Detect SQL island over multiple parameters and conditions
Asked Answered
W

1

3

(PostgreSQL 8.4) I got a great introduction to SQL gaps-and-islands here on Stack Overflow but I still have a question. Many island detection CTEs are based on a running order of a timestamp and some flag which breaks the sequence when it changes. But what if the "break" condition is a little more complex?

CREATE TABLE T1
(
  id SERIAL PRIMARY KEY,
  val INT,   -- some device
  status INT -- 0=OFF, 1=ON
);

INSERT INTO T1 (val, status) VALUES (10, 1);
INSERT INTO T1 (val, status) VALUES (10, 0);
INSERT INTO T1 (val, status) VALUES (11, 1);
INSERT INTO T1 (val, status) VALUES (11, 1);
INSERT INTO T1 (val, status) VALUES (10, 0);
INSERT INTO T1 (val, status) VALUES (12, 1);
INSERT INTO T1 (val, status) VALUES (13, 1);
INSERT INTO T1 (val, status) VALUES (13, 0);
INSERT INTO T1 (val, status) VALUES (13, 1);

In this case, val represents a device, and status is either ON or OFF. I want to select records 1, 3, 6, 7 and 9 with the following logic.

  1. #10 turns ON -- OK, new sequence, include in SELECT
  2. #10 turns OFF -- ends sequence properly, ignore row
  3. #11 turns ON -- OK, new sequence, include in SELECT
  4. #11 turns ON -- duplicate, ignore row
  5. #10 turns OFF -- #10 wasn't ON, ignore
  6. #12 turns ON -- OK, implicitly turns OFF #11, include in SELECT
  7. #13 turns ON -- OK, implicitly turns OFF #12, include in SELECT
  8. #13 turns OFF -- ends sequence properly, ignore row
  9. #13 turns ON -- OK, new sequence, include in SELECT

Basically, only one device can be ON at a time, and the "break" condition is that:

  • new.val = running.val AND new.status = 0
  • new.val <> running.val AND new.status = 1

I'm looking for something in the form of a CTE, no cursors please.

Waves answered 17/9, 2012 at 21:46 Comment(2)
Given the rows you want, that first condition is noise. You don't appear to be interested in the OFF rows at all.Sapient
Added two extra rows (8, 9) along with explanation.Waves
L
2

Answer for updated question

SELECT *
FROM  (
   SELECT *
         ,lag(val, 1, 0) OVER (PARTITION BY status ORDER BY id) last_val
         ,lag(status) OVER (PARTITION BY val ORDER BY id) last_status
   FROM   t1
   ) x
WHERE  status = 1
AND    (last_val <> val OR last_status = 0)

How?

Same as before, but this time combine two window functions. Switching on a device qualifies if ..
1. the last device switched on was a different one.
2. or the same device has been switched off in its last entry. The corner case with NULL for the first row of the partition is irrelevant, because then the row already qualified in 1.


Answer for original version of question.

If your I understand your task correctly, this simple query does the job:

SELECT *
FROM  (
   SELECT *
         ,lag(val, 1, 0) OVER (ORDER BY id) last_on
   FROM   t1
   WHERE  status = 1
   ) x
WHERE  last_on <> val

Returns rows 1, 3, 6, 7 as requested.

How?

The subquery ignores all switching off, as that is just noise, according to your description. Leaves entries where a device is switched on. Among those, only those entries are disqualified, where the same device was on already (the last entry switching on). Use the window function lag() for that. In particular I provide 0 as default to cover the special case of the first row - assuming that there is no device with val = 0.
If there is, pick another impossible number.
If no number is impossible, leave the special case as NULL with lag(val) OVER ... and in the outer query check with:

WHERE last_on IS DISTINCT FROM val
Lusatia answered 17/9, 2012 at 22:8 Comment(2)
That does work great, but I was missing an example to illustrate why I think status is important. Consider the extra two rows, which add an OFF from 13, then another ON. The same device twice in a row, which lag ignores (the 9th row needs to be picked up)Waves
@Jeff: I added another solution. FYI, if you have updates for your question that would invalidate existing answers written in good faith, the recommended way is to start a new question. We may just roll back such edits. I added a new answer this once.Lusatia

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