Thread: Guidance needed on an alternative take on common prefix SQL
Hi, I've seen various Postgres examples here and elsewhere that deal with the old common-prefix problem (i.e. "given 1234 showme the longest match"). I'm in need of a bit of guidance on how best to implement an alternative take. Frankly I don't quite know where to startbut I'm guessing it will probably involve CTEs, which is an area I'm very weak on. So, without further ado, here's the scenario: Given an SQL filtering query output that includes the following column: 87973891 87973970 87973971 87973972 87973973 87973975 87973976 87973977 87973978 87973979 8797400 The final output should be further filtered down to: 87973891 8797397 8797400 i.e. if $last_digit is present 0–9 inclusive, recursively filter until the remaining string is all the same (i.e. in thiscase, when $last_digit[0-9] is removed, 8797397 is the same). So, coming back to the example above: 8797397[0-9] is present so the "nearest common" I would be looking for is 8797397 because once [0-9] is removed, the 7 is the same on the preceedingdigit. The other two rows ( 87973891 and 8797400) are left untouched because $last_digit is not present in [0-9]. Hope this question makes sense ! Laura
On 8/6/19 6:25 PM, Laura Smith wrote:
> Hi,
>
> I've seen various Postgres examples here and elsewhere that deal with the old common-prefix problem (i.e. "given 1234
showme the longest match").
>
> I'm in need of a bit of guidance on how best to implement an alternative take. Frankly I don't quite know where to
startbut I'm guessing it will probably involve CTEs, which is an area I'm very weak on.
>
> So, without further ado, here's the scenario:
>
> Given an SQL filtering query output that includes the following column:
> 87973891
> 87973970
> 87973971
> 87973972
> 87973973
> 87973975
> 87973976
> 87973977
> 87973978
> 87973979
> 8797400
>
> The final output should be further filtered down to:
> 87973891
> 8797397
> 8797400
>
> i.e. if $last_digit is present 0–9 inclusive, recursively filter until the remaining string is all the same (i.e. in
thiscase, when $last_digit[0-9] is removed, 8797397 is the same).
>
> So, coming back to the example above:
> 8797397[0-9] is present
> so the "nearest common" I would be looking for is 8797397 because once [0-9] is removed, the 7 is the same on the
preceedingdigit.
>
> The other two rows ( 87973891 and 8797400) are left untouched because $last_digit is not present in [0-9].
>
> Hope this question makes sense !
>
> Laura
>
>
Hows this?
select distinct
case cc
when 1 then num
else left(num,-1)
end
from (
select
num,
(select count(*) as cc from numbers n2 where left(n2.num, -1) = left(numbers.num, -1))
from numbers
) as tmpx ;
-Andy
On Wednesday, August 7, 2019 2:01 AM, Andy Colson <andy@squeakycode.net> wrote: > On 8/6/19 6:25 PM, Laura Smith wrote: > > > Hi, > > I've seen various Postgres examples here and elsewhere that deal with the old common-prefix problem (i.e. "given 1234show me the longest match"). > > I'm in need of a bit of guidance on how best to implement an alternative take. Frankly I don't quite know where to startbut I'm guessing it will probably involve CTEs, which is an area I'm very weak on. > > So, without further ado, here's the scenario: > > Given an SQL filtering query output that includes the following column: > > 87973891 > > 87973970 > > 87973971 > > 87973972 > > 87973973 > > 87973975 > > 87973976 > > 87973977 > > 87973978 > > 87973979 > > 8797400 > > The final output should be further filtered down to: > > 87973891 > > 8797397 > > 8797400 > > i.e. if $last_digit is present 0–9 inclusive, recursively filter until the remaining string is all the same (i.e. inthis case, when $last_digit[0-9] is removed, 8797397 is the same). > > So, coming back to the example above: > > 8797397[0-9] is present > > so the "nearest common" I would be looking for is 8797397 because once [0-9] is removed, the 7 is the same on the preceedingdigit. > > The other two rows ( 87973891 and 8797400) are left untouched because $last_digit is not present in [0-9]. > > Hope this question makes sense ! > > Laura > > Hows this? > > select distinct > case cc > when 1 then num > else left(num,-1) > end > from ( > select > num, > (select count(*) as cc from numbers n2 where left(n2.num, -1) = left(numbers.num, -1)) > from numbers > ) as tmpx ; > > -Andy Hi Andy, That looks supremely clever ! I have just done a quick test and looks like it works as intended. Will do some more thorough testing with a larger datasetin due course. Thank you very much indeed Laura
On 8/7/19 3:36 AM, Laura Smith wrote:
> On Wednesday, August 7, 2019 2:01 AM, Andy Colson <andy@squeakycode.net> wrote:
>
>> On 8/6/19 6:25 PM, Laura Smith wrote:
>>
>>> Hi,
>>> I've seen various Postgres examples here and elsewhere that deal with the old common-prefix problem (i.e. "given
1234show me the longest match").
>>> I'm in need of a bit of guidance on how best to implement an alternative take. Frankly I don't quite know where to
startbut I'm guessing it will probably involve CTEs, which is an area I'm very weak on.
>>> So, without further ado, here's the scenario:
>>> Given an SQL filtering query output that includes the following column:
>>> 87973891
>>> 87973970
>>> 87973971
>>> 87973972
>>> 87973973
>>> 87973975
>>> 87973976
>>> 87973977
>>> 87973978
>>> 87973979
>>> 8797400
>>> The final output should be further filtered down to:
>>> 87973891
>>> 8797397
>>> 8797400
>>> i.e. if $last_digit is present 0–9 inclusive, recursively filter until the remaining string is all the same (i.e.
inthis case, when $last_digit[0-9] is removed, 8797397 is the same).
>>> So, coming back to the example above:
>>> 8797397[0-9] is present
>>> so the "nearest common" I would be looking for is 8797397 because once [0-9] is removed, the 7 is the same on the
preceedingdigit.
>>> The other two rows ( 87973891 and 8797400) are left untouched because $last_digit is not present in [0-9].
>>> Hope this question makes sense !
>>> Laura
>> Hows this?
>>
>> select distinct
>> case cc
>> when 1 then num
>> else left(num,-1)
>> end
>> from (
>> select
>> num,
>> (select count(*) as cc from numbers n2 where left(n2.num, -1) = left(numbers.num, -1))
>> from numbers
>> ) as tmpx ;
>>
>> -Andy
>
>
> Hi Andy,
>
> That looks supremely clever !
>
> I have just done a quick test and looks like it works as intended. Will do some more thorough testing with a larger
datasetin due course.
>
> Thank you very much indeed
>
> Laura
>
>
If the target field is really an integer type and you have lots of rows
you might be better off with arithmetic functions.
create table short as select id/10 as base, array_agg(mod(id,10)) as
odds from head group by base;
select * from short;
base | odds
---------+---------------------
879740 | {0}
8797389 | {1}
8797397 | {0,1,2,3,5,6,7,8,9}
(3 rows)
select case when array_length(odds,1) = 1 then 10*base + odds[1] else
base end from short;
base
----------
8797400
87973891
8797397
(3 rows)