What would be the most efficient query to calculate the average of time difference between consecutive rows in a table? Note that the table has no primary key.
If the table looks like below:
| tran_end_time |
|-----------------------|
|2022-02-08 07:04:46.610|
|2022-02-08 07:09:47.403|
|2022-02-08 07:14:48.100|
|2022-02-08 07:20:03.973|
Then I need the answer to be:
avg('2022-02-08 07:20:03.973' - '2022-02-08 07:14:48.100',
'2022-02-08 07:14:48.100' - '2022-02-08 07:09:47.403',
'2022-02-08 07:09:47.403' - '2022-02-08 07:04:46.610')
We can use DATEDIFF along with LAG:
WITH cte AS (
SELECT tran_end_time,
LAG(tran_end_time) OVER (ORDER BY tran_end_time) AS tran_end_time_lag
FROM yourTable
)
SELECT AVG(DATEDIFF(minute, tran_end_time_lag, tran_end_time)) AS diff_avg
FROM cte
WHERE tran_end_time_lag IS NOT NULL;
Note that the WHERE clause in the final query above ensures that we do not include any diff involving the earliest record.
Related
I'm trying to get the day difference between 2 dates in Impala but I need to exclude weekends.
I know it should be something like this but I'm not sure how the weekend piece would go...
DATEDIFF(resolution_date,created_date)
Thanks!
One approach at such task is to enumerate each and every day in the range, and then filter out the week ends before counting.
Some databases have specific features to generate date series, while in others offer recursive common-table-expression. Impala does not support recursive queries, so we need to look at alternative solutions.
If you have a table wit at least as many rows as the maximum number of days in a range, you can use row_number() to offset the starting date, and then conditional aggregation to count working days.
Assuming that your table is called mytable, with column id as primary key, and that the big table is called bigtable, you would do:
select
t.id,
sum(
case when dayofweek(dateadd(t.created_date, n.rn)) between 2 and 6
then 1 else 0 end
) no_days
from mytable t
inner join (select row_number() over(order by 1) - 1 rn from bigtable) n
on t.resolution_date > dateadd(t.created_date, n.rn)
group by id
I have a table that I would like to sort by a timestamp desc and then compare all consecutive rows to determine the difference between each row. From there, I would like to find all the rows whose difference is greater than ~2hours.
I'm stuck on how to actually compare consecutive rows in a table. Any help would be much appreciated.
I'm using Oracle SQL Developer 3.2
You didn't show us your table definition, but something like this:
select *
from (
select t.*,
t.timestamp_column,
t.timestamp_column - lag(timestamp_column) over (order by timestamp_column) as diff
from the_table t
) x
where diff > interval '2' hour;
This assumes that timestamp_column is defined as timestamp not date (otherwise the result of the difference wouldn't be an interval)
I work for a sports film analysis company. We have teams with unique team IDs and I would like to find the number of consecutive weeks they have uploaded film to our site moving backwards from today. Each upload also has its own row in a separate table that I can join on teamid and has a unique date of when it was uploaded. So far I put together a simple query that pulls each unique DATEDIFF(week) value and groups on teamid.
Select teamid, MAX(weekdiff)
(Select teamid, DATEDIFF(week, dateuploaded, GETDATE()) as weekdiff
from leroy_events
group by teamid, weekdiff)
What I am given is a list of teamIDs and unique weekly date differences. I would like to then find the max for each teamID without breaking an increment of 1. For example, if my data set is:
Team datediff
11453 0
11453 1
11453 2
11453 5
11453 7
11453 13
I would like the max value for team: 11453 to be 2.
Any ideas would be awesome.
I have simplified your example assuming that I already have a table with weekdiff column. That would be what you're doing with DATEDIFF to calculate it.
First, I'm using LAG() window function to assign previous value (in ordered set) of a weekdiff to the current row.
Then, using a WHERE condition I'm retrieving max(weekdiff) value that has a previous value which is current_value - 1 for consecutive weekdiffs.
Data:
create table leroy_events ( teamid int, weekdiff int);
insert into leroy_events values (11453,0),(11453,1),(11453,2),(11453,5),(11453,7),(11453,13);
Code:
WITH initial_data AS (
Select
teamid,
weekdiff,
lag(weekdiff,1) over (partition by teamid order by weekdiff) as lag_weekdiff
from
leroy_events
)
SELECT
teamid,
max(weekdiff) AS max_weekdiff_consecutive
FROM
initial_data
WHERE weekdiff = lag_weekdiff + 1 -- this insures retrieving max() without breaking your consecutive increment
GROUP BY 1
SQLFiddle with your sample data to see how this code works.
Result:
teamid max_weekdiff_consecutive
11453 2
You can use SQL window functions to probe relationships between rows of the table. In this case the lag() function can be used to look at the previous row relative to a given order and grouping. That way you can determine whether a given row is part of a group of consecutive rows.
You still need overall to aggregate or filter to reduce the number of rows for each group of interest (i.e. each team) to 1. It's convenient in this case to aggregate. Overall, it might look like this:
select
team,
case min(datediff)
when 0 then max(datediff)
else -1
end as max_weeks
from (
select
team,
datediff,
case
when (lag(datediff) over (partition by team order by datediff) != datediff - 1)
then 0
else 1
end as is_consec
from diffs
) cd
where is_consec = 1
group by team
The inline view just adds an is_consec column to the data, marking whether each row is part of a group of consecutive rows. The outer query filters on that column (you cannot filter directly on a window function), and chooses the maximum datediff from the remaining rows for each team.
There are a few subtleties there:
The case expression in the inline view is written as it is to exploit the fact that the lag() computed for the first row of each partition will be NULL, which does not evaluate unequal (nor equal) to any value. Thus the first row in each partition is always marked consecutive.
The case testing min(datediff) in the outer select clause picks up teams that have no record with datediff = 0, and assigns -1 to column max_weeks for them.
It would also have been possible to mark rows non-consecutive if the first in their group did not have datediff = 0, but then you would lose such teams from the results altogether.
I have a table S with time series data like this:
key day delta
For a given key, it's possible but unlikely that days will be missing.
I'd like to construct a cumulative column from the delta values (positive INTs), for the purposes of inserting this cumulative data into another table. This is what I've got so far:
SELECT key, day,
SUM(delta) OVER (PARTITION BY key ORDER BY day asc RANGE BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW),
delta
FROM S
In my SQL flavor, default window clause is RANGE BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW, but I left that in there to be explicit.
This query is really slow, like order of magnitude slower than the old broken query, which filled in 0s for the cumulative count. Any suggestions for other methods to generate the cumulative numbers?
I did look at the solutions here:
Running total by grouped records in table
The RDBMs I'm using is Vertica. Vertica SQL precludes the first subselect solution there, and its query planner predicts that the 2nd left outer join solution is about 100 times more costly than the analytic form I show above.
I think you're essentially there. You may just need to update the syntax a bit:
SELECT s_qty,
Sum(s_price)
OVER(
partition BY NULL
ORDER BY s_qty ASC rows UNBOUNDED PRECEDING ) "Cumulative Sum"
FROM sample_sales;
Output:
S_QTY | Cumulative Sum
------+----------------
1 | 1000
100 | 11000
150 | 26000
200 | 28000
250 | 53000
300 | 83000
2000 | 103000
(7 rows)
reference link:
https://dwgeek.com/vertica-cumulative-sum-average-and-example.html/
Sometimes it's faster to just use a correlated subquery:
SELECT
[key]
, [day]
, delta
, (SELECT SUM(delta) FROM S WHERE [key] < t1.[key]) AS DeltaSum
FROM S t1
I have a table with sequential timestamps:
2011-03-17 10:31:19
2011-03-17 10:45:49
2011-03-17 10:47:49
...
I need to find the average time difference between each of these(there could be dozens) in seconds or whatever is easiest, I can work with it from there. So for example the above inter-arrival time for only the first two times would be 870 (14m 30s). For all three times it would be: (870 + 120)/2 = 445 (7m 25s).
A note, I am using postgreSQL 8.1.22 .
EDIT: The table I mention above is from a different query that is literally just a one-column list of timestamps
Not sure I understood your question completely, but this might be what you are looking for:
SELECT avg(difference)
FROM (
SELECT timestamp_col - lag(timestamp_col) over (order by timestamp_col) as difference
FROM your_table
) t
The inner query calculates the distance between each row and the preceding row. The result is an interval for each row in the table.
The outer query simply does an average over all differences.
i think u want to find avg(timestamptz).
my solution is avg(current - min value). but since result is interval, so add it to min value again.
SELECT avg(target_col - (select min(target_col) from your_table))
+ (select min(target_col) from your_table)
FROM your_table
If you cannot upgrade to a version of PG that supports window functions, you
may compute your table's sequential steps "the slow way."
Assuming your table is "tbl" and your timestamp column is "ts":
SELECT AVG(t1 - t0)
FROM (
-- All this silliness would be moot if we could use
-- `` lead(ts) over (order by ts) ''
SELECT tbl.ts AS t0,
next.ts AS t1
FROM tbl
CROSS JOIN
tbl next
WHERE next.ts = (
SELECT MIN(ts)
FROM tbl subquery
WHERE subquery.ts > tbl.ts
)
) derived;
But don't do that. Its performance will be terrible. Please do what
a_horse_with_no_name suggests, and use window functions.