I have a single record in the dataset but from the method that saves to the database, it is executed twice. What can be caused this? how do I solve it? I save the file in a blob storage.
def saveFile[T](dataset: Dataset[T]): Unit = {
dataset
.write
.mode("overwrite")
.json("...json")
}
val res = df.map(x => {
...
val operation = saveToTable()
...
})
saveFile(res)
I think your method is being executed twice because dataset.write.mode("overwrite").json("...json") will save the dataset to the specified location and if you run the saveFile method again, it will save the dataset again to the specified location. Try using dataset.write.mode("append").json("...json") which will save the dataset to the specified location if it doesn't exist, otherwise it will append the dataset to the specified location.
Related
I have a dataframe var cache :DataFrame = _. As an initial run i have given, cache = existingDF, the existingdf is read from an excel using crealytics.spark.excel.
but in the subsequent run, the existingDF will get another updated excel file, it should be cache = cache.union(existingDF)
But I seem to get only existingDF inside cache. In short whenever i call cache it seems to read the excel. How do i avoid this? This issue is not there while reading it as csv. (It was there when i used .persist on the csv read, but got fixed when i removed .persist
More Simply:
var a = _
while(true){
val b = spark.read.format("com.crealytics.spark.excel")...
if (Option(a).isEmpty){
a = b
}
else if a!=b
a = b.union(a)
}
The variable a is always getting updated along with b, so it never becomes different from b. How do I avoid this?
The folder of hdfs is like :
/test/data/2020-03-01/{multiple inside files csv}
/test/data/2020-03-02/{multiple files csv}
/test/data/2020-03-03/{multiple files csv }
i want to read data inside folder one by one not whole by
spark.read.csv("/test/data/*") //not in such manner
Not in above manner , i want to read file one by one; so that i can make the log entry in some database that date folder is read ; so that on next time i can skip that folder in next day or same day if program run accidentally:
val conf = new Configuration()
val iterate = org.apache.hadoop.fs.FileSystem.get(new URI(strOutput), conf).listLocatedStatus(new org.apache.hadoop.fs.Path(strOutput))
while (iterate.hasNext) {
val pathStr = iterate.next().getPath.toString
println("log---->"+pathStr)
val df = spark.read.text(pathStr)
}
Try something like above and read as data frame, if you want you can union new date df with old df.
I have multiple parquet files (around 1000). I need to load each one of them, process it and save the result to a Hive table. I have a for loop but it only seems to work with 2 or 5 files, but not with 1000, as it seems Sparks tries to load them all at the same time, and I need it do it individually in the same Spark session.
I tried using a for loop, then a for each, and I ussed unpersist() but It fails anyway.
val ids = get_files_IDs()
ids.foreach(id => {
println("Starting file " + id)
var df = load_file(id)
var values_df = calculate_values(df)
values_df.write.mode(SaveMode.Overwrite).saveAsTable("table.values_" + id)
df.unpersist()
})
def get_files_IDs(): List[String] = {
var ids = sqlContext.sql("SELECT CAST(id AS varchar(10)) FROM table.ids WHERE id IS NOT NULL")
var ids_list = ids.select("id").map(r => r.getString(0)).collect().toList
return ids_list
}
def calculate_values(df:org.apache.spark.sql.DataFrame): org.apache.spark.sql.DataFrame ={
val values_id = df.groupBy($"id", $"date", $"hr_time").agg(avg($"value_a") as "avg_val_a", avg($"value_b") as "avg_value_b")
return values_id
}
def load_file(id:String): org.apache.spark.sql.DataFrame = {
val df = sqlContext.read.parquet("/user/hive/wh/table.db/parquet/values_for_" + id + ".parquet")
return df
}
What I would expect is for Spark to load file ID 1, process the data, save it to the Hive table and then dismiss that date and cotinue with the second ID and so on until it finishes the 1000 files. Instead of it trying to load everything at the same time.
Any help would be very appreciated! I've been stuck on it for days. I'm using Spark 1.6 with Scala Thank you!!
EDIT: Added the definitions. Hope it can help to get a better view. Thank you!
Ok so after a lot of inspection I realised that the process was working fine. It processed each file individualy and saved the results. The issue was that in some very specific cases the process was taking way way way to long.
So I can tell that with a for loop or for each you can process multiple files and save the results without problem. Unpersisting and clearing cache do helps on performance.
We are currently facing a performance issue in sparksql written in scala language. Application flow is mentioned below.
Spark application reads a text file from input hdfs directory
Creates a data frame on top of the file using programmatically specifying schema. This dataframe will be an exact replication of the input file kept in memory. Will have around 18 columns in the dataframe
var eqpDF = sqlContext.createDataFrame(eqpRowRdd, eqpSchema)
Creates a filtered dataframe from the first data frame constructed in step 2. This dataframe will contain unique account numbers with the help of distinct keyword.
var distAccNrsDF = eqpDF.select("accountnumber").distinct().collect()
Using the two dataframes constructed in step 2 & 3, we will get all the records which belong to one account number and do some Json parsing logic on top of the filtered data.
var filtrEqpDF =
eqpDF.where("accountnumber='" + data.getString(0) + "'").collect()
Finally the json parsed data will be put into Hbase table
Here we are facing performance issues while calling the collect method on top of the data frames. Because collect will fetch all the data into a single node and then do the processing, thus losing the parallel processing benefit.
Also in real scenario there will be 10 billion records of data which we can expect. Hence collecting all those records in to driver node will might crash the program itself due to memory or disk space limitations.
I don't think the take method can be used in our case which will fetch limited number of records at a time. We have to get all the unique account numbers from the whole data and hence I am not sure whether take method, which takes
limited records at a time, will suit our requirements
Appreciate any help to avoid calling collect methods and have some other best practises to follow. Code snippets/suggestions/git links will be very helpful if anyone have had faced similar issues
Code snippet
val eqpSchemaString = "acoountnumber ....."
val eqpSchema = StructType(eqpSchemaString.split(" ").map(fieldName =>
StructField(fieldName, StringType, true)));
val eqpRdd = sc.textFile(inputPath)
val eqpRowRdd = eqpRdd.map(_.split(",")).map(eqpRow => Row(eqpRow(0).trim, eqpRow(1).trim, ....)
var eqpDF = sqlContext.createDataFrame(eqpRowRdd, eqpSchema);
var distAccNrsDF = eqpDF.select("accountnumber").distinct().collect()
distAccNrsDF.foreach { data =>
var filtrEqpDF = eqpDF.where("accountnumber='" + data.getString(0) + "'").collect()
var result = new JSONObject()
result.put("jsonSchemaVersion", "1.0")
val firstRowAcc = filtrEqpDF(0)
//Json parsing logic
{
.....
.....
}
}
The approach usually take in this kind of situation is:
Instead of collect, invoke foreachPartition: foreachPartition applies a function to each partition (represented by an Iterator[Row]) of the underlying DataFrame separately (the partition being the atomic unit of parallelism of Spark)
the function will open a connection to HBase (thus making it one per partition) and send all the contained values through this connection
This means the every executor opens a connection (which is not serializable but lives within the boundaries of the function, thus not needing to be sent across the network) and independently sends its contents to HBase, without any need to collect all data on the driver (or any one node, for that matter).
It looks like you are reading a CSV file, so probably something like the following will do the trick:
spark.read.csv(inputPath). // Using DataFrameReader but your way works too
foreachPartition { rows =>
val conn = ??? // Create HBase connection
for (row <- rows) { // Loop over the iterator
val data = parseJson(row) // Your parsing logic
??? // Use 'conn' to save 'data'
}
}
You can ignore collect in your code if you have large set of data.
Collect Return all the elements of the dataset as an array at the driver program. This is usually useful after a filter or other operation that returns a sufficiently small subset of the data.
Also this can cause the driver to run out of memory, though, because collect() fetches the entire RDD/DF to a single machine.
I have just edited your code, which should work for you.
var distAccNrsDF = eqpDF.select("accountnumber").distinct()
distAccNrsDF.foreach { data =>
var filtrEqpDF = eqpDF.where("accountnumber='" + data.getString(0) + "'")
var result = new JSONObject()
result.put("jsonSchemaVersion", "1.0")
val firstRowAcc = filtrEqpDF(0)
//Json parsing logic
{
.....
.....
}
}
I have 20 million files in S3 spanning roughly 8000 days.
The files are organized by timestamps in UTC, like this: s3://mybucket/path/txt/YYYY/MM/DD/filename.txt.gz. Each file is UTF-8 text containing between 0 (empty) and 100KB of text (95th percentile, although there are a few files that are up to several MBs).
Using Spark and Scala (I'm new to both and want to learn), I would like to save "daily bundles" (8000 of them), each containing whatever number of files were found for that day. Ideally I would like to store the original filenames as well as their content. The output should reside in S3 as well and be compressed, in some format that is suitable for input in further Spark steps and experiments.
One idea was to store bundles as a bunch of JSON objects (one per line and '\n'-separated), e.g.
{id:"doc0001", meta:{x:"blah", y:"foo", ...}, content:"some long string here"}
{id:"doc0002", meta:{x:"foo", y:"bar", ...}, content: "another long string"}
Alternatively, I could try the Hadoop SequenceFile, but again I'm not sure how to set that up elegantly.
Using the Spark shell for example, I saw that it was very easy to read the files, for example:
val textFile = sc.textFile("s3n://mybucket/path/txt/1996/04/09/*.txt.gz")
// or even
val textFile = sc.textFile("s3n://mybucket/path/txt/*/*/*/*.txt.gz")
// which will take for ever
But how do I "intercept" the reader to provide the file name?
Or perhaps I should get an RDD of all the files, split by day, and in a reduce step write out K=filename, V=fileContent?
You can use this
First You can get a Buffer/List of S3 Paths :
import scala.collection.JavaConverters._
import java.util.ArrayList
import com.amazonaws.services.s3.AmazonS3Client
import com.amazonaws.services.s3.model.ObjectListing
import com.amazonaws.services.s3.model.S3ObjectSummary
import com.amazonaws.services.s3.model.ListObjectsRequest
def listFiles(s3_bucket:String, base_prefix : String) = {
var files = new ArrayList[String]
//S3 Client and List Object Request
var s3Client = new AmazonS3Client();
var objectListing: ObjectListing = null;
var listObjectsRequest = new ListObjectsRequest();
//Your S3 Bucket
listObjectsRequest.setBucketName(s3_bucket)
//Your Folder path or Prefix
listObjectsRequest.setPrefix(base_prefix)
//Adding s3:// to the paths and adding to a list
do {
objectListing = s3Client.listObjects(listObjectsRequest);
for (objectSummary <- objectListing.getObjectSummaries().asScala) {
files.add("s3://" + s3_bucket + "/" + objectSummary.getKey());
}
listObjectsRequest.setMarker(objectListing.getNextMarker());
} while (objectListing.isTruncated());
//Removing Base Directory Name
files.remove(0)
//Creating a Scala List for same
files.asScala
}
Now Pass this List object to the following piece of code, note : sc is an object of SQLContext
var df: DataFrame = null;
for (file <- files) {
val fileDf= sc.textFile(file)
if (df!= null) {
df= df.unionAll(fileDf)
} else {
df= fileDf
}
}
Now you got a final Unified RDD i.e. df
Optional, And You can also repartition it in a single BigRDD
val files = sc.textFile(filename, 1).repartition(1)
Repartitioning always works :D
have you tried something along the lines of sc.wholeTextFiles?
It creates an RDD where the key is the filename and the value is the byte array of the whole file. You can then map this so the key is the file date, and then groupByKey?
http://spark.apache.org/docs/latest/programming-guide.html
At your scale, elegant solution would be a stretch.
I would recommend against using sc.textFile("s3n://mybucket/path/txt/*/*/*/*.txt.gz") as it takes forever. What you can do is use AWS DistCp or something similar to move files into HDFS. Once its in HDFS, spark is quite fast in ingesting the information in whatever way suits you.
Note that most of these processes require some sort of file list so you'll need to generate that somehow. for 20 mil files, this creation of file list will be a bottle neck. I'd recommend creating a file that get appended with the file path, every-time a file gets uploaded to s3.
Same for output, put into hdfs and then move to s3 (although direct copy might be equally efficient).