How to decide number of buckets in Spark - apache-spark

I have read quite a few articles on bucketing in Spark but haven't still been able to get a clear picture of it. But moreover what i have understood till now is that "Bucketing is like partition inside a partition, and it is used for the candidates having very high cardinality which helps in avoiding reshuffling operation".
Even in Spark documentation, can't find enough explanation. Pasting an example from the documentation
peopleDF.write.bucketBy(42, "name").sortBy("age").saveAsTable("people_bucketed")
Unable to understand here, how the number '42' is decided for bucketing. Kindly help to understand the same. Also, any clearer explanation on bucketing would also be great.

42 is just like what is the meaning of life? An example therefore.
Spark Bucketing is handy for ETL in Spark whereby Spark Job A writes out the data for t1 according to Bucketing def and Spark Job B writes out data for t2 likewise and Spark Job C joins t1 and t2 using Bucketing definitions avoiding shuffles aka exchanges. Optimization.
There is no general formula. It depends on volumes, available executors, etc. The main point is avoiding shuffling. As a guideline defaults for JOINing and AGGr are set to 200, so 200 or greater could be an approach, but again how many resources do you have on your cluster?
But to satisfy your quest for knowledge one could argue that the 42 should really be set to number of Executors (= 1 core) you have allocated to the Spark Job/ App, leaving aside the issue of skewness.

Related

Why can coalesce lead to too few nodes for processing?

I am trying to understand spark partitions and in a blog I come across this passage
However, you should understand that you can drastically reduce the parallelism of your data processing — coalesce is often pushed up further in the chain of transformation and can lead to fewer nodes for your processing than you would like. To avoid this, you can pass shuffle = true. This will add a shuffle step, but it also means that the reshuffled partitions will be using full cluster resources if possible.
I understand that coalesce means to take the data on some of the least data containing executors and shuffle them to already existing executors via a hash partitioner. I am not able to understand what the author is trying to say in this para though. Can somebody please explain to me what is being said in this paragraph?
Coalesce has some not so obvious effects due to Spark
Catalyst.
E.g.
Let’s say you had a parallelism of 1000, but you only wanted to write
10 files at the end. You might think you could do:
load().map(…).filter(…).coalesce(10).save()
However, Spark’s will effectively push down the coalesce operation to
as early a point as possible, so this will execute as:
load().coalesce(10).map(…).filter(…).save()
You can read in detail here an excellent article, that I quote from, that I chanced upon some time ago: https://medium.com/airbnb-engineering/on-spark-hive-and-small-files-an-in-depth-look-at-spark-partitioning-strategies-a9a364f908
In summary: Catalyst treatment of coalesce can reduce concurrency early in the pipeline. This I think is what is being alluded to, though of course each case is different and JOIN and aggregating are not subject to such effects in general due to 200 default partitioning that applies for such Spark operations.
As what you have said in your question "coalesce means to take the data on some of the least data containing executors and shuffle them to already existing executors via a hash practitioner". This effectively means the following
The number of partitions have reduced
The main difference between repartition and coalesce is that in coalesce the movement of the data across the partitions is fewer than in repartition thus reducing the level of shuffle thus being more efficient.
Adding the property shuffle=true is just to distribute the data evenly across the nodes which is the same as using repartition(). You can use shuffle=true if you feel that your data might get skewed in the nodes after performing coalesce.
Hope this answers your question

How to spark partitionBy/bucketBy correctly?

Q1. Will adhoc (dynamic) repartition of the data a line before a join help to avoid shuffling or will the shuffling happen anyway at the repartition and there is no way to escape it?
Q2. should I repartition/partitionBy/bucketBy? what is the right approach if I will join according to column day and user_id in the future? (I am saving the results as hive tables with .write.saveAsTable). I guess to partition by day and bucket by user_id but that seems to create thousands of files (see Why is Spark saveAsTable with bucketBy creating thousands of files?)
Some 'guidance' off the top of my head, noting that title and body of text differ to a degree:
Question 1:
A JOIN will do any (hash) partitioning / repartitioning required automatically - if needed and if not using a Broadcast JOIN. You may
set the number of partitions for shuffling or use the default - 200.
There are more parties (DF's) to consider.
repartition is a transformation, so any up-front repartition may not be executed at all due to Catalyst optimization - see the physical plan generated from the .explain. That's the deal with lazy
evaluation - determining if something is necessary upon Action
invocation.
Question 2:
If you have a use case to JOIN certain input / output regularly, then using Spark's bucketBy is a good approach. It obviates shuffling. The
databricks docs show this clearly.
A Spark schema using bucketBy is NOT compatible with Hive. so these remain Spark only tables, unless this changed recently.
Using Hive partitioning as you state depend on push-down logic, partition pruning etc. It should work as well but you may have have
different number of partitions inside Spark framework after the read.
It's a bit more complicated than saying I have N partitions so I will
get N partitions on the initial read.

is it more efficient to cache a dataframe in on partition or more partitions

I'm persisting a dataFrame, and in the spark interface i see that this dataframe is partitioned in my 7 nodes.
My spark job have transformations with wide dependencies.
Could it be more performant to force the cache in only 1 partition ?
To avoid shuffle?
Thanks
There is a balance between number of partitions and therefore concurrency. Dare I say it, you are a little off-beam here. Meaning:
Too much partitioning makes no sense --> too much overhead.
Just one partition would mean a coalesce or re-partition and would lack parallel processing of what Spark offers to get the job done quicker, e.g. many workers in parallel loading supermarket shelves is faster than just you and I doing it on our own.
The truth is somewhere in between in terms of number of partitions which at scale needs to be estimated and trialled, and, shuffling can rarely be avoided unless you base the partitioning on what you read in from HDFS/Hadoop Source (e.g. KUDU) or S3, or from JDBC.

Spark Decision tree fit runs in 1 task

I am trying to "train" a DecisionTreeClassifier using Apache Spark running in a cluster in Amazon EMR. Even though I can see that there are around 50 Executors added and that the features are created by querying a Postgres database using SparkSQL and stored in a DataFrame.
The DesisionTree fit method takes for many hours even though the Dataset is not that big (10.000 db entries with a couple of hundreds of bytes each row).I can see that there is only one task for this so I assume this is the reason that it's been so slow.
Where should I look for the reason that this is running in one task?
Is it the way that I retrieve the data?
I am sorry if this is a bit vague but I don't know if the code that retrieves the data is relevant, or is it a parameter in the algorithm (although I didn't find anything online), or is it just Spark tuning?
I would appreciate any direction!
Thanks in advance.
Spark relies on data locality. It seems that all the data is located in a single place. Hence spark uses a single partition to process it. You could apply a repartition or state the number of partitions you would like to use at load time. I would also look into the decision tree Api and see if you can set the number of partitions for it specifically.
Basically, partitions are your level of parallelism.

How is task distributed in spark

I am trying to understand that when a job is submitted from the spark-submit and I have spark deployed system with 4 nodes how is the work distributed in spark. If there is large data set to operate on, I wanted to understand exactly in how many stages are the task divided and how many executors run for the job. Wanted to understand how is this decided for every stage.
It's hard to answer this question exactly, because there are many uncertainties.
Number of stages depends only on described workflow, which includes different kind of maps, reduces, joins, etc. If you understand it, you basically can read that right from the code. But most importantly that helps you to write more performant algorithms, because it's generally known the one have to avoid shuffles. For example, when you do a join, it requires shuffle - it's a boundary stage. This is pretty simple to see, you have to print rdd.toDebugString() and then look at indentation (look here), because indentation is a shuffle.
But with number of executors that's completely different story, because it depends on number of partitions. It's like for 2 partitions it requires only 2 executors, but for 40 ones - all 4, since you have only 4. But additionally number of partitions might depend on few properties you can provide at the spark-submit:
spark.default.parallelism parameter or
data source you use (f.e. for HDFS and Cassandra it is different)
It'd be a good to keep all of the cores in cluster busy, but no more (meaning single process only just one partition), because processing of each partition takes a bit of overhead. On the other hand if your data is skewed, then some cores would require more time to process bigger partitions, than others - in this case it helps to split data to more partitions so that all cores are busy roughly same amount of time. This helps with balancing cluster and throughput at the same time.

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