Re: Counting the number of repeated phrases in a column

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От Merlin Moncure
Тема Re: Counting the number of repeated phrases in a column
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Msg-id CAHyXU0x=EbguRAuBPj++gR_RWvWsdwRTsm4s5+8omEZy9wy23Q@mail.gmail.com
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Ответ на Re: Counting the number of repeated phrases in a column  (Rob Sargent <robjsargent@gmail.com>)
Ответы Re: Counting the number of repeated phrases in a column
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On Thu, Jan 27, 2022 at 11:09 AM Rob Sargent <robjsargent@gmail.com> wrote:
>
> On 1/27/22 10:03, Merlin Moncure wrote:
>
> On Wed, Jan 26, 2022 at 5:23 PM Merlin Moncure <mmoncure@gmail.com> wrote:
>
> with s as (select 'Hello World Hello World' as sentence)
> select
>   phrase,
>   array_upper(string_to_array((select sentence from s), phrase), 1) -
> 1 as occurrances
> from
> (
>   select array_to_string(x, ' ') as phrase
>   from
>   (
>     select distinct v[a:b]  x
>     from regexp_split_to_array((select sentence from s), ' ') v
>     cross join lateral generate_series(1, array_upper(v, 1)) a
>     cross join lateral generate_series(a + 1, array_upper(v, 1)) b
>   ) q
> ) q;
>
> Simplified to:
> select distinct array_to_string(v[a:b], ' ') phrase, count(*) as occurrences
> from regexp_split_to_array('Hello World Hello World', ' ') v
> cross join lateral generate_series(1, array_upper(v, 1)) a
> cross join lateral generate_series(a + 1, array_upper(v, 1)) b
> group by 1;
>
>          phrase          │ occurances
> ─────────────────────────┼────────────
>  World Hello             │          1
>  Hello World Hello       │          1
>  Hello World             │          2
>  Hello World Hello World │          1
>  World Hello World       │          1
>
> merlin
>
>
> And since we're looking for repeated phrases maybe add
>
> having count(*) > 1

thanks.  also, testing on actual data, I noticed that a couple other
things are mandatory, mainly doing a bit of massaging before
tokenizing:

select distinct array_to_string(v[a:b], ' ') phrase, count(*) as occurrences
from
(
  select array_agg(t) v
  from
  (
    select trim(replace(unnest(v), E'\n', '')) t
    from regexp_split_to_array(<sentence>, ' ') v
  ) q
  where length(t) > 1
) q
cross join lateral generate_series(1, array_upper(v, 1)) a
cross join lateral generate_series(a + 1, array_upper(v, 1)) b
group by 1
having count(*) > 1;

We are definitely in N^2 space here, so look for things to start
breaking down for sentences > 1000 words.

merlin



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