match words in all rows in same column - oracle

Question

There is a column in table 'mytable' named 'Description'.
+----+-------------------------------+
| ID | Description |
+----+-------------------------------+
| 1 | My NAME is Sajid KHAN |
| 2 | My Name is Ahmed Khan |
| 3 | MY friend name is Salman Khan |
+----+-------------------------------+
I need to write an Oracle SQL query/procedure/function to list the distinct words in the column.
The output should be:
+------------------+-------+
| Word | Count |
+------------------+-------+
| MY | 3 |
| NAME | 3 |
| IS | 3 |
| SAJID | 1 |
| KHAN | 3 |
| AHMED | 1 |
| FRIEND | 1 |
| SALMAN | 1 |
+------------------+-------+
Word matching should be case-insensitive.
I am using Oracle 12.1.

Answer 1

Let's suppose we would somehow manage to split every description in words.
So, instead of single row with Id = 1 and Description = 'My NAME is Sajid KHAN' we'd have 5 rows like this
ID | Description
--- | ------------
1 | My
1 | NAME
1 | is
1 | Sajid
1 | KHAN
in this form it'd be trivial, something like
select Description, count(*) from data_in_new_form group by Description
So, let's do this using recursive query.
create table mytable
as
select 1 as ID, 'My NAME is Sajid KHAN' as Description from dual
union all
select 2, 'My Name is Ahmed Khan' from dual
union all
select 3, 'MY friend name is Salman Khan' from dual
union all
select 4, 'test, punctuation! it is' from dual
;
with
rec (id, str, depth, element_value) as
(
-- Anchor member.
select id, upper(Description) as str, 1 as depth, REGEXP_SUBSTR( upper(Description), '(.*?)( |$)', 1, 1, NULL, 1 ) AS element_value
from mytable
UNION ALL
-- Recursive member.
select id, str, depth + 1, REGEXP_SUBSTR( str ,'(.*?)( |$)', 1, depth+1, NULL, 1 ) AS element_value
from rec
where depth < regexp_count(str, ' ')+1
)
, data as (
select * from rec
--order by id, depth
)
select element_value, count(*) from data
group by element_value
order by element_value
;
Please notice this version doesn't do anything about punctuation assuming words are separated with spaces.
UPDATE alternative way using hierarchic query
with rec as
(
SELECT id, LEVEL AS depth,
REGEXP_SUBSTR( upper(description) ,'(.*?)( |$)', 1, LEVEL, NULL, 1 ) AS element_value
FROM mytable
CONNECT BY LEVEL <= regexp_count(description, ' ')+1
and prior id = id
and prior SYS_GUID() is not null
)
, data as (
select * from rec
--order by id, depth
)
select element_value, count(*) from data
group by element_value
order by 2 desc
;

Answer 2

This query will work. The ordering of the words may be different. However, frequent words come at the beginning as you have listed.
SELECT word,
COUNT(*)
FROM
(SELECT TRIM (REGEXP_SUBSTR (Description, '[^ ]+', 1, ROWNUM) ) AS Word
FROM
(SELECT LISTAGG(UPPER(Description),' ') within GROUP(
ORDER BY ROWNUM ) AS Description
FROM mytable
)
CONNECT BY LEVEL <= REGEXP_COUNT ( Description, '[^ ]+')
)
GROUP BY WORD
ORDER BY 2 DESC;