partitioned | Postgres database table partitioning support for Rails | Database library

The new divisions are chosen so that the total memory of the files in each partition doesn't exceed 1000 MB.

If the main consideration for repartitioning is memory, it might be a good idea to use .repartition(partition_size='1000MB'). The script looks like:

We can use ROW_NUMBER here along with an aggregation query:

I don't think there is a safe and supported way to do that (without catalog modifications). Fortunately, there is nothing special about sequences that would make dropping them a problem. So take a short down time and:

• remove the default value that uses the identity sequence

• record the current value of the sequence

• drop the table

• create a new sequence with an appropriate START value

• use the new sequence to set new default values

If you want an identity column, you should define it on the partitioned table, not on one of the partitions.

I would argue it's a disadvantage! Because splitting the date parts makes it much harder to do date filtering. For example say you want to query the last 10 days of data which may cross month boundaries? With a single date value you can just run simple queries like

...where date >= current_date() - interval 10 days

and Spark will figure out the right partitions for you. Spark can also handle other date functions, like year(date) = 2019 or month(date) = 2 and again it will properly do the partition pruning for you.

I always encourage using a single date column for partitioning. Let Spark do the work.

Also, an important thing to keep in mind is that date format should be yyyy-MM-dd.

Find the parent_id record with item_id = 1, and then update all records whose parent_id equals parent_id.

Use boolean indexing:

For this answer, I'll assume you have correctly understood the reasons why you should and should not do Hive-style partitioning and won't be covering the backing theory.

In this case, it's important to ensure we not only correctly calculate the number of files needed inside our splits but also repartition our dataset based on these calculations. Failure to do repartitions before write-out on Hive-style partition datasets may result in your job attempting to write out millions of tiny files which will kill your performance.

In our case, the strategy we will use will be to create files that are at most N rows per file, which will bound the size of each file. We can't easily limit the exact size of each file inside the splits, but we can use row counts as a good approximation.

The methodology we will use to accomplish this will be to create a synthetic column that describes which 'batch' a row will belong to, repartition the final dataset on both the Hive split column and this synthetic column, and use this result on write.

In order to ensure our synthetic column indicates the proper batch a row belongs to, we need to determine the number of rows inside each hive split, and 'sprinkle' the rows inside this split into the proper number of files.

The strategy in total will look something like this:

1. Determine number of rows per Hive value
2. Join this count against main dataframe
3. Determine number of files in split by dividing row count per split by rows per file
4. Create random index between 0 and the file count, essentially 'picking' the file the row will belong to
5. Calculate number of unique combinations of Hive split columns and our synthetic column
6. Repartition output dataset over both Hive column and synthetic column into the number of unique combinations. i.e. one file per combination, exactly what we want

Let's start by considering the following dataset:

Problem

When you do this

Basically it's the same as for a single line. You could split by gear, use lapply to compute the spline for each split and then bind back together using e.g. bind_rows: