views:

2444

answers:

6

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#

+1  A: 

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
ephemient
This would not give the cumulative totals that was asked for, dropping the "period.start <= times.time" statement would achieve this, as my answer showed.
ManiacZX
True. I misread the question.
ephemient
+3  A: 

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

Ron

Ron Savage
not cumulative...
KristoferA - Huagati.com
Yep, missed that part. :-) Ron
Ron Savage
...but the sentence 'You have to grok to the "intent" of the user requirement, not the literal "reading" of it.' almost deserves an upvote... :)
KristoferA - Huagati.com
A: 

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.

Cade Roux
+1  A: 

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;
ManiacZX
+6  A: 
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
KristoferA - Huagati.com
+1  A: 

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
);
dland
Yes, but what about those periods which had no entries? If there are no entries in a specific 5-minute time frame, I'd like to see a 0 count there. How can you achieve that?
dubek
You need a table with the desired time intervals over a day. Left join it with the sec_to_time column above. You obtain the counts with coalesce(count(*), 0). I think @ManiacZX's suggestion does this.
dland