Of course I know the spark.sql.shuffle.partitionsconfig,
but for example, when I set this config 300 on the small dataset which just has 200 rows, the config is not valid, the actual partition number is just 2,
anthor example, I set this config 3000 on the dataset which has 30 billion rows, the config is not valid too, the actual partition number is just 600,
we see that when we set a big value partitions config on a small dataset, the config would be not valid,
So I just want to know How does Spark decide the partitions number of the next stage when shuffle in SparkSQL? Or How to force this config to be valid ?
My Spark SQL is just like below:
set spark.sql.shuffle.partitions=3000;
with base_data as (
select
device_id
from
table_name
where
dt = '20210621'
distribute by
rand()
)
select count(1) from base_data
In general Narrow transformation does not change number of partitions .
Wide transformations transformation does not change number of partitions.
Narrow transformation In Narrow transformation, all the elements that are required to compute the records in single partition live in the single partition of parent RDD. A limited subset of partition is used to calculate the result. Narrow transformations are the result of map(), filter().
Wide transformation — In wide transformation, all the elements that are required to compute the records in the single partition may live in many partitions of parent RDD. The partition may live in many partitions of parent RDD. Wide transformations are the result of groupbyKey and reducebyKey.
Update after question change:
you can assume "spark.sql.shuffle.partitions" as a query hint where we are forcing executors that make that number of partitions for joins or aggregations in my view we should not play with this value unless we are very sure that what are no of grouping key would be.
This will make unnecessary shuffling of data over the network.
Related
What I understand is
When we repartition any dataframe with value n, data will continue to remain on those n partitions, until you hit any shuffle stages or other value of repartition or coalesce.
For Shuffle, it only comes into the play when you hit any shuffle stages and data will continue to remain on those partitions until you hit coalesce or repartition.
I am right ?
If yes then, can any one point out a striking difference?
TLDR - Repartition is invoked as per developer's need but shuffle is done when there is a logical demand
I assume you're talking about config property spark.sql.shuffle.partitions and method .repartition.
As data distribution is an important aspect in any distributed environment, which not only governs parallelism but can also create adverse impacts if the distribution is uneven. However, repartitioning itself is a costly operation as it involves heavy movement of data (i.e. Shuffling). The .repartition method is used to explicitly repartition the data into new partitions - meaning to increase or decrease the number of partitions in the program based on your need. You can invoke this whenever you want.
As opposed to this, spark.sql.shuffle.partitions is a configuration property that governs the number of partitions created when a data movement happens as a result of operations like aggregations and joins.
Configures the number of partitions to use when shuffling data for
joins or aggregations.
When you're performing transformations other than join or aggregation, the above configuration won't have any impact on the number of partitions the new Dataframe will have.
Your confusion between the two is due to both operations involving shuffling. While that is true, the former (i.e. repartition) is an explicit operation where the user is dictating the framework to increase or decrease the number of partitions - which in turn causes shuffling, while in case of joins/aggregation - the shuffling is caused by the operation itself.
Basically -
Joins/Aggregations cause shuffling which causes repartitioning
repartition is asked thus, shuffling has to be done
Another method coalesce make the difference clearer.
For reference, coalesce is a variant of repartition which can only lower the number of partitions, not necessarily equal in size. As it already knows the number of partitions are only to be decreased, it can perform it with minimal shuffling (just join two adjacent partitions until the number is met).
Consider your dataframe has 4 partitions but has data only in 2 of them, thus you decide to reduce the number of partitions to 2. When using coalesce spark tries to achieve this without shuffling or with minimal shuffling.
df.rdd().getNumPartitions(); // Returns 4 with size 0, 0, 2, 4
df=df.coalesce(2); // Decrease partitions to 2
df.rdd().getNumPartitions(); // Returns 2 now with size 2, 4
So there was no shuffling involved. While the following
df1.rdd().getNumPartitions() // Returns 4
df2.rdd().getNumPartitions() // Returns 8
df1.join(df2).rdd().getNumPartitions() // Returns 200
As you've performed a join it'll always return the number of partitions based on spark.sql.shuffle.partitions
I am a newbie in spark and I am trying to understand shuffle partition and repartition function. But i still dont understand how they are different. Both reduces the number of partition??
Thank you
The biggest difference between shuffle partition and repartition is when things are defined.
The configuration spark.sql.shuffle.partitions is a property and according to the documentation
Configures the number of partitions to use when shuffling data for joins or aggregations.
That means, every time you run a Join or any type of aggregation in spark that will shuffle the data according to the configuration, where the default value is 200. So if you join two datasets the number of partitions in the shuffle will be 200.
The repartition(numPartitions, *cols) function is applied during an execution, where you can define how many partitions you will write, that usually is for output writing based in partition columns or just number. The example in the documentation is pretty good to show.
So in general, Shuffle Partition is for Joins and Aggregations during the execution. Repartition is for number of output files, based in number or partition column.
Lets say I have a dataset with 20 partitions when I was going to read some data. Then I do aggregate operation on that dataset , which would make no of partitions to be 200(because of default shuffle partitions size). Now without calling any action on that dataset so far , I apply coalesce on that same data set giving 30 partitions in coalesce operation and then call some spark action on that dataset.
So my question is, how many partitions will be in action while that dataset would be having its aggregate operation ? Will it be 30 partitions(because that was the coalesce partitions given ) only or 200 shuffle partitions ?
Editing to provide more clarification on my question:
I understand that coalesce operation in itself will not do shuffle unless we drastically changed no of partitions. I also understand that final dataset will have numPartitions size only , but my question is if I change no of partitions before calling any action on that dataframne , would that resulting action will operate on the final no of partitions we had given(in my case 30) or it will also honor intermediate partitions size that we had given in aggregate operation. So in all, I am mainly looking whether aggregation will be done with 200 partitions and then coalesce will be applied or aggregation will also be performed with 30(in my case) partitions only.
Yes, your final action will operate on partitions generated by coalesce, like in your case it's 30.
As we know there is two types of transformation narrow and wide.
Narrow transformation don't do shuffling and don't do repartitioning but wide shuffling shuffle the data between node and generate new partition.
So if you check coalesce is a wide transformation and it will create a new stage before proceeding for next transformation or action and next stage will work on shuffle partition generated by coalesce.
So yes, your actions will going to work on 30 partitions.
https://www.google.com/amp/s/data-flair.training/blogs/spark-rdd-operations-transformations-actions/amp/
Coalesce
Returns a new SparkDataFrame that has exactly numPartitions
partitions. This operation results in a narrow dependency, e.g. if you
go from 1000 partitions to 100 partitions, there will not be a
shuffle, instead each of the 100 new partitions will claim 10 of the
current partitions. If a larger number of partitions is requested, it
will stay at the current number of partitions.
However, if you're doing a drastic coalesce on a SparkDataFrame, e.g.
to numPartitions = 1, this may result in your computation taking place
on fewer nodes than you like (e.g. one node in the case of
numPartitions = 1). To avoid this, call repartition. This will add a
shuffle step, but means the current upstream partitions will be
executed in parallel (per whatever the current partitioning is).
https://spark.apache.org/docs/2.2.1/api/R/coalesce.html
Coalesce: Shuffle the data into existing number of partitions.
https://medium.com/#mrpowers/managing-spark-partitions-with-coalesce-and-repartition-4050c57ad5c4#.36o8a7b5j
In many posts there is the statement - as shown below in some form or another - due to some question on shuffling, partitioning, due to JOIN, AGGR, whatever, etc.:
... In general whenever you do a spark sql aggregation or join which shuffles data this is the number of resulting partitions = 200.
This is set by spark.sql.shuffle.partitions. ...
So, my question is:
Do we mean that if we have set partitioning at 765 for a DF, for example,
That the processing occurs against 765 partitions, but that the output is coalesced / re-partitioned standardly to 200 - referring here to word resulting?
Or does it do the processing using 200 partitions after coalescing / re-partitioning to 200 partitions before JOINing, AGGR?
I ask as I never see a clear viewpoint.
I did the following test:
// genned a DS of some 20M short rows
df0.count
val ds1 = df0.repartition(765)
ds1.count
val ds2 = df0.repartition(765)
ds2.count
sqlContext.setConf("spark.sql.shuffle.partitions", "765")
// The above not included on 1st run, the above included on 2nd run.
ds1.rdd.partitions.size
ds2.rdd.partitions.size
val joined = ds1.join(ds2, ds1("time_asc") === ds2("time_asc"), "outer")
joined.rdd.partitions.size
joined.count
joined.rdd.partitions.size
On the 1st test - not defining sqlContext.setConf("spark.sql.shuffle.partitions", "765"), the processing and num partitions resulted was 200. Even though SO post 45704156 states it may not apply to DFs - this is a DS.
On the 2nd test - defining sqlContext.setConf("spark.sql.shuffle.partitions", "765"), the processing and num partitions resulted was 765. Even though SO post 45704156 states it may not apply to DFs - this is a DS.
It is a combination of both your guesses.
Assume you have a set of input data with M partitions and you set shuffle partitions to N.
When executing a join, spark reads your input data in all M partitions and re-shuffle the data based on the key to N partitions. Imagine a trivial hashpartitioner, the hash function applied on the key pretty much looks like A = hashcode(key) % N, and then this data is re-allocated to the node in charge of handling the Ath partition. Each node can be in charge of handling multiple partitions.
After shuffling, the nodes will work to aggregate the data in partitions they are in charge of. As no additional shuffling needs to be done here, the nodes can produce the output directly.
So in summary, your output will be coalesced to N partitions, however it is coalesced because it is processed in N partitions, not because spark applies one additional shuffle stage to specifically repartition your output data to N.
Spark.sql.shuffle.partitions is the parameter which decides the number of partitions while doing shuffles like joins or aggregation i.e where data movement is there across the nodes. The other part spark.default.parallelism will be calculated on basis of your data size and max block size, in HDFS it’s 128mb. So if your job does not do any shuffle it will consider the default parallelism value or if you are using rdd you can set it by your own. While shuffling happens it will take 200.
Val df = sc.parallelize(List(1,2,3,4,5),4).toDF()
df.count() // this will use 4 partitions
Val df1 = df
df1.except(df).count // will generate 200 partitions having 2 stages
I am writing my dataframe like below
df.write().format("com.databricks.spark.avro").save("path");
However I am getting around 200 files where around 30-40 files are empty.I can understand that it might be due to empty partitions. I then updated my code like
df.coalesce(50).write().format("com.databricks.spark.avro").save("path");
But I feel it might impact performance. Is there any other better approach to limit number of output files and remove empty files
You can remove the empty partitions in your RDD before writing by using repartition method.
The default partition is 200.
The suggested number of partition is number of partitions = number of cores * 4
repartition your dataframe using this method. To eliminate skew and ensure even distribution of data choose column(s) in your dataframe with high cardinality (having unique number of values in the columns) for the partitionExprs argument to ensure even distribution.
As default no. of RDD partitions is 200; you have to do shuffle to remove skewed partitions.
You can either use repartition method on the RDD; or make use of DISTRIBUTE BY clause on dataframe - which will repartition along with distributing data among partitions evenly.
def repartition(numPartitions: Int, partitionExprs: Column*): Dataset[T]
Returns dataset instance with proper partitions.
You may use repartitionAndSortWithinPartitions - which can improve compression ratio.