I had a question that is related to pyspark's repartitionBy() function which I originally posted in a comment on this question. I was asked to post it as a separate question, so here it is:
I understand that df.partitionBy(COL) will write all the rows with each value of COL to their own folder, and that each folder will (assuming the rows were previously distributed across all the partitions by some other key) have roughly the same number of files as were previously in the entire table. I find this behavior annoying. If I have a large table with 500 partitions, and I use partitionBy(COL) on some attribute columns, I now have for example 100 folders which each contain 500 (now very small) files.
What I would like is the partitionBy(COL) behavior, but with roughly the same file size and number of files as I had originally.
As demonstration, the previous question shares a toy example where you have a table with 10 partitions and do partitionBy(dayOfWeek) and now you have 70 files because there are 10 in each folder. I would want ~10 files, one for each day, and maybe 2 or 3 for days that have more data.
Can this be easily accomplished? Something like df.write().repartition(COL).partitionBy(COL) seems like it might work, but I worry that (in the case of a very large table which is about to be partitioned into many folders) having to first combine it to some small number of partitions before doing the partitionBy(COL) seems like a bad idea.
Any suggestions are greatly appreciated!
You've got several options. In my code below I'll assume you want to write in parquet, but of course you can change that.
(1) df.repartition(numPartitions, *cols).write.partitionBy(*cols).parquet(writePath)
This will first use hash-based partitioning to ensure that a limited number of values from COL make their way into each partition. Depending on the value you choose for numPartitions, some partitions may be empty while others may be crowded with values -- for anyone not sure why, read this. Then, when you call partitionBy on the DataFrameWriter, each unique value in each partition will be placed in its own individual file.
Warning: this approach can lead to lopsided partition sizes and lopsided task execution times. This happens when values in your column are associated with many rows (e.g., a city column -- the file for New York City might have lots of rows), whereas other values are less numerous (e.g., values for small towns).
(2) df.sort(sortCols).write.parquet(writePath)
This options works great when you want (1) the files you write to be of nearly equal sizes (2) exact control over the number of files written. This approach first globally sorts your data and then finds splits that break up the data into k evenly-sized partitions, where k is specified in the spark config spark.sql.shuffle.partitions. This means that all values with the same values of your sort key are adjacent to each other, but sometimes they'll span a split, and be in different files. This, if your use-case requires all rows with the same key to be in the same partition, then don't use this approach.
There are two extra bonuses: (1) by sorting your data its size on disk can often be reduced (e.g., sorting all events by user_id and then by time will lead to lots of repetition in column values, which aids compression) and (2) if you write to a file format the supports it (like Parquet) then subsequent readers can read data in optimally by using predicate push-down, because the parquet writer will write the MAX and MIN values of each column in the metadata, allowing the reader to skip rows if the query specifies values outside of the partition's (min, max) range.
Note that sorting in Spark is more expensive than just repartitioning and requires an extra stage. Behind the scenes Spark will first determine the splits in one stage, and then shuffle the data into those splits in another stage.
(3) df.rdd.partitionBy(customPartitioner).toDF().write.parquet(writePath)
If you're using spark on Scala, then you can write a customer partitioner, which can get over the annoying gotchas of the hash-based partitioner. Not an option in pySpark, unfortunately. If you really want to write a custom partitioner in pySpark, I've found this is possible, albeit a bit awkward, by using rdd.repartitionAndSortWithinPartitions:
df.rdd \
.keyBy(sort_key_function) \ # Convert to key-value pairs
.repartitionAndSortWithinPartitions(numPartitions=N_WRITE_PARTITIONS,
partitionFunc=part_func) \
.values() # get rid of keys \
.toDF().write.parquet(writePath)
Maybe someone else knows an easier way to use a custom partitioner on a dataframe in pyspark?
df.repartition(COL).write().partitionBy(COL)
will write out one file per partition. This will not work well if one of your partition contains a lot of data. e.g. if one partition contains 100GB of data, Spark will try to write out a 100GB file and your job will probably blow up.
df.repartition(2, COL).write().partitionBy(COL)
will write out a maximum of two files per partition, as described in this answer. This approach works well for datasets that are not very skewed (because the optimal number of files per partition is roughly the same for all partitions).
This answer explains how to write out more files for the partitions that have a lot of data and fewer files for the small partitions.
Related
I have a case where i am trying to write some results using dataframe write into S3 using the below query with input_table_1 size is 13 Gb and input_table_2 as 1 Mb
input_table_1 has columns account, membership and
input_table_2 has columns role, id , membership_id, quantity, start_date
SELECT
/*+ BROADCASTJOIN(input_table_2) */
account,
role,
id,
quantity,
cast(start_date AS string) AS start_date
FROM
input_table_1
INNER JOIN
input_table_2
ON array_contains(input_table_1.membership, input_table_2.membership_id)
where membership array contains list of member_ids
This dataset write using Spark dataframe is generating around 1.1TiB of data in S3 with around 700 billion records.
We identified that there are duplicates and used dataframe.distinct.write.parquet("s3path") to remove the duplicates . The record count is reduced to almost 1/3rd of the previous total count with around 200 billion rows but we observed that the output size in S3 is now 17.2 TiB .
I am very confused how this can happen.
I have used the following spark conf settings
spark.sql.shuffle.partitions=20000
I have tried to do a coalesce and write to s3 but it did not work.
Please suggest if this is expected and when can be done ?
There's two sides to this:
1) Physical translation of distinct in Spark
The Spark catalyst optimiser turns a distinct operation into an aggregation by means of the ReplaceDeduplicateWithAggregate rule (Note: in the execution plan distinct is named Deduplicate).
This basically means df.distinct() on all columns is translated into a groupBy on all columns with an empty aggregation:
df.groupBy(df.columns:_*).agg(Map.empty).
Spark uses a HashPartitioner when shuffling data for a groupBy on respective columns. Since the groupBy clause in your case contains all columns (well, implicitly, but it does), you're more or less randomly shuffling data to different nodes in the cluster.
Increasing spark.sql.shuffle.partitions in this case is not going to help.
Now on to the 2nd side, why does this affect the size of your parquet files so much?
2) Compression in parquet files
Parquet is a columnar format, will say your data is organised in columns rather than row by row. This allows for powerful compression if data is adequately laid-out & ordered. E.g. if a column contains the same value for a number of consecutive rows, it is enough to write that value just once and make a note of the number of repetitions (a strategy called run length encoding). But Parquet also uses various other compression strategies.
Unfortunately, data ends up pretty randomly in your case after shuffling to remove duplicates. The original partitioning of input_table_1 was much better fitted.
Solutions
There's no single answer how to solve this, but here's a few pointers I'd suggest doing next:
What's causing the duplicates? Could these be removed upstream? Or is there a problem with the join condition causing duplicates?
A simple solution is to just repartition the dataset after distinct to match the partitioning of your input data. Adding a secondary sorting (sortWithinPartition) is likely going to give you even better compression. However, this comes at the cost of an additional shuffle!
As #matt-andruff pointed out below, you can also achieve this in SQL using cluster by. Obviously, that also requires you to move the distinct keyword into your SQL statement.
Write your own deduplication algorithm as Spark Aggregator and group / shuffle the data just once in a meaningful way.
I have a Spark DataFrame consisting of many double columns that are measurements, but I want a way of annotating each unique row by computing a hash of several other non-measurement columns. This hash results in garbled strings that are highly unique, and I've noticed my dataset size increases substantially when this column is present. How can I sort / lay out my data to decrease the overall dataset size?
I know that the Snappy compression protocol used on my parquet files executes best upon runs of similar data, so I think a sort over the primary key could be useful, but I also can't coalesce() the entire dataset into a single file (it's hundreds of GB in total size before the primary key creation step).
My hashing function is SHA2(128) FYI.
If you have a column that can be computed from the other columns, then simply omit that column before compression, and reconstruct it after decompression.
I have a pretty straightforward pyspark SQL application (spark 2.4.4, EMR 5.29) that reads a dataframe of the schema topic, year, count:
df.show()
+--------+----+------+
| topic|year| count|
+--------+----+------+
|covid-19|2017|606498|
|covid-19|2016|454678|
|covid-19|2011| 10517|
|covid-19|2008| 6193|
|covid-19|2015|510391|
|covid-19|2013| 29551|
I then need to sort by year and collect counts to a list so that they be in ascending order, by year:
df.orderBy('year').groupBy('topic').agg(collect_list('count').alias('counts'))
The issue is, since I order by year, the number of partitions used for this stage is the number of years in my dataset. I thus get a crazy bottleneck stage where 15 out of 300 executors are utilised, leading to obvious memory spills and disk spills, eventually failing the stage due to no space left on device for the overpopulated partitions.
Even more interesting is that I found a way to circumvent this which intuitively appears to be much less efficient, but actually does work, since no bottlenecks are created:
df.groupBy('topic').pivot('year', values=range(START, FINISH)).agg(first('count')) \
.select('topic', array([col(c) for c in range(START, FINISH)]).alias('counts'))
This leads to my desired output, which is an array of counts sorted by year.
Anyone with an explanation or idea why this happens, or how best to prevent this?
I found this answer which and this jira where it is basically suggested to 'add noise' to the sort by key to avoid these skew related issues.
I think it is worth mentioning that the pivot method is a better resolution than adding noise, and to my knowledge whenever sorting by a column that has a small range of values. would appreciate any info on this and alternate implementations.
Range Partitioning is used for Sorting, ordering, under water by Spark.
From the docs it is clear that the calculation for determining the number of partitions that will contain ranges of data for sorting subsequently via mapPartitions,
is based on sampling from the existing partitions prior to computing some heuristically optimal number of partitions for these computed ranges.
These ranges which are partitions may decrease the number of partitions as a range must be contained with a single partition - for the order by / sort to work. Via mapPartitions type approach.
This:
df.repartitionByRange(100, 'some_col1', 'some_colN')...
can help or of you order by more columns I suspect. But here it appears not to be the case based on your DF.
The question has nothing to do with pyspark, BTW.
Interesting point, but explainable: reduced partitions needing to hold more data via collect_list based on year, there are obviously more topics than years.
I'm processing a file each day with PySpark for contaning information about device navigation through the web. At the end of each month I want to use window functions in order to have the navigation journey for each device. It's a very slow processing, even with many nodes, so I'm looking for ways to speed it up.
My idea was to partition the data but I have 2 billion distinct keys, so partitionBy does not seem appropriate. Even bucketBy might not be a good choice because I create n buckets each day, so the files are not appended but for each day there are x parts of files that are created.
Does anyone have a solution ?
So here is an example of the export for each day (inside of each parquet file we find 9 partitions):
And here is the partitionBy query that we launch at the beggining of each month (compute_visit_number and compute_session_number are two udf that i've created on the notebook):
You want to ensure that each devices data is in the same partition to prevent exchanges when you do your window function. Or at least minimise the number of partitions the data could be in.
To do this I would create a column called partitionKey when you write the data - which contained a mod on the mc_device column - where the number you mod by is the number of partitions you want. Base this number of the size of the cluster that will run the end of month query. (If mc_device is not an integer then create a checksum first).
You can create a secondary partition on the date column if still needed.
Your end of month query should change:
w = Windows.partitionBy('partitionKey', 'mc_device').orderBy(event_time')
If you kept the date as a secondary partition column then repartition the dataframe to partitionKey only:
df = df.repartition('partitionKey')
At this point each devices data will be in the same partition and no exchanges should be needed. The sort should be faster and your query will hopefully complete in a sensible time.
If it is still slow you need more partitions when writing the data.
I am in the process of learning Cassandra as an alternative to SQL databases for one of the projects I am working for, that involves Big Data.
For the purpose of learning, I've been watching the videos offered by DataStax, more specifically DS220 which covers modeling data in Cassandra.
While watching one of the videos in the course series I was introduced to the concept of splitting partitions to manage partition size.
My current understanding is that Cassandra has a max logical capacity of 2B entries per partition, but a suggested max of a couple 100s MB per partition.
I'm currently dealing with large amounts of real-time financial data that I must store (time series), meaning I can easily fill out GBs worth of data in a day.
The video course talks about introducing an additional partition key in order to split a partition with the purpose or reducing the size per partition requirement.
The video pointed out to using either a time based key or an arbitrary "bucket" key that gets incremented when the number of manageable rows has been reached.
With that in mind, this led me to the following problem: given that partition keys are only used as equality criteria (ie. point to the partition to find records), how do I find all the records that end up being spread across multiple partitions without having to specify either the bucket or timestamp key?
For example, I may receive 1M records in a single day, which would likely go over the 100-500Mb partition limit, so I wouldn't be able to set a partition on a per date basis, that means that my daily data would be broken down into hourly partitions, or alternatively, into "bucketed" partitions (for balanced partition sizes). This means that all my daily data would be spread across multiple partitions splits.
Given this scenario, how do I go about querying for all records for a given day? (additional clustering keys could include a symbol for which I want to have the results for, or I want all the records for that specific day)
Any help would be greatly appreciated.
Thank you.
Basically this goes down to choosing right resolution for your data. I would say first step for you would be to determinate what is best fit for your data. Lets for sake of example take 1 hour as something that is good and question is how to fetch all records for particular date.
Your application logic will be slightly more complicated since you are trading simplicity for ability to store large amounts of data in distributed fashion. You take date which you need and issue 24 queries in a loop and glue data on application level. However when you glue that in can be huge (I do not know your presentation or export requirements so this can pull 1M to memory).
Other idea can be having one table as simple lookup table which has key of date and values of partition keys having financial data for that date. Than when you read you go first to lookup table to get keys and then to partitions having results. You can also store counter of values per partition key so you know what amount of data you expect.
All in all it is best to figure out some natural bucket in your data set and add it to date (organization, zip code etc.) and you can use trick with additional lookup table. This approach can be used for symbol you mentioned. You can have symbols as partition keys, clustering per date and values of partitions having results for that date as values. Than you query for symbol # on 29-10-2015 and you see partitions A, D and Z have results so you go to those partitions and get financial data from them and glue it together on application level.