SQL find average time difference between rows for a given category - sql

Question

I browsed SO but could not quite find the exact answer or maybe it was for a different language.
Let's say I have a table, where each row is a record of a trade:
trade_id customer trade_date
1 A 2013-05-01 00:00:00
2 B 2013-05-01 10:00:00
3 A 2013-05-02 00:00:00
4 A 2013-05-05 00:00:00
5 B 2013-05-06 12:00:00
I would like to have the average time between trades, in days or fraction of days, for each customer, and the number of days since last trade. So for instance for customer A, time between trades 1 and 3 is 1 day and between trades 3 and 4 is 3 days, for an average of 2. So the end table would look like something like this (assuming today it's the 2013-05-10):
customer avg_time_btw_trades time_since_last_trade
A 2.0 5.0
B 5.08 3.5
If a customer has only got 1 trade I guess NULL is fine as output.
Not even sure SQL is the best way to do this (I am working with SQL server), but any help is appreciated!

Answer 1

SELECT
customer,
DATEDIFF(second, MIN(trade_date), MAX(trade_date)) / (NULLIF(COUNT(*), 1) - 1) / 86400.0,
DATEDIFF(second, MAX(trade_date), GETDATE() ) / 86400.0
FROM
yourTable
GROUP BY
customer
http://sqlfiddle.com/#!6/eb46e/7
EDIT: Added final field that I didn't notice, apologies.

Answer 2

The following SQL script uses your data and gives the expected results.
DECLARE #temp TABLE
( trade_id INT,
customer CHAR(1),
trade_date DATETIME );
INSERT INTO #temp VALUES (1, 'A', '20130501');
INSERT INTO #temp VALUES (2, 'B', '20130501 10:00');
INSERT INTO #temp VALUES (3, 'A', '20130502');
INSERT INTO #temp VALUES (4, 'A', '20130505');
INSERT INTO #temp VALUES (5, 'B', '20130506 12:00');
DECLARE #getdate DATETIME
-- SET #getdate = getdate();
SET #getdate = '20130510';
SELECT s.customer
, AVG(s.days_btw_trades) AS avg_time_between_trades
, CAST(DATEDIFF(hour, MAX(s.trade_date), #getdate) AS float)
/ 24.0 AS time_since_last_trade
FROM (
SELECT CAST(DATEDIFF(HOUR, t2.trade_date, t.trade_date) AS float)
/ 24.0 AS days_btw_trades
, t.customer
, t.trade_date
FROM #temp t
LEFT JOIN #temp t2 ON t2.customer = t.customer
AND t2.trade_date = ( SELECT MAX(t3.trade_date)
FROM #temp t3
WHERE t3.customer = t.customer
AND t3.trade_date < t.trade_date)
) s
GROUP BY s.customer

Answer 3

You need a date difference between every trade and average them.
select
a.customer
,avg(datediff(a.trade_date, b.trade_date))
,datediff(now(),max(a.trade_date))
from yourTable a, yourTable b
where a.customer = b.customer
and b.trade_date = (
select max(trade_date)
from yourTable c
where c.customer = a.customer
and a.trade_date > c.trade_date)
#gets the one earlier date for every trade
group by a.customer

Answer 4

Just for grins I added a solution that would use CTE's. You could probably use a temp table if the first query is too large. I used #MatBailie creation script for the table:
CREATE TABLE customer_trades (
id INT IDENTITY(1,1),
customer_id INT,
trade_date DATETIME,
PRIMARY KEY (id),
INDEX ix_user_trades (customer_id, trade_date)
)
INSERT INTO
customer_trades (
customer_id,
trade_date
)
VALUES
(1, '2013-05-01 00:00:00'),
(2, '2013-05-01 10:00:00'),
(1, '2013-05-02 00:00:00'),
(1, '2013-05-05 00:00:00'),
(2, '2013-05-06 12:00:00')
;
;WITH CTE as(
select customer_id, trade_date, datediff(hour,trade_date,ISNULL(LEAD(trade_date,1) over (partition by customer_id order by trade_date),GETDATE())) Trade_diff
from customer_trades
)
, CTE2 as
(SELECT customer_id, trade_diff, LAST_VALUE(trade_diff) OVER(Partition by customer_id order by trade_date) Curr_Trade from CTE)
SELECT Customer_id, AVG(trade_diff) AV, Max(Curr_Trade) Curr_Trade
FROM CTE2
GROUP BY customer_id