I have a large DataFrame (billions of records) that I am applying a filter to which reduces the size of the DataFrame considerably (tens of millions of records). Of course this leads to highly skewed partitions which is destroying the database write performance.
I know that the data needs to be repartitioned. I created a column consisting of random integers between 1 and 36. I have 36 cores available and 36 corresponding partitions. The idea was that I could repartition on this random number column to end up with 36 relatively even partitions. The issue is that the repartition alone takes over 90 minutes. I call the repartition operation right before the database write. Am I missing anything here?
Thank you!
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 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 am trying to remove duplicates in spark dataframes by using dropDuplicates() on couple of columns. But job is getting hung due to lots of shuffling involved and data skew. I have used 5 cores and 30GB of memory to do this. Data on which I am performing dropDuplicates() is about 12 million rows.
Please suggest me the most optimal way to remove duplicates in spark, considering data skew and shuffling involved.
Delete duplicate operations is an expensive operation as it compare values from one RDD to all other RDDs and tries to consolidate the results. Considering the size of your data results can time consuming.
I would recommend groupby transformation on the columns of your dataframe followed by commit action. This way only the consolidated results from your RDD will be compared with other RDD that too lazily and then you can request the result through any of the action like commit / show etc
transactions.groupBy("col1”,”col2").count.sort($"count".desc).show
distance():
df.select(['id', 'name']).distinct().show()
dropDuplicates()
df.dropDuplicates(['id', 'name']).show()
dropDuplicates() is the way to go if you want to drop duplicates over a subset of columns, but at the same time you want to keep all the columns of the original structure.
Suppose there is a dataset with some number of rows.
I need to find out the Heterogeneity i.e.
distinct number of rows divide by total number of rows.
Please help me with spark query to execute the same.
Dataset and dataframe supports distinct function which finds distinct rows in the dataset.
So essentially you need to do
val heterogeneity = dataset.distinct.count / dataset.count
Only thing is if the dataset is big the distinct could be expensive and you might need to set the spark shuffle partition correctly.
In my code, I have a sequence of dataframes where I want to filter out the dataframe's which are empty. I'm doing something like:
Seq(df1, df2).map(df => df.count() > 0)
However, this is taking extremely long and is consuming around 7 minutes for approximately 2 dataframe's of 100k rows each.
My question: Why is Spark's implementation of count() is slow. Is there a work-around?
Count is a lazy operation. So it does not matter how big is your dataframe. But if you have too many costly operations on the data to get this dataframe, then once the count is called spark would actually do all the operations to get these dataframe.
Some of the costly operations may be operations which needs shuffling of data. Like groupBy, reduce etc.
So my guess is you have some complex processing to get these dataframes or your initial data which you used to get this dataframe is too huge.