Partitioning by timestamps close to each other (say 30min)
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I have a dataset where I want to partition it by timestamp close to each other(say less than 30min)
Driver | Timestamp A | 10/30/2019 05:02:28 A | 10/30/2019 05:05:28 A | 10/30/2019 05:09:28 A | 10/30/2019 05:12:28 A | 10/30/2019 07:54:28 A | 10/30/2019 07:57:28 A | 10/30/2019 08:02:28 A | 10/30/2019 12:14:28 A | 10/30/2019 12:17:28 A | 10/30/2019 12:22:28
How can we partition it like below:
id | Driver | Timestamp 1 | A | 10/30/2019 05:02:28 1 | A | 10/30/2019 05:05:28 1 | A | 10/30/2019 05:09:28 1 | A | 10/30/2019 05:12:28 2 | A | 10/30/2019 07:54:28 2 | A | 10/30/2019 07:57:28 2 | A | 10/30/2019 08:02:28 3 | A | 10/30/2019 12:14:28 3 | A | 10/30/2019 12:17:28 3 | A | 10/30/2019 12:22:28
Any help would be highly appreciated, thank you so much!
It depends on what you exactly want.
If you want to break into a new group when there is a 30+ minutes gap between two consecutive timestamps, you can use
lag() and a cumulative
select sum(case when timestamp < lag_timestamp + interval '30' minute then 0 else 1 end ) id, driver, timestamp from ( select t.*, lag(timestamp) over(partition by driver order by timestamp) lag_timestamp from mytable t ) t
Highest Voted 'partition' Questions - Page 10, Use this tag for questions about code that partitions data, memory, virtual machines, databases or Partitioning by timestamps close to each other (say 30min). It depends on what you exactly want. If you want to break into a new group when there is a 30+ minutes gap between two consecutive timestamps, you can use lag () and a cumulative sum (): select sum(case when timestamp < lag_timestamp + interval '30' minute then 0 else 1 end ) id, driver, timestamp from ( select t.*, lag(timestamp) over(partition by driver order by timestamp) lag_timestamp from mytable t ) t.
Check if your version supports the
sessionize table operator:
SELECT * FROM Sessionize ( ON ( SELECT * FROM tab ) PARTITION BY driver ORDER BY ts USING TimeColumn('ts') Timeout(1800) )
Get total amount of time between records when changed state , We can use conditional aggregation and an OVER clause to get the 1, 0) OVER (PARTITION BY device_owner ORDER BY timestamp)� In this example, we truncate the timestamp column to a yearly table, resulting in about 20 million rows per year. If all of our queries specify a date(s), or date range(s), and those specified usually cover data within a single year, this may be a great starting strategy for partitioning, as it would result in a single table per year, with a manageable number of rows per table.
I think you're looking to sessionize you data per driver. Try this method. It appends the session_id to its respective driver to create driver specific session_id.
select driver||sum(session_code) over (partition by driver order by timestamp) as session_id, driver, timestamp from (select driver, timestamp, case when timestamp > lag(timestamp) over (partition by driver order by timestamp) + interval '1800' second then 1 else 0 end as session_code from your_table) a
Introduction to partitioned tables | BigQuery, If the table is partitioned on a TIMESTAMP column, you can create There might be other query cost savings when the query actually runs. 0 Partitioning by timestamps close to each other (say 30min) Nov 26 '19. Badges (4) Gold
Ask TOM "dropping and adding partitions periodically", Now the criteria for dropping the partitions is sysdate-30. when others than null; , this will see that the procedure executes successfully. 2 ( timestamp date, What if you have multiple partitions for a single day i.e. if a single day has say partition since the data on table A is still required to be there until the next month). I'm looking into partitioning a table in my InnoDB database. I have a column corresponding to a UTC timestamp, and I want to partition around that. The basic idea that I want to capture is a
Partition Maintenance, With partitioning, first the partition is chosen and "opened", then a smaller BTree ( of say 4 RANGE PARTITIONing by day (or other unit of time) lets you do a nearly Being the first field in some index lets the engine find the 'next' value when opening the table. CREATE TABLE tbl ( dt DATETIME NOT NULL, -- or DATE . use_deprecated_int96_timestamps: Write timestamps to INT96 Parquet format. Default FALSE. coerce_timestamps: Cast timestamps a particular resolution. Can be NULL, "ms" or "us". Default NULL (no casting) allow_truncated_timestamps: Allow loss of data when coercing timestamps to a particular resolution.
Multi-Store Tables, Every multi-store table contains at least two partitions: one of which is in the in- memory DEFAULT STORAGE and the other is in disk based EXTENDED STORAGE. single physical partition, we can't tell how much data is in each logical partition. Group. Beginner. 1 hr. 30 min. Get Started with SAP HANA Dynamic Tiering. I have all timestamps in between a period, but I need to group it every 10 minutes and I have no idea how. I tried date_trunc which does not have the precision I need. ex: between 2013-04-07 15:30:00, 2013-04-07 15:40:00 5 results .
- Please tag your question with the RDBMS that you are using: mysql, sql-server, postgresql...?
- because when it comes to date/time, many products are far from ANSI SQL compliant.
- What if you have one record every minute during 1 hour, will they belong to the same group or you want to break into a new group 30 minutes after the timestamp of the first record?
- For the purpose, 30 minutes after the timestamp will be good.
- @AyushKumar: 30 minutes between 2 consecutive timestamps, or 30 minutes after the first timestamp? FYI, the first on is easier to solve.
- This won't work if I have three cases like it, right? I want something which can work like a row_number() over (partition by..)
- @AyushKumar: your question is unclear to me. However as far as concerns, this query will work for your sample data and produce the results that you expect. Please try it against your real data.
- @AyushKumar . . . This answers your question. You should accept the answer. Yes, it works for "3 cases", whatever you mean by that.
- Perfect, I did the above and then cumulative of id, worked perfectly. Thank you so much @GMB