How to Find the Longest Consecutive Series of Events in SQL
lukaseder
A very interesting problem that can be solved very easily with SQL is to find consecutive series of events in a time series. But what is a consecutive series of events in a time series?
Take Stack Overflow, for example. Stack Overflow has a cool reputation system that uses badges to reward certain behaviour. As a social website, they encourage users to visit the platform every day. As such, two distinct badges are awarded:
Informally, it is obvious what this means. You’ll have to log in on day 1. Then again on day 2. Then again (perhaps several times, it doesn’t matter) on day 3. Forgot to log in on day 4? Ooops. We’ll start counting again.
How to do this in SQL?
On this blog, every problem will find its solution in SQL. So does this. And in order to solve this problem, we’re going to use the awesome Stack Exchange Data Explorer, which exposes a lot of Stack Exchange’s publicly available usage information.
Note that we won’t query the consecutive days of visits, as this information is not made available publicly. Instead, let’s query the consecutive days of posts a user has made.
The backing database is SQL Server, so we can run the following statement:
SELECT DISTINCT CAST(CreationDate AS DATE) AS date
FROM Posts
WHERE OwnerUserId = ##UserId##
ORDER BY 1
… which, for my own UserId generates something like:
date
--------------------------------------
2010-11-26
2010-11-27 <---- Gap here after 2 days
2010-11-29
2010-11-30
2010-12-01
2010-12-02
2010-12-03 <---- Gap here after 5 days
2010-12-05
2010-12-06
2010-12-07
2010-12-08
2010-12-09 <---- Gap here after 5 days
2010-12-13
2010-12-14
...
Visually, it is very easy to see how many days in a row there were posts without any gaps. But how to do it with SQL?
To simplify the problem, let’s “store” individual queries in common table expressions. The above query, we’ll call dates:
WITH
-- This table contains all the distinct date
-- instances in the data set
dates(date) AS (
SELECT DISTINCT CAST(CreationDate AS DATE)
FROM Posts
WHERE OwnerUserId = ##UserId##
)
...
Now, the goal of the resulting query is to put all consecutive dates in the same group, such that we can aggregate over this group. The following query is what we want to write:
SELECT
COUNT(*) AS consecutiveDates,
MIN(date) AS minDate,
MAX(date) AS maxDate
FROM groups
GROUP BY grp -- This "grp" value will be explained later
ORDER BY 1 DESC, 2 DESC
We’d like to aggregate each group “grp” and count the number of dates in the group, as well as find the lowest and the highest date within each group.
Generating groups for consecutive dates
Let’s look at the data again, and to illustrate the idea, we’ll add consecutive row numbers, regardless of the gaps in dates:
row number date
--------------------------------
1 2010-11-26
2 2010-11-27
3 2010-11-29 <-- gap before this row
4 2010-11-30
5 2010-12-01
6 2010-12-02
7 2010-12-03
8 2010-12-05 <-- gap before this row
As you can see, regardless whether there is a gap between dates (two dates are not consecutive), their row numbers will still be consecutive. We can do this with the ROW_NUMBER() window function, very easily:
SELECT
ROW_NUMBER() OVER (ORDER BY date) AS [row number],
date
FROM dates
Now, let’s check out the following, interesting query:
WITH
-- This table contains all the distinct date
-- instances in the data set
dates(date) AS (
SELECT DISTINCT CAST(CreationDate AS DATE)
FROM Posts
WHERE OwnerUserId = ##UserId##
),
-- Generate "groups" of dates by subtracting the
-- date's row number (no gaps) from the date itself
-- (with potential gaps). Whenever there is a gap,
-- there will be a new group
groups AS (
SELECT
ROW_NUMBER() OVER (ORDER BY date) AS rn,
dateadd(day, -ROW_NUMBER() OVER (ORDER BY date), date) AS grp,
date
FROM dates
)
SELECT *
FROM groups
ORDER BY rn
(run the statement yourself, here)
All we did is subtract the row number from the date to get a new date “grp“. The actual date obtained this way is irrelevant. It’s just an auxiliary value.
What we can guarantee, though, is that for consecutive dates, the value of grp will be the same because for all consecutive dates, the following two equations yield true:
date2 - date1 = 1 // difference in days between dates
rn2 - rn1 = 1 // difference in row numbers
Yet, for non-consecutive dates, while the difference in row numbers is still 1, the difference in days is no longer 1. The groups can now be seen easily:
WITH
-- This table contains all the distinct date
-- instances in the data set
dates(date) AS (
SELECT DISTINCT CAST(CreationDate AS DATE)
FROM Posts
WHERE OwnerUserId = ##UserId##
),
-- Generate "groups" of dates by subtracting the
-- date's row number (no gaps) from the date itself
-- (with potential gaps). Whenever there is a gap,
-- there will be a new group
groups AS (
SELECT
ROW_NUMBER() OVER (ORDER BY date) AS rn,
dateadd(day, -ROW_NUMBER() OVER (ORDER BY date), date) AS grp,
date
FROM dates
)
SELECT
COUNT(*) AS consecutiveDates,
MIN(date) AS minDate,
MAX(date) AS maxDate
FROM groups
GROUP BY grp
ORDER BY 1 DESC, 2 DESC
The fact that we chose the granularity of days in the above query is a random choice. We simply took the timestamp from our time series and “collapsed” it to the desired granularity using a CAST function:
SELECT DISTINCT CAST(CreationDate AS DATE)
If we want to know the consecutive weeks, we’ll simply change that function to a different expression, e.g.
This new expression takes the year and the week and generates values like 201503 for week 03 in the year 2015. The rest of the statement remains exactly the same:
WITH
weeks(week) AS (
SELECT DISTINCT datepart(year, CreationDate) * 100
+ datepart(week, CreationDate)
FROM Posts
WHERE OwnerUserId = ##UserId##
),
groups AS (
SELECT
ROW_NUMBER() OVER (ORDER BY week) AS rn,
dateadd(day, -ROW_NUMBER() OVER (ORDER BY week), week) AS grp,
week
FROM weeks
)
SELECT
COUNT(*) AS consecutiveWeeks,
MIN(week) AS minWeek,
MAX(week) AS maxWeek
FROM groups
GROUP BY grp
ORDER BY 1 DESC, 2 DESC
The above has been one very useful example of using window functions (ROW_NUMBER()) in SQL. Learn more about window functions in any of the following articles:
