When a resilient distributed dataset (RDD) is created from a text file or collection (or from another RDD), do we need to call "cache" or "persist" explicitly to store the RDD data into memory? Or is the RDD data stored in a distributed way in the memory by default?
val textFile = sc.textFile("/user/emp.txt")
As per my understanding, after the above step, textFile is a RDD and is available in all/some of the node's memory.
If so, why do we need to call "cache" or "persist" on textFile RDD then?
Most RDD operations are lazy. Think of an RDD as a description of a series of operations. An RDD is not data. So this line:
val textFile = sc.textFile("/user/emp.txt")
It does nothing. It creates an RDD that says "we will need to load this file". The file is not loaded at this point.
RDD operations that require observing the contents of the data cannot be lazy. (These are called actions.) An example is RDD.count — to tell you the number of lines in the file, the file needs to be read. So if you write textFile.count, at this point the file will be read, the lines will be counted, and the count will be returned.
What if you call textFile.count again? The same thing: the file will be read and counted again. Nothing is stored. An RDD is not data.
So what does RDD.cache do? If you add textFile.cache to the above code:
val textFile = sc.textFile("/user/emp.txt")
textFile.cache
It does nothing. RDD.cache is also a lazy operation. The file is still not read. But now the RDD says "read this file and then cache the contents". If you then run textFile.count the first time, the file will be loaded, cached, and counted. If you call textFile.count a second time, the operation will use the cache. It will just take the data from the cache and count the lines.
The cache behavior depends on the available memory. If the file does not fit in the memory, for example, then textFile.count will fall back to the usual behavior and re-read the file.
I think the question would be better formulated as:
When do we need to call cache or persist on a RDD?
Spark processes are lazy, that is, nothing will happen until it's required.
To quick answer the question, after val textFile = sc.textFile("/user/emp.txt") is issued, nothing happens to the data, only a HadoopRDD is constructed, using the file as source.
Let's say we transform that data a bit:
val wordsRDD = textFile.flatMap(line => line.split("\\W"))
Again, nothing happens to the data. Now there's a new RDD wordsRDD that contains a reference to testFile and a function to be applied when needed.
Only when an action is called upon an RDD, like wordsRDD.count, the RDD chain, called lineage will be executed. That is, the data, broken down in partitions, will be loaded by the Spark cluster's executors, the flatMap function will be applied and the result will be calculated.
On a linear lineage, like the one in this example, cache() is not needed. The data will be loaded to the executors, all the transformations will be applied and finally the count will be computed, all in memory - if the data fits in memory.
cache is useful when the lineage of the RDD branches out. Let's say you want to filter the words of the previous example into a count for positive and negative words. You could do this like that:
val positiveWordsCount = wordsRDD.filter(word => isPositive(word)).count()
val negativeWordsCount = wordsRDD.filter(word => isNegative(word)).count()
Here, each branch issues a reload of the data. Adding an explicit cache statement will ensure that processing done previously is preserved and reused. The job will look like this:
val textFile = sc.textFile("/user/emp.txt")
val wordsRDD = textFile.flatMap(line => line.split("\\W"))
wordsRDD.cache()
val positiveWordsCount = wordsRDD.filter(word => isPositive(word)).count()
val negativeWordsCount = wordsRDD.filter(word => isNegative(word)).count()
For that reason, cache is said to 'break the lineage' as it creates a checkpoint that can be reused for further processing.
Rule of thumb: Use cache when the lineage of your RDD branches out or when an RDD is used multiple times like in a loop.
Do we need to call "cache" or "persist" explicitly to store the RDD data into memory?
Yes, only if needed.
The RDD data stored in a distributed way in the memory by default?
No!
And these are the reasons why :
Spark supports two types of shared variables: broadcast variables, which can be used to cache a value in memory on all nodes, and accumulators, which are variables that are only “added” to, such as counters and sums.
RDDs support two types of operations: transformations, which create a new dataset from an existing one, and actions, which return a value to the driver program after running a computation on the dataset. For example, map is a transformation that passes each dataset element through a function and returns a new RDD representing the results. On the other hand, reduce is an action that aggregates all the elements of the RDD using some function and returns the final result to the driver program (although there is also a parallel reduceByKey that returns a distributed dataset).
All transformations in Spark are lazy, in that they do not compute their results right away. Instead, they just remember the transformations applied to some base dataset (e.g. a file). The transformations are only computed when an action requires a result to be returned to the driver program. This design enables Spark to run more efficiently – for example, we can realize that a dataset created through map will be used in a reduce and return only the result of the reduce to the driver, rather than the larger mapped dataset.
By default, each transformed RDD may be recomputed each time you run an action on it. However, you may also persist an RDD in memory using the persist (or cache) method, in which case Spark will keep the elements around on the cluster for much faster access the next time you query it. There is also support for persisting RDDs on disk, or replicated across multiple nodes.
For more details please check the Spark programming guide.
Below are the three situations you should cache your RDDs:
using an RDD many times
performing multiple actions on the same RDD
for long chains of (or very expensive) transformations
Adding another reason to add (or temporarily add) cache method call.
for debug memory issues
with cache method, spark will give debugging informations regarding the size of the RDD. so in the spark integrated UI, you will get RDD memory consumption info. and this proved very helpful diagnosing memory issues.
Related
There is the following trick how to trim Apache Spark dataframe lineage, especially for iterative computations:
def getCachedDataFrame(df: DataFrame): DataFrame = {
val rdd = df.rdd.cache()
df.sqlContext.createDataFrame(rdd, df.schema)
}
It looks like some sort of pure magic, but right now I'm wondering why do we need to invoke cache() method on RDD? What is the purpose of having cache in this lineage trimming logic?
To understand the purpose of caching, it helps to understand the different types of RDD operations: transformations and actions. From the docs:
RDDs support two types of operations: transformations, which create a new dataset from an existing one, and actions, which return a value to the driver program after running a computation on the dataset.
Also consider this bit:
All transformations in Spark are lazy, in that they do not compute their results right away. Instead, they just remember the transformations applied to some base dataset (e.g. a file). The transformations are only computed when an action requires a result to be returned to the driver program. This design enables Spark to run more efficiently. For example, we can realize that a dataset created through map will be used in a reduce and return only the result of the reduce to the driver, rather than the larger mapped dataset.
So Spark's transformations (like map for example) are all lazy, because this helps Spark in being smart about which calculations need to be done while creating a query plan.
What does this have to do with caching?
Consider the following code:
// Reading in some data
val df = spark.read.parquet("some_big_file.parquet")
// Applying some transformations on the data (lazy operations here)
val cleansedDF = df
.filter(filteringFunction)
.map(cleansingFunction)
// Executing an action, triggering the transformations to be calculated
cleansedDF.write.parquet("output_file.parquet")
// Executing another action, triggering the transformations to be calculated
// again
println(s"You have ${cleansedDF.count} rows in the cleansed data")
In here, we reading in some file, applying some transformations and applying 2 actions to the same dataframe: cleansedDF.write.parquet and cleansedDF.count.
As the comments in the code explain, if we run the code like this we will be actually computing those transformations twice. Since the transformations are lazy, they will only get executed if an action requires them to be executed.
How can we prevent this double calculation? With caching: we can tell Spark to keep "save" result of some transformations so that they don't have to be calculated multiple times. This can be either on disk/memory/....
So with this knowledge, our code could be something like this:
// Reading in some data
val df = spark.read.parquet("some_big_file.parquet")
// Applying some transformations on the data (lazy operations here) AND caching the result of this calculation
val cleansedDF = df
.filter(filteringFunction)
.map(cleansingFunction)
.cache
// Executing an action, triggering the transformations to be calculated AND the results to be cached
cleansedDF.write.parquet("output_file.parquet")
// Executing another action, reusing the cached data
println(s"You have ${cleansedDF.count} rows in the cleansed data")
I've adjusted the comments in this code block to highlight the difference with the previous block.
Note that .persist also exists. With .cache you use the default storage level, with .persist you can specify which storage level as this SO answer nicely explains.
Hope this helps!
I have a large parquet dataset that I am reading with Spark. Once read, I filter for a subset of rows which are used in a number of functions that apply different transformations:
The following is similar but not exact logic to what I'm trying to accomplish:
df = spark.read.parquet(file)
special_rows = df.filter(col('special') > 0)
# Thinking about adding the following line
special_rows.cache()
def f1(df):
new_df_1 = df.withColumn('foo', lit(0))
return new_df_1
def f2(df):
new_df_2 = df.withColumn('foo', lit(1))
return new_df_2
new_df_1 = f1(special_rows)
new_df_2 = f2(special_rows)
output_df = new_df_1.union(new_df_2)
output_df.write.parquet(location)
Because a number of functions might be using this filtered subset of rows, I'd like to cache or persist it in order to potentially speed up execution speed / memory consumption. I understand that in the above example, there is no action called until my final write to parquet.
My questions is, do I need to insert some sort of call to count(), for example, in order to trigger the caching, or if Spark during that final write to parquet call will be able to see that this dataframe is being used in f1 and f2 and will cache the dataframe itself.
If yes, is this an idiomatic approach? Does this mean in production and large scale Spark jobs that rely on caching, random operations that force an action on the dataframe pre-emptively are frequently used, such as a call to count?
there is no action called until my final write to parquet.
and
Spark during that final write to parquet call will be able to see that this dataframe is being used in f1 and f2 and will cache the dataframe itself.
are correct. If you do output_df.explain(), you will see the query plan, which will show that what you said is correct.
Thus, there is no need to do special_rows.cache(). Generally, cache is only necessary if you intend to reuse the dataframe after forcing Spark to calculate something, e.g. after write or show. If you see yourself intentionally calling count(), you're probably doing something wrong.
You might want to repartition after running special_rows = df.filter(col('special') > 0). There can be a large number of empty partitions after running a filtering operation, as explained here.
The new_df_1 will make cache special_rows which will be reused by new_df_2 here new_df_1.union(new_df_2). That's not necessarily a performance optimization. Caching is expensive. I've seen caching slow down a lot of computations, even when it's being used in a textbook manner (i.e. caching a DataFrame that gets reused several times downstream).
Counting does not necessarily make sure the data is cached. Counts avoid scanning rows whenever possible. They'll use the Parquet metadata when they can, which means they don't cache all the data like you might expect.
You can also "cache" data by writing it to disk. Something like this:
df.filter(col('special') > 0).repartition(500).write.parquet("some_path")
special_rows = spark.read.parquet("some_path")
To summarize, yes, the DataFrame will be cached in this example, but it's not necessarily going to make your computation run any faster. It might be better to have no cache or to "cache" by writing data to disk.
I am curious to know when i need to persist my dataframe in spark and when not. Cases:-
If i need data from file ( Do i need to persist it? if i apply repetitive count like:-
val df=spark.read.json("file://root/Download/file.json")
df.count
df.count
Do i need to persist df?? because according to me it should store df in memory after first count and use same df in second count. Record in file is 4 , Because when i practically check it , it read file again and again, So why spark doesn't store it in memory
Second question is in spark read is an action or transformation?
DataFrames by design are immutable, so every transformation done on them would create a new data frame altogether. A spark pipeline generally involves multiple transformations leading to multiple data frames being created. If spark stores all of these data frames, the memory requirement would be huge. So spark leaves the responsibility of persisting data frames to the user. Whichever data frame you are planning on re using, you can persist them and later unpersist them when done.
I don't think we can define spark read as an action or a transformation. Action/Transformation is applied over a data frame. To identify the difference, you should remember that the transformation operation will return a new dataframe while action will return some value/s.
Problem: I want to import data into Spark EMR from S3 using:
data = sqlContext.read.json("s3n://.....")
Is there a way I can set the number of nodes that Spark uses to load and process the data? This is an example of how I process the data:
data.registerTempTable("table")
SqlData = sqlContext.sql("SELECT * FROM table")
Context: The data is not too big, takes a long time to load into Spark and also to query from. I think Spark partitions the data into too many nodes. I want to be able to set that manually. I know when dealing with RDDs and sc.parallelize I can pass the number of partitions as an input. Also, I have seen repartition(), but I am not sure if it can solve my problem. The variable data is a DataFrame in my example.
Let me define partition more precisely. Definition one: commonly referred to as "partition key" , where a column is selected and indexed to speed up query (that is not what i want). Definition two: (this is where my concern is) suppose you have a data set, Spark decides it is going to distribute it across many nodes so it can run operations on the data in parallel. If the data size is too small, this may further slow down the process. How can i set that value
By default it partitions into 200 sets. You can change it by using set command in sql context sqlContext.sql("set spark.sql.shuffle.partitions=10");. However you need to set it with caution based up on your data characteristics.
You can call repartition() on dataframe for setting partitions. You can even set spark.sql.shuffle.partitions this property after creating hive context or by passing to spark-submit jar:
spark-submit .... --conf spark.sql.shuffle.partitions=100
or
dataframe.repartition(100)
Number of "input" partitions are fixed by the File System configuration.
1 file of 1Go, with a block size of 128M will give you 10 tasks. I am not sure you can change it.
repartition can be very bad, if you have lot of input partitions this will make lot of shuffle (data traffic) between partitions.
There is no magic method, you have to try, and use the webUI to see how many tasks are generated.
Looking at the combination of MapReduce and HBase from a data-flow perspective, my problem seems to fit. I have a large set of documents which I want to Map, Combine and Reduce. My previous SQL implementation was to split the task into batch operations, cumulatively storing what would be the result of the Map into table and then performing the equivalent of a reduce. This had the benefit that at any point during execution (or between executions), I had the results of the Map at that point in time.
As I understand it, running this job as a MapReduce would require all of the Map functions to run each time.
My Map functions (and indeed any function) always gives the same output for a given input. There is simply no point in re-calculating output if I don't have to. My input (a set of documents) will be continually growing and I will run my MapReduce operation periodically over the data. Between executions I should only really have to calculate the Map functions for newly added documents.
My data will probably be HBase -> MapReduce -> HBase. Given that Hadoop is a whole ecosystem, it may be able to know that a given function has been applied to a row with a given identity. I'm assuming immutable entries in the HBase table. Does / can Hadoop take account of this?
I'm made aware from the documentation (especially the Cloudera videos) that re-calculation (of potentially redundant data) can be quicker than persisting and retrieving for the class of problem that Hadoop is being used for.
Any comments / answers?
If you're looking to avoid running the Map step each time, break it out as its own step (either by using the IdentityReducer or setting the number of reducers for the job to 0) and run later steps using the output of your map step.
Whether this is actually faster than recomputing from the raw data each time depends on the volume and shape of the input data vs. the output data, how complicated your map step is, etc.
Note that running your mapper on new data sets won't append to previous runs - but you can get around this by using a dated output folder. This is to say that you could store the output of mapping your first batch of files in my_mapper_output/20091101, and the next week's batch in my_mapper_output/20091108, etc. If you want to reduce over the whole set, you should be able to pass in my_mapper_output as the input folder, and catch all of the output sets.
Why not apply your SQL workflow in a different environment? Meaning, add a "processed" column to your input table. When time comes to run a summary, run a pipeline that goes something like:
map (map_function) on (input table filtered by !processed); store into map_outputs either in hbase or simply hdfs.
map (reduce function) on (map_outputs); store into hbase.
You can make life a little easier, assuming you are storing your data in Hbase sorted by insertion date, if you record somewhere timestamps of successful summary runs, and open the filter on inputs that are dated later than last successful summary -- you'll save some significant scanning time.
Here's an interesting presentation that shows how one company architected their workflow (although they do not use Hbase):
http://www.scribd.com/doc/20971412/Hadoop-World-Production-Deep-Dive-with-High-Availability