find pairs of NULL records without resorting to CURSORS

Question

find pairs of NULL records without resorting to CURSORS

aaron.reese

SSChampion

Points: 13498
More actions
May 7, 2013 at 5:43 am

#293108

create table #tmp
(
ref int,
Date datetime,
time int,
seqno int
)
insert into #tmp (ref,date,time,seqno) values(34227664,'20130227',0,399850)
insert into #tmp (ref,date,time,seqno) values(34227664,'20130304',0,399860)
insert into #tmp (ref,date,time,seqno) values(34227664,'20130313',0,399870)
insert into #tmp (ref,date,time,seqno) values(34227664,'20130326',40820,NULL)
insert into #tmp (ref,date,time,seqno) values(34227664,'20130326',41201,NULL)
insert into #tmp (ref,date,time,seqno) values(34227664,'20130326',41287,NULL)
insert into #tmp (ref,date,time,seqno) values(34227664,'20130326',44797,NULL)
insert into #tmp (ref,date,time,seqno) values(34227664,'20130326',44809,5158850)
insert into #tmp (ref,date,time,seqno) values(34227664,'20130326',44820,NULL)
insert into #tmp (ref,date,time,seqno) values(34227664,'20130326',46147,5158870)
insert into #tmp (ref,date,time,seqno) values(34227664,'20130328',53698,NULL)
insert into #tmp (ref,date,time,seqno) values(34227664,'20130328',60518,NULL)
SELECT * from #tmp order by date,time,seqno
I am trying to identify the 2nd record in pairs of NULL sequence nos. E.g.
'20130326',41201 is the 2nd in a pair,
'20130326',44797 is the 2nd in a pair (although they come as a block of 4)
'20130328',60518 is the 2nd in a pair.
'20130326',44820 is not part of a pair so I am not interested in it.
So far I have tried:
Rank partition by seqno order by ref,date,time and this gives me 1 for each record which has a seqno and an incrementing counter for each that has null.
mod (rnak % 2) to give me 0 or 1 and take the 0 as being the 2nd in the pair but it goes wrong for '20130326',44820 because this is in the NULL ranks but is not paired and it throws the last pair out.
Also tried referencing the data to itself using rank = rank -1 in the join but this doesnt help unless we also go for min() and max() to create islands of records which is hidden RBAR and very slow on 1M records.
anyone got any bright ideas?

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ChrisM@Work SSC Guru Points: 186127 More actions · Answer 1

Why isn't 41287 the second of a pair?

^{“Write the query the simplest way. If through testing it becomes clear that the performance is inadequate, consider alternative query forms.” - Gail Shaw}

For fast, accurate and documented assistance in answering your questions, please read this article.
Understanding and using APPLY, (I) and (II) Paul White
Hidden RBAR: Triangular Joins / The "Numbers" or "Tally" Table: What it is and how it replaces a loop Jeff Moden

Lowell SSC Guru Points: 323518 More actions · Answer 2

my first guess looks pretty good;

just using row number, and joining the data against itself:

WITH MySampleData

AS

(

SELECT row_number() OVER(partition by date order by date,time,seqno) AS RW,

*

from #tmp

)

SELECT

T1.*,

T2.*

FROM MySampleData T1

LEFT OUTER JOIN MySampleData T2 ON T1.RW + 1 = T2.RW

WHERE T1.seqno IS NULL AND T2.seqno IS NULL

Lowell

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aaron.reese SSChampion Points: 13498 More actions · Answer 3

@lowell Good effort but not quite.

Your results set picks up the recod with a time stamp of 41287 which is the first in a pair.

@chrism-2

Same answer as above: a pair is defined as a discreet pair; i.e. if you are the second in a pair, you cannot be the first in the next pair even when the records are consecutive.

so

40820 and 41201 are a pair

41287 and 44797 are a pair

53698 and 60518 are a pair

ChrisM@Work SSC Guru Points: 186127 More actions · Answer 4

aaron.reese (5/7/2013)
@Lowell Good effort but not quite.
Your results set picks up the recod with a time stamp of 41287 which is the first in a pair.
@chrism-2
Same answer as above: a pair is defined as a discreet pair; i.e. if you are the second in a pair, you cannot be the first in the next pair even when the records are consecutive.
so
40820 and 41201 are a pair
41287 and 44797 are a pair
53698 and 60518 are a pair

Thanks. Try this:

SELECT *,

Filter = CASE WHEN seqno IS NULL

THEN ROW_NUMBER() OVER (PARTITION BY rnThingy ORDER BY date, time, seqno)

ELSE NULL END%2

FROM (

SELECT *,

rnThingy = CASE WHEN seqno IS NULL

THEN (ROW_NUMBER() OVER (ORDER BY date, time, seqno) - ROW_NUMBER() OVER (ORDER BY seqno, date, time))

ELSE NULL END

FROM #tmp

) d

ORDER BY date, time, seqno

^{“Write the query the simplest way. If through testing it becomes clear that the performance is inadequate, consider alternative query forms.” - Gail Shaw}

For fast, accurate and documented assistance in answering your questions, please read this article.
Understanding and using APPLY, (I) and (II) Paul White
Hidden RBAR: Triangular Joins / The "Numbers" or "Tally" Table: What it is and how it replaces a loop Jeff Moden

J Livingston SQL SSC Guru Points: 51550 More actions · Answer 5

;WITH jls

AS

(

SELECT ROW_NUMBER ( ) OVER ( PARTITION BY date ORDER BY date ) AS RN

, *

FROM #tmp

WHERE (seqno IS NULL)

)

select * from jls where RN %2 = 0

________________________________________________________________
you can lead a user to data....but you cannot make them think
and remember....every day is a school day

aaron.reese SSChampion Points: 13498 More actions · Answer 6

@JLS

Thanks but I needed to be more precise over the ordering than just using date.

@chrism-2

Genius!

Definately is giving me the right results for the #tmp sample data set. As you would expect the 'real' data is much more complicated. I will run against this later and see what the performance is like and whether any other anomolies crawl out of the dataset.

Obiron

aaron.reese SSChampion Points: 13498 More actions · Answer 7

@chrism-2

I managed to run that against my entire dataset (some 800K records) and it took less than 30 seconds so a big thanks for that code.

Obiron

ChrisM@Work SSC Guru Points: 186127 More actions · Answer 8

aaron.reese (5/8/2013)
@ChrisM
I managed to run that against my entire dataset (some 800K records) and it took less than 30 seconds so a big thanks for that code.
Obiron

That's a result! Thank you very much for the feedback 😀

^{“Write the query the simplest way. If through testing it becomes clear that the performance is inadequate, consider alternative query forms.” - Gail Shaw}

For fast, accurate and documented assistance in answering your questions, please read this article.
Understanding and using APPLY, (I) and (II) Paul White
Hidden RBAR: Triangular Joins / The "Numbers" or "Tally" Table: What it is and how it replaces a loop Jeff Moden