How to Find Patterns in Your Data with SQL

Copyright © 2017, Oracle and/or its affiliates. All rights reserved. |
How to Find Patterns in Your Data with SQL
Chris Saxon, @ChrisRSaxon & @SQLDaily
blogs.oracle.com/sql
youtube.com/c/TheMagicofSQL
asktom.oracle.com

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This presentation contains <regular expressions>!

Copyright © 2017, Oracle and/or its affiliates. All rights reserved. | blogs.oracle.com/sql www.youtube.com/c/TheMagicOfSQL @ChrisRSaxon
I thought
this was
about SQL!
Ryan McGuire / Gratisography

* => zero or more matches
+ => one or more matches
{n,m} => N through M matches
(either optional)

Am I Improving?
Can Beat My PB?
Am I Training Regularly?

Am I running
every day?

RUN_DATE TIME_IN_S DISTANCE_IN_MILES
01 Jan 2018 420 1
02 Jan 2018 2,400 5
03 Jan 2018 4,932 10
06 Jan 2018 2,350 5
07 Jan 2018 410 1
10 Jan 2018 400 1
13 Jan 2018 2,300 5
14 Jan 2018 425 1
15 Jan 2018 422 1

01 Jan 2018 420 1
02 Jan 2018 2,400 5
03 Jan 2018 4,932 10
06 Jan 2018 2,350 5
07 Jan 2018 410 1
10 Jan 2018 400 1
13 Jan 2018 2,300 5
14 Jan 2018 425 1
15 Jan 2018 422 1
#1
#3
#2
#4

How I know if rows are consecutive?

current value = previous value + 1

lag ( run_date ) over
( order by run_date )
Get the previous row's date

RUN_DATE RN TIME_IN_S DISTANCE_IN_MILES
01 Jan 2018 1 420 1
02 Jan 2018 2 2,400 5
03 Jan 2018 3 4,932 10
06 Jan 2018 4 2,350 5
07 Jan 2018 5 410 1
10 Jan 2018 6 400 1
13 Jan 2018 7 2,300 5
14 Jan 2018 8 425 1
15 Jan 2018 9 422 1
consecutive =>
constant gap

RUN_DATE RN TIME_IN_S DISTANCE_IN_MILES
01 Jan 2018 1 420 1
02 Jan 2018 2 2,400 5
03 Jan 2018 3 4,932 10
06 Jan 2018 4 2,350 5
07 Jan 2018 5 410 1
10 Jan 2018 6 400 1
13 Jan 2018 7 2,300 5
14 Jan 2018 8 425 1
15 Jan 2018 9 422 1
-
-
-
-
-
-
-
-
-

RUN_DATE RN RUN_DATE - RN TIME_IN_S DISTANCE_IN_MILES
01 Jan 2018 1 31 Dec 2017 420 1
02 Jan 2018 2 31 Dec 2017 2,400 5
03 Jan 2018 3 31 Dec 2017 4,932 10
06 Jan 2018 4 02 Jan 2018 2,350 5
07 Jan 2018 5 02 Jan 2018 410 1
10 Jan 2018 6 04 Jan 2018 400 1
13 Jan 2018 7 06 Jan 2018 2,300 5
14 Jan 2018 8 06 Jan 2018 425 1
15 Jan 2018 9 06 Jan 2018 422 1
-
-
-
-
-
-
-
-
-

Tabibitosan Method

row_number ()
over ( order by run_date )

run_date -
row_number ()
over ( order by run_date ) grp

with grps as (
select run_date ,
run_date -
row_number ()
over ( order by run_date ) grp
from running_log r
)
select min ( run_date ), count (*)
from grps
group by grp

12c Pattern Matching

01 Jan 2018 420 1
02 Jan 2018 2,400 5
03 Jan 2018 4,932 10
06 Jan 2018 2,350 5
07 Jan 2018 410 1
10 Jan 2018 400 1
13 Jan 2018 2,300 5
14 Jan 2018 425 1
15 Jan 2018 422 1
this = prev + 1

01 Jan 2018 420 1
02 Jan 2018 2,400 5
03 Jan 2018 4,932 10
06 Jan 2018 2,350 5
07 Jan 2018 410 1
10 Jan 2018 400 1
13 Jan 2018 2,300 5
14 Jan 2018 425 1
15 Jan 2018 422 1
this = prev + 1
this = prev + 3

01 Jan 2018 420 1
02 Jan 2018 2,400 5
03 Jan 2018 4,932 10
06 Jan 2018 2,350 5
07 Jan 2018 410 1
10 Jan 2018 400 1
13 Jan 2018 2,300 5
14 Jan 2018 425 1
15 Jan 2018 422 1
this = prev + 1
this = prev + 3
this ≠ prev + 1

define
consecutive as
run_date = prev ( run_date ) + 1

pattern ( strt consecutive* )
define
consecutive as

define
consecutive as
"Always true"
> 0 matches
Any row <>

RUN_DATE VARIABLE TIME_IN_S DISTANCE_IN_MILES
01 Jan 2018 STRT 420 1
02 Jan 2018 CONSECUTIVE 2,400 5
03 Jan 2018 CONSECUTIVE 4,932 10
06 Jan 2018 STRT 2,350 5
07 Jan 2018 CONSECUTIVE 410 1
10 Jan 2018 STRT 400 1
13 Jan 2018 STRT 2,300 5

define
consecutive as
Which row is prev?!

order by run_date
define
consecutive as

match_recognize (
order by run_date
measures
first ( run_date ) as start_date,
count (*) as days
define
consecutive as
);
How many consecutive rows?
From first row in group

START_DATE DAYS
01 Jan 2018 3
06 Jan 2018 2
10 Jan 2018 1
13 Jan 2018 3

So which is
better?
Pixabay
pattern
matching

Performance ~ similar
Availability MR 12c vs 8i*
Readability personal pref
Match_recognize vs. Tabibitosan

Am I running >= 3
times/week?
Pixabay

How I know if runs are in the same week?

last Monday = prev last Monday

trunc ( run_date , 'iw' )
Return the start of the ISO week…
…Monday!

RUN_DATE TRUNC(RUN_DATE, 'IW') TIME_IN_S DISTANCE_IN_MILES
01 Jan 2018 01 Jan 2018 420 1
02 Jan 2018 01 Jan 2018 2,400 5
03 Jan 2018 01 Jan 2018 4,932 10
06 Jan 2018 01 Jan 2018 2,350 5
07 Jan 2018 01 Jan 2018 410 1
10 Jan 2018 08 Jan 2018 400 1
13 Jan 2018 08 Jan 2018 2,300 5
14 Jan 2018 08 Jan 2018 425 1
15 Jan 2018 15 Jan 2018 422 1

select trunc ( run_date , 'iw' ),
count(*)
from running_log
group by trunc ( run_date , 'iw' )

select trunc ( run_date , 'iw' ),
count(*)
from running_log
group by trunc ( run_date , 'iw' )
having count (*) >= 3

define
same_week as
trunc ( run_date, 'iw' ) =
prev ( trunc ( run_date, 'iw' ) )

match_recognize (
order by run_date
measures
count (*) as days
pattern ( strt same_week* )
define
same_week as
);

match_recognize (
order by run_date
measures
count (*) as days
pattern ( strt same_week {2, } )
define
same_week as
);
Two or more matches

match_recognize (
order by run_date
measures
count (*) as days
define
same_week as
);

match_recognize (
order by run_date
measures
count (*) as days
define
consecutive as
);

How I can I find consecutive weeks
with >= 3 runs?

Find weeks >= 3 runs
then
check these are consecutive!

with min_three_runs as (
select * from running_log
match_recognize (
order by run_date
measures
first ( trunc ( run_date, 'iw' ) )
as week_start,
count(*) as days
define
same_week as
)

How I can I find consecutive weeks?

current Monday = prev Monday + 7

WEEK_START RN WEEK_START – RN RUNS
01 Jan 2018 1 31 Dec 2017 5
08 Jan 2018 2 06 Jan 2018 3
15 Jan 2018 3 12 Jan 2018 4
05 Feb 2018 4 01 Feb 2018 3
12 Feb 2018 5 07 Feb 2018 3
05 Mar 2018 6 28 Feb 2018 3
12 Mar 2018 7 07 Mar 2018 4

WEEK_START RN * 7 RUNS
01 Jan 2018 7 5
08 Jan 2018 14 3
15 Jan 2018 21 4
05 Feb 2018 28 3
12 Feb 2018 35 3
05 Mar 2018 42 3
12 Mar 2018 49 4

WEEK_START RN * 7 WEEK_START – ( RN * 7 ) RUNS
01 Jan 2018 7 25 Dec 2017 5
08 Jan 2018 14 25 Dec 2017 3
15 Jan 2018 21 25 Dec 2017 4
05 Feb 2018 28 08 Jan 2018 3
12 Feb 2018 35 08 Jan 2018 3
05 Mar 2018 42 22 Jan 2018 3
12 Mar 2018 49 22 Jan 2018 4

select run_date - ( 7 *
row_number()
) grp
from min_three_runs

match_recognize (
order by week_start
measures
first ( week_start ) as start_date,
count (*) as weeks
pattern ( strt consecutive_weeks* )
define
consecutive_weeks as
week_start = prev ( week_start ) + 7
);

Pixabay
Am I running >= 3
times in 7 days?

current day < first day + 7

01 Jan 2018 420 1
02 Jan 2018 2,400 5
03 Jan 2018 4,932 10
06 Jan 2018 2,350 5
07 Jan 2018 410 1
10 Jan 2018 400 1
13 Jan 2018 2,300 5
14 Jan 2018 425 1
15 Jan 2018 422 1
#1
#2

01 Jan 2018 01 – 07 Jan 2018 420 1
02 Jan 2018 2,400 5
03 Jan 2018 4,932 10
06 Jan 2018 2,350 5
07 Jan 2018 410 1
10 Jan 2018 08 – 14 Jan 2018 400 1
13 Jan 2018 2,300 5
14 Jan 2018 425 1
15 Jan 2018 15 – 21 Jan 2018 422 1

01 Jan 2018 01 – 07 Jan 2018 420 1
02 Jan 2018 2,400 5
03 Jan 2018 4,932 10
06 Jan 2018 2,350 5
07 Jan 2018 410 1
10 Jan 2018 10 – 17 Jan 2018 400 1
13 Jan 2018 2,300 5
14 Jan 2018 425 1
15 Jan 2018 422 1

select r.*, min ( run_date ) over (
order by run_date
) mn_date
from running_log r

RUN_DATE MN_DATE TIME_IN_S DISTANCE_IN_MILES
01 Jan 2018 01 Jan 2018 420 1
02 Jan 2018 01 Jan 2018 2,400 5
03 Jan 2018 01 Jan 2018 4,932 10
06 Jan 2018 01 Jan 2018 2,350 5
07 Jan 2018 01 Jan 2018 410 1
10 Jan 2018 01 Jan 2018 400 1
13 Jan 2018 01 Jan 2018 2,300 5
14 Jan 2018 01 Jan 2018 425 1
15 Jan 2018 01 Jan 2018 422 1

Windowing clause to the rescue…

select r.*, min ( run_date ) over (
order by run_date
range between 7 preceding
and current row
) mn_date
from running_log r
Consider rows with values
in the previous 7 days

01 Jan 2018 01 Jan 2018 420 1
02 Jan 2018 01 Jan 2018 2,400 5
03 Jan 2018 01 Jan 2018 4,932 10
06 Jan 2018 01 Jan 2018 2,350 5
07 Jan 2018 01 Jan 2018 410 1
10 Jan 2018 400 1
13 Jan 2018 2,300 5
14 Jan 2018 425 1
15 Jan 2018 422 1

01 Jan 2018 01 Jan 2018 420 1
02 Jan 2018 01 Jan 2018 2,400 5
03 Jan 2018 01 Jan 2018 4,932 10
06 Jan 2018 01 Jan 2018 2,350 5
07 Jan 2018 01 Jan 2018 410 1
10 Jan 2018 03 Jan 2018 400 1
13 Jan 2018 2,300 5
14 Jan 2018 425 1
15 Jan 2018 422 1

01 Jan 2018 01 Jan 2018 420 1
02 Jan 2018 01 Jan 2018 2,400 5
03 Jan 2018 01 Jan 2018 4,932 10
06 Jan 2018 01 Jan 2018 2,350 5
07 Jan 2018 01 Jan 2018 410 1
10 Jan 2018 03 Jan 2018 400 1
13 Jan 2018 06 Jan 2018 2,300 5
14 Jan 2018 07 Jan 2018 425 1
15 Jan 2018 10 Jan 2018 422 1

Windowing clause to the rescue!

11.2 Recursive With

with rws as (
select r.*, row_number() over ( order by run_date ) rn
from running_log r
)

with rws as (
from running_log r
), within_7 (
run_date, time_in_s, distance_in_miles, rn, grp_start
) as (
select run_date, time_in_s, distance_in_miles, rn, run_date grp_start
from rws
where rn = 1

with rws as (
from running_log r
), within_7 (
) as (
from rws
where rn = 1
union all
select r.run_date, r.time_in_s, r.distance_in_miles, r.rn,
from within_7 w
join rws r
on w.rn + 1 = r.rn
)

with rws as (
from running_log r
), within_7 (
) as (
from rws
where rn = 1
union all
case
when r.run_date < w.grp_start + 7 then grp_start
else r.run_date
end grp_start
from within_7 w
join rws r
on w.rn + 1 = r.rn
)

with rws as (
from running_log r
), within_7 (
) as (
from rws
where rn = 1
union all
case
when r.run_date < w.grp_start + 7 then grp_start
else r.run_date
end grp_start
from within_7 w
join rws r
on w.rn + 1 = r.rn
)
select grp, w.*
from within_7 w

10g Model

select * from running_log
model
dimension by ( row_number() over ( order by run_date ) rn )
measures ( run_date, 1 grp, run_date grp_start )
rules (
grp_start[1] = run_date[cv()],
grp_start[any] =
case
when run_date[cv()] < grp_start[cv()-1] + 7 then
grp_start[cv() - 1]
else run_date[cv()]
end ,
grp[any] =
case
when run_date[cv()] < grp_start[cv()-1] + 7 then
grp[cv() - 1]
else nvl(grp[cv() - 1] + 1, 1)
end
);

define
within7 as
run_date < first ( run_date ) + 7

pattern ( within7+ )
define
within7 as
> 1 matches

match_recognize (
order by run_date
measures
count (*) as days
pattern ( within7+ )
define
within7 as
);

Am I getting
faster? stocksnap.io

current time < prev time

Analytic Functions

lag ( time_in_s )

select r.*, case
when time_in_s <
lag ( time_in_s )
then 'FASTER'
else 'SLOWER'
end faster
from running_log r

define
faster as
time_in_s < prev ( time_in_s )

pattern ( slower faster* )
define
faster as

match_recognize (
order by run_date
measures
classifier () as faster
define
faster as
);

FASTER
SLOWER
SLOWER
FASTER
FASTER

match_recognize (
order by run_date
measures
one row per match
define
faster as
);

match_recognize (
order by run_date
measures
all rows per match
define
faster as
);

RUN_DATE FASTER TIME_IN_S DISTANCE_IN_MILES
01 Jan 2018 SLOWER 420 1
02 Jan 2018 SLOWER 2,400 5
03 Jan 2018 SLOWER 4,932 10
06 Jan 2018 FASTER 2,350 5
07 Jan 2018 FASTER 410 1
10 Jan 2018 FASTER 400 1
13 Jan 2018 SLOWER 2,300 5
14 Jan 2018 FASTER 425 1
15 Jan 2018 FASTER 422 1

01 Jan 2018 SLOWER 420 1
02 Jan 2018 SLOWER 2,400 5
03 Jan 2018 SLOWER 4,932 10
06 Jan 2018 FASTER 2,350 5
07 Jan 2018 FASTER 410 1
10 Jan 2018 FASTER 400 1
13 Jan 2018 SLOWER 2,300 5
14 Jan 2018 FASTER 425 1
15 Jan 2018 FASTER 422 1
SLOWER!

01 Jan 2018 420 1
07 Jan 2018 410 1
10 Jan 2018 400 1
14 Jan 2018 425 1
15 Jan 2018 422 1
02 Jan 2018 2,400 5
06 Jan 2018 2,350 5
13 Jan 2018 2,300 5
03 Jan 2018 4,932 10

match_recognize (
partition by distance_in_miles
order by run_date
measures
all rows per match
define
faster as
);

01 Jan 2018 SLOWER 420 1
07 Jan 2018 FASTER 410 1
10 Jan 2018 FASTER 400 1
14 Jan 2018 SLOWER 425 1
15 Jan 2018 FASTER 422 1
02 Jan 2018 SLOWER 2,400 5
06 Jan 2018 FASTER 2,350 5
13 Jan 2018 FASTER 2,300 5
03 Jan 2018 SLOWER 4,932 10

Can I run 10k in
< 50 minutes?

Sum the total distance for runs
with a total time < 50 minutes

cumulative time <= 3,000 seconds

pattern ( fifty_minutes+ )
define
fifty_minutes as
sum ( time_in_s ) <= 3000
Returns the running total

match_recognize (
order by run_date
measures
first ( run_date ) as strt ,
sum ( time_in_s ) as total_time,
sum ( distance_in_miles ) as dist
pattern ( fifty_minutes+ )
define
fifty_minutes as
sum ( time_in_s ) <= 3000
);

STRT TOTAL_TIME DIST
01 Jan 2018 2,820 6
06 Jan 2018 2,760 6
10 Jan 2018 2,700 6
14 Jan 2018 847 2
Where's my 10 mile run?

any runs cumulative time < 3,000
and
one run cumulative time >= 3,000

pattern ( )

pattern ( under_fifty* over_fifty )

define
under_fifty as
sum ( time_in_s ) < 3000,
over_fifty as
sum ( time_in_s ) >= 3000
);
Includes under_fifty values

match_recognize (
order by run_date
measures
define
under_fifty as
over_fifty as
);

01 Jan 2018 7,752 16
06 Jan 2018 3,160 7
13 Jan 2018 3,147 7
Hmmm….

match_recognize (
order by run_date
measures
after match skip past last row
define
under_fifty as
over_fifty as
);

match_recognize (
order by run_date
measures
after match skip to next row
define
under_fifty as
over_fifty as
);

01 Jan 2018 7,752 16
02 Jan 2018 7,332 15
03 Jan 2018 4,932 10
06 Jan 2018 3,160 7
07 Jan 2018 3,110 7
10 Jan 2018 3,125 7
13 Jan 2018 3,147 7

Pixabay

How often did I run 5 miles
Followed by 2+ 1 mile runs
Within 7 days?

pattern ( five_mile one_mile {2,} )
define

define
five_mile as distance_in_miles = 5,

define
one_mile as distance_in_miles = 1

define
and run_date < first ( run_date ) + 7

match_recognize (
order by run_date
measures
count (*) as total_runs
define
);

STRT TOTAL_RUNS
06 Jan 2018 3
13 Jan 2018 3

Why would I want to do that?!

How do
I debug it?
Gratisography

(Regular) [exprsion]+ are easy to missteak

regex101.comregex101.com

classifier
=> Which variable matched?

classifier
match_number
=> Which group is this?

classifier
match_number
all rows per match

classifier
match_number
all rows per match with unmatched rows
=> Show me everything!

match_recognize (
order by run_date
measures
classifier () as var,
match_number () as grp
all rows per match with unmatched rows
define
);

RUN_DATE VAR GRP TIME_IN_S DISTANCE_IN_MILES
01 Jan 2018 420 1
02 Jan 2018 2,400 5
03 Jan 2018 4,932 10
06 Jan 2018 FIVE_MILE 1 2,350 5
07 Jan 2018 ONE_MILE 1 410 1
10 Jan 2018 ONE_MILE 1 400 1
13 Jan 2018 FIVE_MILE 2 2,300 5
14 Jan 2018 ONE_MILE 2 425 1
15 Jan 2018 ONE_MILE 2 422 1

Want more?
Pixabay

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Keith Laker
Analytic SQL PM

How to Find Patterns in Your Data with SQL

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