SQL- Calculating average of differences between times - sql-server

Question

I have an sql table that has transaction history of all the clients. I want to find what is the average difference in time between two transactions.
ClientCode Date
DL2xxx 2016-04-18 00:00:00.000
DL2xxx 2016-04-18 00:00:00.000
E19xxx 2016-04-18 00:00:00.000
E19xxx 2016-04-18 00:00:00.000
E19xxx 2016-04-18 00:00:00.000
JDZxxx 2016-04-18 00:00:00.000
Given above are the first few lines of the table the date given is the date transaction happened. I want to take an average of difference in days when successive transactions happen. Say for a client he makes transactions of Day 1, Day 3, Day 10, and Day 15. So differences are {2, 7, 5} average of which is 4.66. If only one transaction takes place this should be 0.
ClientCode AverageDays
DL2xxx <float_value>
DL2xxx <float_value>
E19xxx <float_value>
This is what the output should look like where each unique client code occurs only once.

Answer 1

You can use a query like below if you table name is T
see live demo
select
ClientCode,
AvgDays =ISNULL(AVG(d),0)
from
(
select
*,
d=DATEDIFF(
d,
dateofT,
LEAD(DateofT) over(
partition by ClientCode
order by DateofT asc ))
from t
)t
group by ClientCode

Answer 2

If Windowing functions aren't available to you, here's an alternative
--CREATE SAMPLE DATA
CREATE TABLE #TMP(ClientID INT, EventDate DATE)
GO
INSERT INTO #TMP VALUES
(1,DATEADD(DD,RAND()*365,'20180101'))
,(2,DATEADD(DD,RAND()*365,'20180101'))
,(3,DATEADD(DD,RAND()*365,'20180101'))
,(4,DATEADD(DD,RAND()*365,'20180101'))
,(5,DATEADD(DD,RAND()*365,'20180101'))
GO 50
--PRE SQL 2012 Compatible
SELECT A.ClientID
,AVG(DATEDIFF(DD,C.EventDate,A.Eventdate)) AS ClientAvg
FROM #TMP A
CROSS APPLY (SELECT ClientID, MAX(EventDate) EventDate FROM #TMP B
WHERE A.ClientID = B.ClientID AND A.EventDate > B.EventDate
GROUP BY ClientID) C
GROUP BY A.ClientID
ORDER BY A.ClientID

Answer 3

You can use LAG() function to compare a date to it's previous date by client, then group by client and calculate the average.
IF OBJECT_ID('tempdb..#Transactions') IS NOT NULL
DROP TABLE #Transactions
CREATE TABLE #Transactions (
ClientCode VARCHAR(100),
Date DATE)
INSERT INTO #Transactions (
ClientCode,
Date)
VALUES
('DL2', '2016-04-18'),
('DL2', '2016-04-19'),
('DL2', '2016-04-26'),
('E19', '2016-01-01'),
('E19', '2016-01-11'),
('E19', '2016-01-12')
;WITH DayDifferences AS
(
SELECT
T.ClientCode,
T.Date,
DayDifference = DATEDIFF(
DAY,
LAG(T.Date) OVER (PARTITION BY T.ClientCode ORDER BY T.Date ASC),
T.Date)
FROM
#Transactions AS T
)
SELECT
D.ClientCode,
AverageDayDifference = AVG(ISNULL(CONVERT(FLOAT, D.DayDifference), 0))
FROM
DayDifferences AS D
GROUP BY
D.ClientCode

Answer 4

Using the observation that the sum of differences within a group is simply the max - min of that group, you can use the simple group by select:
select IIF(COUNT(*) > 1,
(CAST(DATEDIFF(day, MIN(DateofT), MAX(DateofT)) AS FLOAT)) / (COUNT(*) - 1), 0.0)
AS AVGDays, ClientCode
FROM t GROUP BY ClientCode