We have an AWS S3 bucket with millions of documents in a complex hierarchy, and a CSV file with (among other data) links to a subset of those files, I estimate this file will be about 1000 to 10.000 rows. I need to join the data from the CSV file with the contents of the documents for further processing in Spark. In case it matters, we're using Scala and Spark 2.4.4 on an Amazon EMR 6.0.0 cluster.
I can think of two ways to do this. First is to add a transformation on the CSV DataFrame that adds the content as a new column:
val df = spark.read.format("csv").load("<csv file>")
val attempt1 = df.withColumn("raw_content", spark.sparkContext.textFile($"document_url"))
or variations thereof (for example, wrapping it in a udf) don't seem to work, I think because sparkContext.textFile returns an RDD, so I'm not sure it's even supposed to work this way? Even if I get it working, is the best way to keep it performant in Spark?
An alternative I tried to think of is to use spark.sparkContext.wholeTextFiles upfront and then join the two dataframes together:
val df = spark.read.format("csv").load("<csv file>")
val contents = spark.sparkContext.wholeTextFiles("<s3 bucket>").toDF("document_url", "raw_content");
val attempt2 = df.join(contents, df("document_url") === contents("document_url"), "left")
but wholeTextFiles doesn't go into subdirectories and the needed paths are hard to predict, and I'm also unsure of the performance impact of trying to build an RDD of the entire bucket of millions of files if I only need a small fraction of it, since the S3 API probably doesn't make it very fast to list all the objects in the bucket.
Any ideas? Thanks!
I did figure out a solution in the end:
val df = spark.read.format("csv").load("<csv file>")
val allS3Links = df.map(row => row.getAs[String]("document_url")).collect()
val joined = allS3Links.mkString(",")
val contentsDF = spark.sparkContext.wholeTextFiles(joined).toDF("document_url", "raw_content");
The downside to this solution is that it pulls all the urls to the driver, but it's workable in my case (100,000 * ~100 char length strings is not that much) and maybe even unavoidable.
Related
I am currently working on 11,000 files. Each file will generate a data frame which will be Union with the previous one. Below is the code:
var df1 = sc.parallelize(Array(("temp",100 ))).toDF("key","value").withColumn("Filename", lit("Temp") )
files.foreach( filename => {
val a = filename.getPath.toString()
val m = a.split("/")
val name = m(6)
println("FILENAME: " + name)
if (name == "_SUCCESS") {
println("Cannot Process '_SUCCSS' Filename")
} else {
val freqs=doSomething(a).toDF("key","value").withColumn("Filename", lit(name) )
df1=df1.unionAll(freqs)
}
})
First, i got an error of java.lang.StackOverFlowError on 11,000 files. Then, i add a following line after df1=df1.unionAll(freqs):
df1=df1.cache()
It resolves the problem but after each iteration, it is getting slower. Can somebody please suggest me what should be done to avoid StackOverflowError with no decrease in time.
Thanks!
The issue is that spark manages a dataframe as a set of transformations. It begins with the "toDF" of the first dataframe, then perform the transformations on it (e.g. withColumn), then unionAll with the previous dataframe etc.
The unionAll is just another such transformation and the tree becomes very long (with 11K unionAll you have an execution tree of depth 11K). The unionAll when building the information can get to a stack overflow situation.
The caching doesn't solve this, however, I imagine you are adding some action along the way (otherwise nothing would run besides building the transformations). When you perform caching, spark might skip some of the steps and therefor the stack overflow would simply arrive later.
You can go back to RDD for iterative process (your example actually is not iterative but purely parallel, you can simply save each separate dataframe along the way and then convert to RDD and use RDD union).
Since your case seems to be join unioning a bunch of dataframes without true iterations, you can also do the union in a tree manner (i.e. union pairs, then union pairs of pairs etc.) this would change the depth from O(N) to O(log N) where N is the number of unions.
Lastly, you can read and write the dataframe to/from disk. The idea is that after every X (e.g. 20) unions, you would do df1.write.parquet(filex) and then df1 = spark.read.parquet(filex). When you read the lineage of a single dataframe would be the file reading itself. The cost of course would be the writing and reading of the file.
I work with Spark 1.6.1 in Scala.
I have one dataframe, and I want to create different dataframe and only want to read 1 time.
For example one dataframe have two columns ID and TYPE, and I want to create two dataframe one with the value of type = A and other with type value = B.
I've checked another posts on stackoverflow, but found only the option to read the dataframe 2 times.
However, I would like another solution with the best performance possible.
Kinds regards.
Spark will read from the data source multiple times if you perform multiple actions on the data. The way to aviod this is to use cache(). In this way, the data will be saved to memory after the first action, which will make subsequent actions using the data faster.
Your two dataframes can be created in this way, requiring only one read of the data source.
val df = spark.read.csv(path).cache()
val dfA = df.filter($"TYPE" === "A").drop("TYPE")
val dfB = df.filter($"TYPE" === "B").drop("TYPE")
The "TYPE" column is dropped as it should be unnecessary after the separation.
I am new to spark. I have some json data that comes as an HttpResponse. I'll need to store this data in hive tables. Every HttpGet request returns a json which will be a single row in the table. Due to this, I am having to write single rows as files in the hive table directory.
But I feel having too many small files will reduce the speed and efficiency. So is there a way I can recursively add new rows to the Dataframe and write it to the hive table directory all at once. I feel this will also reduce the runtime of my spark code.
Example:
for(i <- 1 to 10){
newDF = hiveContext.read.json("path")
df = df.union(newDF)
}
df.write()
I understand that the dataframes are immutable. Is there a way to achieve this?
Any help would be appreciated. Thank you.
You are mostly on the right track, what you want to do is to obtain multiple single records as a Seq[DataFrame], and then reduce the Seq[DataFrame] to a single DataFrame by unioning them.
Going from the code you provided:
val BatchSize = 100
val HiveTableName = "table"
(0 until BatchSize).
map(_ => hiveContext.read.json("path")).
reduce(_ union _).
write.insertInto(HiveTableName)
Alternatively, if you want to perform the HTTP requests as you go, we can do that too. Let's assume you have a function that does the HTTP request and converts it into a DataFrame:
def obtainRecord(...): DataFrame = ???
You can do something along the lines of:
val HiveTableName = "table"
val OtherHiveTableName = "other_table"
val jsonArray = ???
val batched: DataFrame =
jsonArray.
map { parameter =>
obtainRecord(parameter)
}.
reduce(_ union _)
batched.write.insertInto(HiveTableName)
batched.select($"...").write.insertInto(OtherHiveTableName)
You are clearly misusing Spark. Apache Spark is analytical system, not a database API. There is no benefit of using Spark to modify Hive database like this. It will only bring a severe performance penalty without benefiting from any of the Spark features, including distributed processing.
Instead you should use Hive client directly to perform transactional operations.
If you can batch-download all of the data (for example with a script using curl or some other program) and store it in a file first (or many files, spark can load an entire directory at once) you can then load that file(or files) all at once into spark to do your processing. I would also check to see it the webapi as any endpoints to fetch all the data you need instead of just one record at a time.
Usually I load csv files and then I run different kind of aggregations like for example "group by" with Spark. I was wondering if it is possible to start this sort of operations during the file loading (typically a few millions of rows) instead of sequentialize them and if it can be worthy (as time saving).
Example:
val csv = sc.textFile("file.csv")
val data = csv.map(line => line.split(",").map(elem => elem.trim))
val header = data.take(1)
val rows = data.filter(line => header(0) != "id")
val trows = rows.map(row => (row(0), row))
trows.groupBy(//row(0) etc.)
For my understanding of how Spark works, the groupBy (or aggregate) will be "postponed" to the loading in memory of the whole file csv. If this is correct, can the loading and the grouping run at the "same" time instead of sequencing the two steps?
the groupBy (or aggregate) will be "postponed" to the loading in memory of the whole file csv.
It is not the case. At the local (single partition) level Spark operates on lazy sequences so operations belonging to a single task (this includes map side aggregation) can squashed together.
In other words when you have chain of methods operations are performed line-by-line not transformation-by-transformation. In other words the first line will be mapped, filtered, mapped once again and passed to aggregator before the next one is accessed.
To start a group by on load operation You could proceed with 2 options:
Write your own loader and make your own group by inside that + aggregationByKey. The cons of that is writting more code & more maintanance.
Use Parquet format files as input + DataFrames, due it's columnar it will read only desired columns used in your groupBy. so it should be faster. - DataFrameReader
df = spark.read.parquet('file_path')
df = df.groupBy('column_a', 'column_b', '...').count()
df.show()
Due Spark is Lazy it won't load your file until you call action methods like show/collect/write. So Spark will know which columns read and which ignore on the load process.
I am working on a project in apache Spark and the requirement is to write the processed output from spark into a specific format like Header -> Data -> Trailer. For writing to HDFS I am using the .saveAsHadoopFile method and writing the data to multiple files using the key as a file name. But the issue is the sequence of the data is not maintained files are written in Data->Header->Trailer or a different combination of three. Is there anything I am missing with RDD transformation?
Ok so after reading from StackOverflow questions, blogs and mail archives from google. I found out how exactly .union() and other transformation works and how partitioning is managed. When we use .union() the partition information is lost by the resulting RDD and also the ordering and that's why My output sequence was not getting maintained.
What I did to overcome the issue is numbering the Records like
Header = 1, Body = 2, and Footer = 3
so using sortBy on RDD which is union of all three I sorted it using this order number with 1 partition. And after that to write to multiple file using key as filename I used HashPartitioner so that same key data should go into separate file.
val header: RDD[(String,(String,Int))] = ... // this is my header RDD`
val data: RDD[(String,(String,Int))] = ... // this is my data RDD
val footer: RDD[(String,(String,Int))] = ... // this is my footer RDD
val finalRDD: [(String,String)] = header.union(data).union(footer).sortBy(x=>x._2._2,true,1).map(x => (x._1,x._2._1))
val output: RDD[(String,String)] = new PairRDDFunctions[String,String](finalRDD).partitionBy(new HashPartitioner(num))
output.saveAsHadoopFile ... // and using MultipleTextOutputFormat save to multiple file using key as filename
This might not be the final or most economical solution but it worked. I am also trying to find other ways to maintain the sequence of output as Header->Body->Footer. I also tried .coalesce(1) on all three RDD's and then do the union but that was just adding three more transformation to RDD's and .sortBy function also take partition information which I thought will be same, but coalesceing the RDDs first also worked. If Anyone has some another approach please let me know, or add more to this will be really helpful as I am new to Spark
References:
Write to multiple outputs by key Spark - one Spark job
Ordered union on spark RDDs
http://apache-spark-user-list.1001560.n3.nabble.com/Union-of-2-RDD-s-only-returns-the-first-one-td766.html -- this one helped a lot