I am new to spark and scala. I have below requirement. I need to process all the files under a path which have sub directories. I guess, I need to write a for-loop logic to process across all the files.
Below is the example of my case:
src/proj_fldr/dataset1/20170624/file1.txt
src/proj_fldr/dataset1/20170624/file2.txt
src/proj_fldr/dataset1/20170624/file3.txt
src/proj_fldr/dataset1/20170625/file1.txt
src/proj_fldr/dataset1/20170625/file2.txt
src/proj_fldr/dataset1/20170625/file3.txt
src/proj_fldr/dataset1/20170626/file1.txt
src/proj_fldr/dataset1/20170626/file2.txt
src/proj_fldr/dataset1/20170626/file3.txt
src/proj_fldr/dataset2/20170624/file1.txt
src/proj_fldr/dataset2/20170624/file2.txt
src/proj_fldr/dataset2/20170624/file3.txt
src/proj_fldr/dataset2/20170625/file1.txt
src/proj_fldr/dataset2/20170625/file2.txt
src/proj_fldr/dataset2/20170625/file3.txt
src/proj_fldr/dataset2/20170626/file1.txt
src/proj_fldr/dataset2/20170626/file2.txt
src/proj_fldr/dataset2/20170626/file3.txt
I need the code to iterate the files like
In src
loop (proj_fldr
loop(dataset
loop(datefolder
loop(file1 then, file2....))))
Since you have a regular file structure you can use the wildcard * when reading the files. You can do the following to read all the files into a single RDD:
val spark = SparkSession.builder.getOrCreate()
val rdd = spark.sparkContext.wholeTextFiles("src/*/*/*/*.txt")
The result will be a RDD[(String, String)] with the path and the content in a tuple for each processed file.
To explicitly set if you want to use local or HDFS files you can append "hdfs://" or "file://" to the beginning of the path.
Related
I'm reading from a path say /json//myfiles_.json
I'm then flattening the json using explode. This causes an error since I have some empty files. How do I tell it to ignore empty files of somehow filter them out?
I can detect individual files checking if the head is empty but I need to do this on the collection of files iterated in the dataframe with the use of the wildcard path.
So the answer seems to be that I need to provide a schema explicitly because it can't infer one from empty file - as you would expect!
e.g.
val schemadf = sqlContext.read.json(schemapath) //infer schema from file with data or do manually
val schema = schemadf.schema
val raw = sqlContext.read.schema(schema).json(monthfile)
val prep = raw.withColumn("MyArray", explode($"MyArray"))
.select($"ID", $"name", $"CreatedAt")
display(prep)
This question already has answers here:
Spark - load CSV file as DataFrame?
(14 answers)
Closed 4 years ago.
How do I read the data from hdfs data sets using scala language? data is any "CSV" file with limited records.
You tagged the question with Spark, so I'm assuming you are trying to use that. I would recommend you start by reading through the Spark documentation here to get an idea of how to use Spark to interact with your data.
https://spark.apache.org/docs/latest/quick-start.html
https://spark.apache.org/docs/latest/sql-programming-guide.html
But, to answer your specific question, in Spark you would read in the CSV file using code like this:
val csvDf = spark.read.format("csv")
.option("sep", ",")
.option("header", "true")
.load("hdfs://some/path/to/data.csv/")
The path your provide will be to a CSV file on HDFS, or a folder containing multiple CSV files. Also, Spark will accept other types of file systems. For example you could also use "file://" to access the local file system, or "s3://" to use S3. Once you have loaded the data you will have a Spark DataFrame object with SQL like methods available to interact with it.
Note, I provided an option for separator just to show you how to do it, but it defaults to "," anyways, so it is not required. Also, if your CSV files do not include a header, you will need to specify the Schema yourself and set header to false instead.
You can read data from HDFS by following this approach :-
val hdfs = FileSystem.get(new URI("hdfs://hdfsUrl:port/"), new Configuration())
val path = new Path("/pathOfTheFileInHDFS/")
val stream = hdfs.open(path)
def readLines = Stream.cons(stream.readLine, Stream.continually( stream.readLine))
//This example checks line for null and prints every existing line consequentally
readLines.takeWhile(_ != null).foreach(line => println(line))
Also please have a look at this article https://blog.matthewrathbone.com/2013/12/28/reading-data-from-hdfs-even-if-it-is-compressed
Please let me know if this answers your question.
So I have a scala program that iterates through a graph and writes out data line by line to a text file. It is essentially an edge list file for use with graphx.
The biggest slow down is actually creating this text file, were talking maybe million records it writes to this text file. Is there a way I can somehow parallel this task or making faster in any way by somehow storing it in memory or anything?
More info:
I am using a hadoop cluster to iterate through a graph and here is my code snippet for my text file creation im doing now to write to HDFS:
val fileName = dbPropertiesFile + "-edgelist-" + System.currentTimeMillis()
val path = new Path("/home/user/graph/" + fileName + ".txt")
val conf = new Configuration()
conf.set("fs.defaultFS", "hdfs://host001:8020")
val fs = FileSystem.newInstance(conf)
val os = fs.create(path)
while (edges.hasNext) {
val current = edges.next()
os.write(current.inVertex().id().toString.getBytes())
os.write(" ".getBytes())
os.write(current.outVertex().id().toString.getBytes())
os.write("\n".toString.getBytes())
}
fs.close()
Writing files to HDFS is never fast. Your tags seem to suggest that you are already using spark anyway, so you could as well, take advantage of it.
sparkContext
.makeRDD(20, edges.toStream)
.map(e => e.inVertex.id -> e.outVertex.id)
.toDF
.write
.delimiter(" ")
.csv(path)
This splits your input into 20 partitions (you can control that number with the numeric parameter to makeRDD above), and writes them in parallel to 20 different chunks in hdfs, that represent your resulting file.
This question is related to this.
I am processing an S3 folder containing csv.gz files in Spark. Each csv.gz file has a header that contains column names. This has been solved by the above SO link and the solution looks like this:
val rdd = sc.textFile("s3://.../my-s3-path").mapPartitions(_.drop(1))
The problem now is that it looks like some of the files have newline ('\n') at the end (we assume we are not sure which file). So when converting the RDD to DataFrame, I'm getting some error. The question now is:
How do I get rid of the last line of each file if it is '\n'?
Why not a simple filter:
val rdd = sc.textFile("s3...").filter(line => !line.equalsIgnoreCase("\n")).mapPartition...
Or filter any empty line:
val rdd = sc.textFile("s3...").filter(line => !line.trim().isEmpty)...
In a pig script I saved a table using PigStorage('|').
I have in the corresponding hadoop folder files like
part-r-00000
etc.
What is the best way to load it in Spark/Scala ? In this table I have 3 fields: Int, String, Float
I tried:
text = sc.hadoopFile("file", classOf[TextInputFormat], classOf[LongWritable], classOf[Text], sc.defaultMinPartitions)
But then I would need somehow to split each line. Is there a better way to do it?
If I were coding in python I would create a Dataframe indexed by the first field and whose columns are the values found in the string field and coefficients the float values. But I need to use scala to use the pca module. And the dataframes don't seem that close to python's ones
Thanks for the insight
PigStorage creates a text file without schema information so you need to do that work yourself something like
sc.textFile("file") // or directory where the part files are
val data = csv.map(line => {
vals=line.split("|")
(vals(0).toInt,vals(1),vals(2).toDouble)}
)