Find the longest sequence of consecutive increasing numbers in SQL - sql

Question

For this example say I have a table with two fields, AREA varchar(30) and OrderNumber INT.
The table has the following data
AREA | OrderNumber
Fontana | 32
Fontana | 42
Fontana | 76
Fontana | 12
Fontana | 3
Fontana | 99
RC | 32
RC | 1
RC | 8
RC | 9
RC | 4
I would like to return
The results I would like to return is for each area the longest length of increasing consecutive values. For Fontana it is 3 (32, 42, 76). For RC it is 2 (8,9)
AREA | LongestLength
Fontana | 3
RC | 2
How would I do this on MS Sql 2005?

Answer 1

One way is to use a recursive CTE that steps over each row. If the row meets the criteria (increasing order number for the same area), you increase the chain length by one. If it doesn't, you start a new chain:
; with numbered as
(
select row_number() over (order by area, eventtime) rn
, *
from Table1
)
, recurse as
(
select rn
, area
, OrderNumber
, 1 as ChainLength
from numbered
where rn = 1
union all
select cur.rn
, cur.area
, cur.OrderNumber
, case
when cur.area = prev.area
and cur.OrderNumber > prev.OrderNumber
then prev.ChainLength + 1
else 1
end
from recurse prev
join numbered cur
on prev.rn + 1 = cur.rn
)
select area
, max(ChainLength)
from recurse
group by
area
Live example at SQL Fiddle.
An alternative way is to use a query to find "breaks", that is, rows that end a sequence of increasing order numbers for the same area. The number of rows between breaks is the length.
; with numbered as
(
select row_number() over (order by area, eventtime) rn
, *
from Table1 t1
)
-- Select rows that break an increasing chain
, breaks as
(
select row_number() over (order by cur.rn) rn2
, cur.rn
, cur.Area
from numbered cur
left join
numbered prev
on cur.rn = prev.rn + 1
where cur.OrderNumber <= prev.OrderNumber
or cur.Area <> prev.Area
or prev.Area is null
)
-- Add a final break after the last row
, breaks2 as
(
select *
from breaks
union all
select count(*) + 1
, max(rn) + 1
, null
from breaks
)
select series_start.area
, max(series_end.rn - series_start.rn)
from breaks2 series_start
join breaks2 series_end
on series_end.rn2 = series_start.rn2 + 1
group by
series_start.area
Live example at SQL Fiddle.

Answer 2

You do not explain why RC's longest sequence does not include 1 while Fontana's does include 32. I take it, the 1 is excluded because it is a decrease: it comes after 32. The Fontana's 32, however, is the first ever item in the group, and I've got two ideas how to explain why it is considered an increase. That's either exactly because it's the group's first item or because it is also positive (i.e. as if coming after 0 and, therefore, an increase).
For the purpose of this answer, I'm assuming the latter, i.e. a group's first item is an increase if it is positive. The below script implements the following idea:
Enumerate the rows in every AREA group in the order of the eventtime column you nearly forgot to mention.
Join the enumerated set to itself to link every row with it's predecessor.
Get the sign of the difference between the row and its preceding value (defaulting the latter to 0). At this point the problem turns into a gaps-and-islands one.
Partition every AREA group by the signs determined in #3 and enumerate every subgroup's rows.
Find the difference between the row numbers from #1 and those found in #4. That would be a criterion to identify individual streaks (together with AREA).
Finally, group the results by AREA, the sign from #3 and the result from #5, count the rows and get the maximum count per AREA.
I implemented the above like this:
WITH enumerated AS (
SELECT
*,
row = ROW_NUMBER() OVER (PARTITION BY AREA ORDER BY eventtime)
FROM atable
),
signed AS (
SELECT
this.eventtime,
this.AREA,
this.row,
sgn = SIGN(this.OrderNumber - COALESCE(last.OrderNumber, 0))
FROM enumerated AS this
LEFT JOIN enumerated AS last
ON this.AREA = last.AREA
AND this.row = last.row + 1
),
partitioned AS (
SELECT
AREA,
sgn,
grp = row - ROW_NUMBER() OVER (PARTITION BY AREA, sgn ORDER BY eventtime)
FROM signed
)
SELECT DISTINCT
AREA,
LongestIncSeq = MAX(COUNT(*)) OVER (PARTITION BY AREA)
FROM partitioned
WHERE sgn = 1
GROUP BY
AREA,
grp
;
A SQL Fiddle demo can be found here.

Answer 3

You can do some math by ROW_NUMBER() to figure out where you have consecutive items.
Here's the code sample:
;WITH rownums AS
(
SELECT [area],
ROW_NUMBER() OVER(PARTITION BY [area] ORDER BY [ordernumber]) AS rid1,
ROW_NUMBER() OVER(PARTITION BY [area] ORDER BY [eventtime]) AS rid2
FROM SomeTable
),
differences AS
(
SELECT [area],
[calc] = rid1 - rid2
FROM rownums
),
summation AS
(
SELECT [area], [calc], COUNT(*) AS lengths
FROM differences
GROUP BY [area], [calc]
)
SELECT [area], MAX(lengths) AS LongestLength
FROM differences
JOIN summation
ON differences.[calc] = summation.[calc]
AND differences.area = calc.area
GROUP BY [area]
So if I do one set of row numbers ordered by my ordernumber and another set of row numbers by my event time, the difference between those two numbers will always be the same, so long as their order is the same.
You can then get a count grouped by those differences and then pull the largest count to get what you need.
EDIT: ...
Ignore the first edit, what I get for rushing.