I am trying to get a better understanding of the Spark internals and I am not sure how to interpret the resulting DAG of a job.
Inspired to the example described at http://dev.sortable.com/spark-repartition/,
I run the following code in the Spark shell to obtain the list of prime numbers from 2 to 2 million.
val n = 2000000
val composite = sc.parallelize(2 to n, 8).map(x => (x, (2 to (n / x)))).flatMap(kv => kv._2.map(_ * kv._1))
val prime = sc.parallelize(2 to n, 8).subtract(composite)
prime.collect()
After executing I checked the SparkUI and observed the DAG in figure.
Now my question is: I call the function subtract only once, why does this operation appears
three times in the DAG?
Also, is there any tutorial that explains a bit how Spark creates these DAGs?
Thanks in advance.
subtract is a transformation which requires a shuffle:
First both RDDs have to be repartitioned using the same partitioner Local ("map-side") part of the transformation is marked as subtract in the stages 0 and 1. At this point both RDDs are converted to (item, null) pairs.
substract you see in the stage 2 happens after the shuffle when RDDs have been combined. This where items are filtered.
In general any operation which requires a shuffle will be executed in at least two stages (depending on the number of predecessors) and tasks belonging to each stage will be shown separately.
Related
I'm learning how Spark works inside Databricks. I understand how shuffling causes stages within jobs, but I don't understand what causes jobs. I thought the relationship was one job per action, but sometimes many jobs happen per action.
E.g.
val initialDF = spark
.read
.parquet("/mnt/training/wikipedia/pagecounts/staging_parquet_en_only_clean/")
val someDF = initialDF
.orderBy($"project")
someDF.show
triggers two jobs, one to peek at the schema and one to do the .show.
And the same code with .groupBy instead
val initialDF = spark
.read
.parquet("/mnt/training/wikipedia/pagecounts/staging_parquet_en_only_clean/")
val someDF = initialDF
.groupBy($"project").sum()
someDF.show
...triggers nine jobs.
Replacing .show with .count, the .groupBy version triggers two jobs, and the .orderBy version triggers three.
Sorry I can't share the data to make this reproducible, but was hoping to understand the rules of when jobs are created in abstract. Happy to share the results of .explain if that's helpful.
show without an argument shows the first 20 rows as a result.
When show is triggered on dataset, it gets converted to head(20) action which in turn get converted to limit(20) action .
show -> head -> limit
About limit
Spark executes limit in an incremental fashion until the limit query is satisfied.
In its first attempt, it tries to retrieve the required number of rows from one partition.
If the limit requirement was not satisfied, in the second attempt, it tries to retrieve the required number of rows from 4 partitions (determined by spark.sql.limit.scaleUpFactor, default 4). and after which 16 partitions are processed and so on until either the limit is satisfied or data is exhausted.
In each of the attempts, a separate job is spawned.
code reference: https://github.com/apache/spark/blob/master/sql/core/src/main/scala/org/apache/spark/sql/execution/SparkPlan.scala#L365
Normally it is 1:1 as you state. That is to say, 1 Action results in 1 Job with 1..N Stages with M Tasks per Stage, and Stages which may be skipped.
However, some Actions trigger extra Jobs 'under water'. E.g. pivot: if you pass only the columns as parameter and not the values for the pivot, then Spark has to fetch all the distinct values first so as to generate columns, performing a collect, i.e. an extra Job.
show is also a special case of extra Job(s) being generated.
Recently I run a spark program and find that the take action seems not to trigger all tasks to run in spark.
When I use take, the result on spark UI is like this
While when I use count,
Note the difference between task numbers.
What I know is the action trigger lazy computations, but I can't find any related issues documented. I need some clarification, thanks.
It is normal. take iteratively evaluates partitions until it collects requested number of records. With narrow dependencies (maps, flatMaps, filters) it can evaluate only a minimal number of partitions.
For example:
sc.range(0, 100, 1, 5).map(_ % 11).filter(_ >= 2).take(1)
requires only one partition to be evaluated
Only wide dependencies (operations which require shuffle like any byKey or sorting) require evaluating more partitions:
sc.range(0, 100, 1, 5).map(_ % 11).map((_, 1)).reduceByKey(_ + _).take(1)
will trigger 6 tasks (5 for all upstream partitions and 1 for take).
I'm trying to understand the jobs that get created by Spark for simple first() vs collect() operations.
Given the code:
myRDD = spark.sparkContext.parallelize(['A', 'B', 'C'])
def func(d):
return d + '-foo'
myRDD = myRDD.map(func)
My RDD is split across 16 partitions:
print(myRDD.toDebugString())
(16) PythonRDD[24] at RDD at PythonRDD.scala:48 []
| ParallelCollectionRDD[23] at parallelize at PythonRDD.scala:475 []
If I call:
myRDD.collect()
I get 1 job with 16 tasks created. I assume this is one task per partition.
However, if I call:
myRDD.first()
I get 3 jobs, with 1, 4, and 11 tasks created. Why have 3 jobs been created?
I'm running spark-2.0.1 with a single 16-core executor, provisioned by Mesos.
It is actually pretty smart Spark behaviour. Your map() is transformation (it is lazy-evaluated) and both first() and collect() are actions (terminal operations). All transformations are applied to the data in time you call actions.
When you call first() then spark tries to perform as low number of operations (transformations) as possible. First, it tries one random partition. If there are no results, it takes 4 times partitions more and calculates. Again, if there are no results, spark takes 4 times partitions more (5 * 4) and again tries to get any result.
In your case in this third try you have only 11 untouched partitions (16 - 1 - 4). If you have more data in RDD or less number of partitions, spark probably can find the first() result sooner.
Let's assume for the following that only one Spark job is running at every point in time.
What I get so far
Here is what I understand what happens in Spark:
When a SparkContext is created, each worker node starts an executor.
Executors are separate processes (JVM), that connects back to the driver program. Each executor has the jar of the driver program. Quitting a driver, shuts down the executors. Each executor can hold some partitions.
When a job is executed, an execution plan is created according to the lineage graph.
The execution job is split into stages, where stages containing as many neighbouring (in the lineage graph) transformations and action, but no shuffles. Thus stages are separated by shuffles.
I understand that
A task is a command sent from the driver to an executor by serializing the Function object.
The executor deserializes (with the driver jar) the command (task) and executes it on a partition.
but
Question(s)
How do I split the stage into those tasks?
Specifically:
Are the tasks determined by the transformations and actions or can be multiple transformations/actions be in a task?
Are the tasks determined by the partition (e.g. one task per per stage per partition).
Are the tasks determined by the nodes (e.g. one task per stage per node)?
What I think (only partial answer, even if right)
In https://0x0fff.com/spark-architecture-shuffle, the shuffle is explained with the image
and I get the impression that the rule is
each stage is split into #number-of-partitions tasks, with no regard for the number of nodes
For my first image I'd say that I'd have 3 map tasks and 3 reduce tasks.
For the image from 0x0fff, I'd say there are 8 map tasks and 3 reduce tasks (assuming that there are only three orange and three dark green files).
Open questions in any case
Is that correct? But even if that is correct, my questions above are not all answered, because it is still open, whether multiple operations (e.g. multiple maps) are within one task or are separated into one tasks per operation.
What others say
What is a task in Spark? How does the Spark worker execute the jar file? and How does the Apache Spark scheduler split files into tasks? are similar, but I did not feel that my question was answered clearly there.
You have a pretty nice outline here. To answer your questions
A separate task does need to be launched for each partition of data for each stage. Consider that each partition will likely reside on distinct physical locations - e.g. blocks in HDFS or directories/volumes for a local file system.
Note that the submission of Stages is driven by the DAG Scheduler. This means that stages that are not interdependent may be submitted to the cluster for execution in parallel: this maximizes the parallelization capability on the cluster. So if operations in our dataflow can happen simultaneously we will expect to see multiple stages launched.
We can see that in action in the following toy example in which we do the following types of operations:
load two datasources
perform some map operation on both of the data sources separately
join them
perform some map and filter operations on the result
save the result
So then how many stages will we end up with?
1 stage each for loading the two datasources in parallel = 2 stages
A third stage representing the join that is dependent on the other two stages
Note: all of the follow-on operations working on the joined data may be performed in the same stage because they must happen sequentially. There is no benefit to launching additional stages because they can not start work until the prior operation were completed.
Here is that toy program
val sfi = sc.textFile("/data/blah/input").map{ x => val xi = x.toInt; (xi,xi*xi) }
val sp = sc.parallelize{ (0 until 1000).map{ x => (x,x * x+1) }}
val spj = sfi.join(sp)
val sm = spj.mapPartitions{ iter => iter.map{ case (k,(v1,v2)) => (k, v1+v2) }}
val sf = sm.filter{ case (k,v) => v % 10 == 0 }
sf.saveAsTextFile("/data/blah/out")
And here is the DAG of the result
Now: how many tasks ? The number of tasks should be equal to
Sum of (Stage * #Partitions in the stage)
This might help you better understand different pieces:
Stage: is a collection of tasks. Same process running against
different subsets of data (partitions).
Task: represents a unit of
work on a partition of a distributed dataset. So in each stage,
number-of-tasks = number-of-partitions, or as you said "one task per
stage per partition”.
Each executer runs on one yarn container, and
each container resides on one node.
Each stage utilizes multiple executers, each executer is allocated multiple vcores.
Each vcore can execute exactly one task at a time
So at any stage, multiple tasks could be executed in parallel. number-of-tasks running = number-of-vcores being used.
If I understand correctly there are 2 ( related ) things that confuse you:
1) What determines the content of a task?
2) What determines the number of tasks to be executed?
Spark's engine "glues" together simple operations on consecutive rdds, for example:
rdd1 = sc.textFile( ... )
rdd2 = rdd1.filter( ... )
rdd3 = rdd2.map( ... )
rdd3RowCount = rdd3.count
so when rdd3 is (lazily) computed, spark will generate a task per partition of rdd1 and each task will execute both the filter and the map per line to result in rdd3.
The number of tasks is determined by the number of partitions. Every RDD has a defined number of partitions. For a source RDD that is read from HDFS ( using sc.textFile( ... ) for example ) the number of partitions is the number of splits generated by the input format. Some operations on RDD(s) can result in an RDD with a different number of partitions:
rdd2 = rdd1.repartition( 1000 ) will result in rdd2 having 1000 partitions ( regardless of how many partitions rdd1 had ).
Another example is joins:
rdd3 = rdd1.join( rdd2 , numPartitions = 1000 ) will result in rdd3 having 1000 partitions ( regardless of partitions number of rdd1 and rdd2 ).
( Most ) operations that change the number of partitions involve a shuffle, When we do for example:
rdd2 = rdd1.repartition( 1000 )
what actually happens is the task on each partition of rdd1 needs to produce an end-output that can be read by the following stage so to make rdd2 have exactly 1000 partitions ( How they do it? Hash or Sort ). Tasks on this side are sometimes referred to as "Map ( side ) tasks".
A task that will later run on rdd2 will act on one partition ( of rdd2! ) and would have to figure out how to read/combine the map-side outputs relevant to that partition. Tasks on this side are sometimes referred to as "Reduce ( side ) tasks".
The 2 questions are related: the number of tasks in a stage is the number of partitions ( common to the consecutive rdds "glued" together ) and the number of partitions of an rdd can change between stages ( by specifying the number of partitions to some shuffle causing operation for example ).
Once the execution of a stage commences, its tasks can occupy task slots. The number of concurrent task-slots is numExecutors * ExecutorCores. In general, these can be occupied by tasks from different, non-dependent stages.
I am running Spark on only one node with Parallelism = 1 in order to compare its performance with a single-threaded application. I'm wondering if Spark is still using a Shuffle although it does not run in parallel. So if e.g. the following command is executed:
val counts = text_file.flatMap(line => line.split(" "))
.map(word => (word, 1))
.reduceByKey(_+_)
I get the following output from the Spark Interactive Scala Shell:
counts: org.apache.spark.rdd.RDD[(String, Int)] = ShuffledRDD[10]
at reduceByKey at <console>:41
So should I assume, that a Shuffle was used before reduceByKey? And does [10] actually has any meaning?
I'm wondering if Spark is still using a Shuffle although it does not run in parallel.
Yes it does. It is worth noting that even with a single core number of partitions may be much larger than one. For example if RDD is created using SparkContext.textFile number of partitions depends on a size of the file system block.
So should I assume, that a Shuffle was used before reduceByKey
No, shuffle is a fundamental part of the reduceByKey logic so it was used during reduceByKey, not before. Simplifying things a little bit shuffle is an equivalent of creating a hash table. Assuming only a single partition it doesn't perform any useful task but is still present.
And does [10] actually has any meaning?
It is a unique (in the current SparkContext) ID for a given RDD. For example if RDD is persisted then the number you see should be a key in SparkContext.getPersistentRDDs.