
DISTINCT users per day
To count every user only once per day, use a CTE with DISTINCT ON:
WITH x AS (SELECT DISTINCT ON (1,2) date, user_id FROM uniques)
SELECT date, '1_D' AS time_series, count(user_id) AS cnt
FROM x
GROUP BY 1
UNION ALL
SELECT DISTINCT ON (1)
date, '2_W'
,count(*) OVER (PARTITION BY (date_trunc('week', date + 1)::date - 1)
ORDER BY date)
FROM x
UNION ALL
SELECT DISTINCT ON (1)
date, '3_M'
,count(*) OVER (PARTITION BY date_trunc('month', date) ORDER BY date)
FROM x
ORDER BY 1, 2
DISTINCT users over dynamic period of time
You can always resort to correlated subqueries. Tend to be slow with big tables!
Building on the previous queries:
WITH du AS (SELECT date, user_id FROM uniques GROUP BY 1,2)
,d AS (
SELECT date
,(date_trunc('week', date + 1)::date - 1) AS week_beg
,date_trunc('month', date)::date AS month_beg
FROM uniques
GROUP BY 1
)
SELECT date, '1_D' AS time_series, count(user_id) AS cnt
FROM du
GROUP BY 1
UNION ALL
SELECT date, '2_W', (SELECT count(DISTINCT user_id) FROM du
WHERE du.date BETWEEN d.week_beg AND d.date )
FROM d
GROUP BY date, week_beg
UNION ALL
SELECT date, '3_M', (SELECT count(DISTINCT user_id) FROM du
WHERE du.date BETWEEN d.month_beg AND d.date)
FROM d
GROUP BY date, month_beg
ORDER BY 1,2;
SQL Fiddle for all three solutions.
Faster with dense_rank()
#Clodoaldo came up with a major improvement: use the window function dense_rank(). Here is another idea for an optimized version. It should be even faster to exclude daily duplicates right away. The performance gain grows with the number of rows per day.
Building on a simplified and sanitized data model
- without the redundant columns
- day as column name instead of date
date is a reserved word in standard SQL and a basic type name in PostgreSQL and shouldn't be used as identifier.
CREATE TABLE uniques(
day date -- instead of "date"
,user_id int
);
Improved query:
WITH du AS (
SELECT DISTINCT ON (1, 2)
day, user_id
,date_trunc('week', day + 1)::date - 1 AS week_beg
,date_trunc('month', day)::date AS month_beg
FROM uniques
)
SELECT day, count(user_id) AS d, max(w) AS w, max(m) AS m
FROM (
SELECT user_id, day
,dense_rank() OVER(PARTITION BY week_beg ORDER BY user_id) AS w
,dense_rank() OVER(PARTITION BY month_beg ORDER BY user_id) AS m
FROM du
) s
GROUP BY day
ORDER BY day;
SQL Fiddle demonstrating the performance of 4 faster variants. It depends on your data distribution which is fastest for you.
All of them are about 10x as fast as the correlated subqueries version (which isn't bad for correlated subqueries).
Without correlated subqueries. SQL Fiddle
with u as (
select
"date", user_id,
date_trunc('week', "date" + 1)::date - 1 week_beg,
date_trunc('month', "date")::date month_beg
from uniques
)
select
"date", count(distinct user_id) D,
max(week_dr) W, max(month_dr) M
from (
select
user_id, "date",
dense_rank() over(partition by week_beg order by user_id) week_dr,
dense_rank() over(partition by month_beg order by user_id) month_dr
from u
) s
group by "date"
order by "date"
Try
SELECT
*
FROM
(
SELECT dates, count(user_id), 'D' as timesereis FROM users_data GROUP BY dates
UNION
SELECT max(dates), count(user_id), 'W' FROM users_data GROUP BY date_part('year',dates)+date_part('week',dates)
UNION
SELECT max(dates), count(user_id), 'M' FROM users_data GROUP BY date_part('year',dates)+date_part('week',dates)
) tEMP order by dates, timesereis
SQLFIDDLE
Try queries like this
SELECT count(distinct user_id), date_format(date, '%Y-%m-%d') as date_period
FROM uniques
GROUP By date_period