SQL query for cumulative frequency of list of datetimes - sql

Question

I have a list of times in a database column (representing visits to a website).
I need to group them in intervals and then get a 'cumulative frequency' table of those dates.
For instance I might have:
9:01
9:04
9:11
9:13
9:22
9:24
9:28
and i want to convert that into
9:05 - 2
9:15 - 4
9:25 - 6
9:30 - 7
How can I do that? Can i even easily achieve this in SQL? I can quite easily do it in C#

Answer 1

create table accu_times (time_val datetime not null, constraint pk_accu_times primary key (time_val));
go
insert into accu_times values ('9:01');
insert into accu_times values ('9:05');
insert into accu_times values ('9:11');
insert into accu_times values ('9:13');
insert into accu_times values ('9:22');
insert into accu_times values ('9:24');
insert into accu_times values ('9:28');
go
select rounded_time,
(
select count(*)
from accu_times as at2
where at2.time_val <= rt.rounded_time
) as accu_count
from (
select distinct
dateadd(minute, round((datepart(minute, at.time_val) + 2)*2, -1)/2,
dateadd(hour, datepart(hour, at.time_val), 0)
) as rounded_time
from accu_times as at
) as rt
go
drop table accu_times
Results in:
rounded_time accu_count
----------------------- -----------
1900-01-01 09:05:00.000 2
1900-01-01 09:15:00.000 4
1900-01-01 09:25:00.000 6
1900-01-01 09:30:00.000 7

Answer 2

I should point out that based on the stated "intent" of the problem, to do analysis on visitor traffic - I wrote this statement to summarize the counts in uniform groups.
To do otherwise (as in the "example" groups) would be comparing the counts during a 5 minute interval to counts in a 10 minute interval - which doesn't make sense.
You have to grok to the "intent" of the user requirement, not the literal "reading" of it. :-)
create table #myDates
(
myDate datetime
);
go
insert into #myDates values ('10/02/2008 09:01:23');
insert into #myDates values ('10/02/2008 09:03:23');
insert into #myDates values ('10/02/2008 09:05:23');
insert into #myDates values ('10/02/2008 09:07:23');
insert into #myDates values ('10/02/2008 09:11:23');
insert into #myDates values ('10/02/2008 09:14:23');
insert into #myDates values ('10/02/2008 09:19:23');
insert into #myDates values ('10/02/2008 09:21:23');
insert into #myDates values ('10/02/2008 09:21:23');
insert into #myDates values ('10/02/2008 09:21:23');
insert into #myDates values ('10/02/2008 09:21:23');
insert into #myDates values ('10/02/2008 09:21:23');
insert into #myDates values ('10/02/2008 09:26:23');
insert into #myDates values ('10/02/2008 09:27:23');
insert into #myDates values ('10/02/2008 09:29:23');
go
declare #interval int;
set #interval = 10;
select
convert(varchar(5), dateadd(minute,#interval - datepart(minute, myDate) % #interval, myDate), 108) timeGroup,
count(*)
from
#myDates
group by
convert(varchar(5), dateadd(minute,#interval - datepart(minute, myDate) % #interval, myDate), 108)
retuns:
timeGroup
--------- -----------
09:10 4
09:20 3
09:30 8

Answer 3

ooh, way too complicated all of that stuff.
Normalise to seconds, divide by your bucket interval, truncate and remultiply:
select sec_to_time(floor(time_to_sec(d)/300)*300), count(*)
from d
group by sec_to_time(floor(time_to_sec(d)/300)*300)
Using Ron Savage's data, I get
+----------+----------+
| i | count(*) |
+----------+----------+
| 09:00:00 | 1 |
| 09:05:00 | 3 |
| 09:10:00 | 1 |
| 09:15:00 | 1 |
| 09:20:00 | 6 |
| 09:25:00 | 2 |
| 09:30:00 | 1 |
+----------+----------+
You may wish to use ceil() or round() instead of floor().
Update: for a table created with
create table d (
d datetime
);

Answer 4

Create a table periods describing the periods you wish to divide the day up into.
SELECT periods.name, count(time)
FROM periods, times
WHERE period.start <= times.time
AND times.time < period.end
GROUP BY periods.name

Answer 5

Create a table containing what intervals you want to be getting totals at then join the two tables together.
Such as:
time_entry.time_entry
-----------------------
2008-10-02 09:01:00.000
2008-10-02 09:04:00.000
2008-10-02 09:11:00.000
2008-10-02 09:13:00.000
2008-10-02 09:22:00.000
2008-10-02 09:24:00.000
2008-10-02 09:28:00.000
time_interval.time_end
-----------------------
2008-10-02 09:05:00.000
2008-10-02 09:15:00.000
2008-10-02 09:25:00.000
2008-10-02 09:30:00.000
SELECT
ti.time_end,
COUNT(*) AS 'interval_total'
FROM time_interval ti
INNER JOIN time_entry te
ON te.time_entry < ti.time_end
GROUP BY ti.time_end;
time_end interval_total
----------------------- -------------
2008-10-02 09:05:00.000 2
2008-10-02 09:15:00.000 4
2008-10-02 09:25:00.000 6
2008-10-02 09:30:00.000 7
If instead of wanting cumulative totals you wanted totals within a range, then you add a time_start column to the time_interval table and change the query to
SELECT
ti.time_end,
COUNT(*) AS 'interval_total'
FROM time_interval ti
INNER JOIN time_entry te
ON te.time_entry >= ti.time_start
AND te.time_entry < ti.time_end
GROUP BY ti.time_end;

Answer 6

This uses quite a few SQL tricks (SQL Server 2005):
CREATE TABLE [dbo].[stackoverflow_165571](
[visit] [datetime] NOT NULL
) ON [PRIMARY]
GO
;WITH buckets AS (
SELECT dateadd(mi, (1 + datediff(mi, 0, visit - 1 - dateadd(dd, 0, datediff(dd, 0, visit))) / 5) * 5, 0) AS visit_bucket
,COUNT(*) AS visit_count
FROM stackoverflow_165571
GROUP BY dateadd(mi, (1 + datediff(mi, 0, visit - 1 - dateadd(dd, 0, datediff(dd, 0, visit))) / 5) * 5, 0)
)
SELECT LEFT(CONVERT(varchar, l.visit_bucket, 8), 5) + ' - ' + CONVERT(varchar, SUM(r.visit_count))
FROM buckets l
LEFT JOIN buckets r
ON r.visit_bucket <= l.visit_bucket
GROUP BY l.visit_bucket
ORDER BY l.visit_bucket
Note that it puts all the times on the same day, and assumes they are in a datetime column. The only thing it doesn't do as your example does is strip the leading zeroes from the time representation.