Find a latest gap between Unix timestamps - sql

Question

I currently have two functions that should return the time a device started logging again, the time when the previous row before it was more than 60 seconds away. These functions may work fine but I have to see it work as it takes forever. Is there any shortcuts to make this faster?
CREATE OR REPLACE FUNCTION findNextTime(startt integer)
RETURNS integer AS
$nextTime$
DECLARE
nextTime integer;
BEGIN
select time into nextTime from m01 where time < startt ORDER BY time DESC LIMIT 1;
return nextTime;
END;
$nextTime$ LANGUAGE plpgsql;
CREATE OR REPlACE FUNCTION findStart()
RETURNS integer AS
$lastTime$
DECLARE
currentTime integer;
lastTime integer;
BEGIN
select time into currentTime from m01 ORDER BY time DESC LIMIT 1;
LOOP
RAISE NOTICE 'Current Time: %', currentTime;
select findNextTime(currentTime) into lastTime;
EXIT WHEN ((currentTime - lastTime) > 60);
currentTime := lastTime;
END LOOP;
return lastTime;
END;
$lastTime$ LANGUAGE plpgsql;
To clarify, I want to essentially find the last time there was a break of more than 60 seconds between any two rows.
CREATE TABLE IF NOT EXISTS m01 (
time integer,
value decimal,
id smallint,
driveId smallint
)
Sample Data:
In this case it would return 1520376063 because the next entry (1520375766) is more than 60 seconds apart it.
| time | value | id | driveid |
|------------|--------------------|------|---------|
| 1520376178 | 516.2 | 5116 | 2 |
| 1520376173 | 507.8 | 5116 | 2 |
| 1520376168 | 499.5 | 5116 | 2 |
| 1520376163 | 491.1 | 5116 | 2 |
| 1520376158 | 482.90000000000003 | 5116 | 2 |
| 1520376153 | 474.5 | 5116 | 2 |
| 1520376148 | 466.20000000000005 | 5116 | 2 |
| 1520376143 | 457.8 | 5116 | 2 |
| 1520376138 | 449.5 | 5116 | 2 |
| 1520376133 | 441.20000000000005 | 5116 | 2 |
| 1520376128 | 432.90000000000003 | 5116 | 2 |
| 1520376123 | 424.6 | 5116 | 2 |
| 1520376118 | 416.20000000000005 | 5116 | 2 |
| 1520376113 | 407.8 | 5116 | 2 |
| 1520376108 | 399.5 | 5116 | 2 |
| 1520376103 | 391.20000000000005 | 5116 | 2 |
| 1520376098 | 382.90000000000003 | 5116 | 2 |
| 1520376093 | 374.5 | 5116 | 2 |
| 1520376088 | 366.20000000000005 | 5116 | 2 |
| 1520376083 | 357.8 | 5116 | 2 |
| 1520376078 | 349.5 | 5116 | 2 |
| 1520376073 | 341.20000000000005 | 5116 | 2 |
| 1520376068 | 332.90000000000003 | 5116 | 2 |
| 1520376063 | 324.5 | 5116 | 2 |
| 1520375766 | 102.5 | 5116 | 2 |

Answer 1

This simple query should replace your two functions. Note the window function lead() in the subquery:
SELECT *
FROM (
SELECT time, lead(time) OVER (ORDER BY time DESC) AS last_time
FROM m01
WHERE time < _startt
) sub
WHERE time > last_time + 60
ORDER BY time DESC
LIMIT 1;
Either way, the crucial part for performance is the right index. Ideally on (time DESC).
Assuming time is defined NOT NULL - which it probably should be, but the table definition in the question does not say so. Else you probably want ORDER BY time DESC NULLS LAST - and a matching index. See:
PostgreSQL sort by datetime asc, null first?
I expect this plpgsql function to perform faster, though, if gaps typically show up early:
CREATE OR REPLACE FUNCTION find_gap_before_time(_startt int)
RETURNS int AS
$func$
DECLARE
_current_time int;
_last_time int;
BEGIN
FOR _last_time IN -- single loop is enough!
SELECT time
FROM m01
WHERE time < _startt
ORDER BY time DESC -- NULLS LAST?
LOOP
IF _current_time > _last_time + 60 THEN -- never true for 1st row
RETURN _current_time;
END IF;
_current_time := _last_time;
END LOOP;
END
$func$ LANGUAGE plpgsql;
Call:
SELECT find_gap_before_time(1520376200);
Result as requested.
Aside: You'd typically save a couple of bytes per row in storage by placing the column value last or first, thereby minimizing alignment padding. Like:
CREATE TABLE m01 (
time integer,
id smallint,
driveId smallint,
value decimal
);
Detailed explanation:
Calculating and saving space in PostgreSQL