Re: Counting the number of repeated phrases in a column

On Thu, Jan 27, 2022 at 11:09 AM Rob Sargent <robjsargent(at)gmail(dot)com> wrote:
>
> On 1/27/22 10:03, Merlin Moncure wrote:
>
> On Wed, Jan 26, 2022 at 5:23 PM Merlin Moncure <mmoncure(at)gmail(dot)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