使用LAG / LEAD分析功能优化自联接Oracle SQL查询？

Question

We have an Oracle SQL query to identify records where the value of a table column has changed from one record to another. Relevant columns are (ID, SOME_COLUMN, FROM_DATE, TO_DATE) where the ID is not unique, and FROM_DATE and TO_DATE determine the time interval for which the the particular row for that ID was effective, ie

(ID1, VAL1, 01/01/2016, 03/01/2016)
(ID1, VAL2, 04/01/2016, 09/01/2016)
(ID1, VAL3, 10/01/2016, 19/01/2016) 

etc.

We could implement this using the following self-join

SELECT N.ID
       O.SOME_COLUMN OLD_VALUE,
       N.SOME_COLUMN NEW_VALUE
FROM OUR_TABLE N, OUR_TABLE O
WHERE N.ID = O.ID
  AND N.FROM_DATE - 1 = O.TO_DATE
  AND N.SOME_COLUMN <> O.SOME_COLUMN

however since the table contains 100 millions of records, it quite hits the performance. Is there a more effective way to do this? Someone hinted analytic functions (eg LAG), but we could not figure out a working solution so far. Any ideas would be appreciated

Answer 1

Yes, you can use LEAD() to fetch the last value :

SELECT t.id,
       t.some_column as OLD_VALUE,
       LEAD(t.some_column) OVER(PARTITION BY t.id ORDER BY t.from_date) as NEW_VALUE
FROM YourTable t

If you want only changes, wrap it with another select and filter OLD_VALUE <> NEW_VALUE

Answer 2

If you want the old value and the new value in a single row, then use lag() :

select t.*,
       lag(some_column) over (partition by id order by from_date) as prev_val
from t;

If the values may not change (as suggested by your sample query):

select t.*
from (select t.*,
             lag(some_column) over (partition by id order by from_date) as prev_val
      from t
     ) t
where prev_val <> some_column;

Answer 3

I think this is the LAG() approach you were talking about.

SELECT * 
  FROM (
    SELECT ID
           N.SOME_COLUMN NEW_VALUE,
           N.FROM_DATE,
           lag(N.SOME_COLUMN) over (partition by N.ID order by FROM_DATE) OLD_VALUE,
           lag(N.TO_DATE) over (partition by N.ID order by FROM_DATE) OLD_TO_DATE,
    FROM OUR_TABLE N
) T
WHERE FROM_DATE - 1 = OLD_TO_DATE
  AND NEW_VALUE<> OLD_VALUE;