I'm pretty new with this topic so any help will be much appreciated.
I trying to read a csv file which is stored in a S3 bucket and convert its data to an RDD to work directly with it without the need to create a file locally.
So far I've been able to load the file using AmazonS3ClientBuilder, but the only thing I've got is to have the file content in a S3ObjectInputStream and I'm not able to work with its content.
val bucketName = "bucket-name"
val credentials = new BasicAWSCredentials(
"acessKey",
"secretKey"
);
val s3client = AmazonS3ClientBuilder
.standard()
.withCredentials(new AWSStaticCredentialsProvider(credentials))
.withRegion(Regions.US_EAST_2)
.build();
val s3object = s3client.getObject(bucketName, "file-name.csv")
val inputStream = s3object.getObjectContent()
....
I have also tried to use a BufferedSource to work with it but once done, I don't know how to convert it to a dataframe or RDD to work with it.
val myData = Source.fromInputStream(inputStream)
....
You can do it with S3A file system provided in Hadoop-AWS module:
Add this dependency https://mvnrepository.com/artifact/org.apache.hadoop/hadoop-aws
Either define <property><name>fs.s3.impl</name><value>org.apache.hadoop.fs.s3a.S3AFileSystem</value></property> in core-site.xml or add .config("fs.s3.impl", classOf[S3AFileSystem].getName) to SparkSession builder
Access S3 using spark.read.csv("s3://bucket/key"). If you want the RDD that was asked spark.read.csv("s3://bucket/key").rdd
At the end I was able to get the results I was searching for taking a look at https://gist.github.com/snowindy/d438cb5256f9331f5eec
Related
I am facing a weird issue and I've stuck with that for a while.
Basically in my spark application I have a method, where I create a file in MapR fs and save some content in this file.
The method is like below:
def collector(value: String): Unit = {
val conf = new Configuration()
val fs= FileSystem.get(conf)
val path = getConfig("path")
Try {fs.create(new Path(path + s"${scala.util.Random.alphanumeric take 10 mkString}" +
Calendar.getInstance.getTimeInMillis))
.write(value.getBytes);
fs.close()}.getOrElse(logger.warn("File not saved"))
}
This method is called from different object. When I run it with --master local[*], then the file is created in specific location in the FS with string that is passed in value. However, when I run it on cluster (--master yarn), then only empty file is saved. When I print value, its printing the string. But for some reason not saving it in the file.
I wonder if anybody has any idea why?
Thanks
I have the following piece of scala code that will write text to a txt file sitting locally.
// PrintWriter
import java.io._
val pw = new PrintWriter(new File("resources/myfile.txt"))
pw.write("Test text")
pw.close
How do i get this to work on azure blob storage?
I have tried:
val pw = new PrintWriter(new File("wasb://[my container name]#[My Storage account].blob.core.windows.net/resources/myfile.txt"))
But it doesn't work.
What am i doing wrong? By the way, for the sake of this example, I'm keeping it simple. In reality, I am outputting more meaningful data.
Thanks
Con
You cannot create a file on Azure blog storage using the java.io or nio package. You need to use their REST API or their SDK.
https://learn.microsoft.com/en-us/rest/api/storageservices/create-file
https://learn.microsoft.com/en-us/azure/storage/blobs/storage-java-how-to-use-blob-storage
I am receiving the streaming data myDStream (DStream[String]) that I want to save in S3 (basically, for this question, it doesn't matter where exactly do I want to save the outputs, but I am mentioning it just in case).
The following code works well, but it saves folders with the names like jsonFile-19-45-46.json, and then inside the folders it saves files _SUCCESS and part-00000.
Is it possible to save each RDD[String] (these are JSON strings) data into the JSON files, not the folders? I thought that repartition(1) had to make this trick, but it didn't.
myDStream.foreachRDD { rdd =>
// datetimeString = ....
rdd.repartition(1).saveAsTextFile("s3n://mybucket/keys/jsonFile-"+datetimeString+".json")
}
AFAIK there is no option to save it as a file. Because it's a distributed processing framework and it's not a good practice write on single file rather than each partition writes it's own files in the specified path.
We can pass only output directory where we wanted to save the data. OutputWriter will create file(s)(depends on partitions) inside specified path with part- file name prefix.
As an alternative to rdd.collect.mkString("\n") you can use hadoop Filesystem library to cleanup output by moving part-00000 file into it's place. Below code works perfectly on local filesystem and HDFS, but I'm unable to test it with S3:
val outputPath = "path/to/some/file.json"
rdd.saveAsTextFile(outputPath + "-tmp")
import org.apache.hadoop.fs.Path
val fs = org.apache.hadoop.fs.FileSystem.get(spark.sparkContext.hadoopConfiguration)
fs.rename(new Path(outputPath + "-tmp/part-00000"), new Path(outputPath))
fs.delete(new Path(outputPath + "-tmp"), true)
For JAVA I implemented this one. Hope it helps:
val fs = FileSystem.get(spark.sparkContext().hadoopConfiguration());
File dir = new File(System.getProperty("user.dir") + "/my.csv/");
File[] files = dir.listFiles((d, name) -> name.endsWith(".csv"));
fs.rename(new Path(files[0].toURI()), new Path(System.getProperty("user.dir") + "/csvDirectory/newData.csv"));
fs.delete(new Path(System.getProperty("user.dir") + "/my.csv/"), true);
I have a Spark program (in Scala) and a SparkContext. I am writing some files with RDD's saveAsTextFile. On my local machine I can use a local file path and it works with the local file system. On my cluster it works with HDFS.
I also want to write other arbitrary files as the result of processing. I'm writing them as regular files on my local machine, but want them to go into HDFS on the cluster.
SparkContext seems to have a few file-related methods but they all seem to be inputs not outputs.
How do I do this?
Thanks to marios and kostya, but there are few steps to writing a text file into HDFS from Spark.
// Hadoop Config is accessible from SparkContext
val fs = FileSystem.get(sparkContext.hadoopConfiguration);
// Output file can be created from file system.
val output = fs.create(new Path(filename));
// But BufferedOutputStream must be used to output an actual text file.
val os = BufferedOutputStream(output)
os.write("Hello World".getBytes("UTF-8"))
os.close()
Note that FSDataOutputStream, which has been suggested, is a Java serialized object output stream, not a text output stream. The writeUTF method appears to write plaint text, but it's actually a binary serialization format that includes extra bytes.
Here's what worked best for me (using Spark 2.0):
val path = new Path("hdfs://namenode:8020/some/folder/myfile.txt")
val conf = new Configuration(spark.sparkContext.hadoopConfiguration)
conf.setInt("dfs.blocksize", 16 * 1024 * 1024) // 16MB HDFS Block Size
val fs = path.getFileSystem(conf)
if (fs.exists(path))
fs.delete(path, true)
val out = new BufferedOutputStream(fs.create(path)))
val txt = "Some text to output"
out.write(txt.getBytes("UTF-8"))
out.flush()
out.close()
fs.close()
Using HDFS API (hadoop-hdfs.jar) you can create InputStream/OutputStream for an HDFS path and read from/write to a file using regular java.io classes. For example:
URI uri = URI.create (“hdfs://host:port/file path”);
Configuration conf = new Configuration();
FileSystem file = FileSystem.get(uri, conf);
FSDataInputStream in = file.open(new Path(uri));
This code will work with local files as well (change hdfs:// to file://).
One simple way to write files to HDFS is to use a SequenceFiles. Here you use the native Hadoop APIs and not the ones provided by Spark.
Here is a simple snippet (in Scala):
import org.apache.hadoop.conf.Configuration
import org.apache.hadoop.fs._
import org.apache.hadoop.io._
val conf = new Configuration() // Hadoop configuration
val sfwriter = SequenceFile.createWriter(conf,
SequenceFile.Writer.file(new Path("hdfs://nn1.example.com/file1")),
SequenceFile.Writer.keyClass(LongWritable.class),
SequenceFile.Writer.valueClass(Text.class))
val lw = new LongWritable()
val txt = new Text()
lw.set(12)
text.set("hello")
sfwriter.append(lw, txt)
sfwriter.close()
...
In case you don't have a key you can use NullWritable.class in its place:
SequenceFile.Writer.keyClass(NullWritable.class)
sfwriter.append(NullWritable.get(), new Text("12345"));
This is the first time i am integrating Email service with liftweb
I want to send Email with attachments(Like:- Documents,Images,Pdfs)
my code looking like below
case class CSVFile(bytes: Array[Byte],filename: String = "file.csv",
mime: String = "text/csv; charset=utf8; header=present" )
val attach = CSVFile(fileupload.mkString.getBytes("utf8"))
val body = <p>Please research the enclosed.</p>
val msg = XHTMLPlusImages(body,
PlusImageHolder(attach.filename, attach.mime, attach.bytes))
Mailer.sendMail(
From("vyz#gmail.com"),
Subject(subject(0)),
To(to(0)),
)
this code is taken from LiftCookbook its not working like my requirement
its working but only the Attached file name is coming(file.csv) no data in it(i uploaded this file (gsy.docx))
Best Regards
GSY
You don't specify what type fileupload is, but assuming it is of type net.liftweb.http. FileParamHolder then the issue is that you can't just call mkString and expect it to have any data since there is no data in the object, just a fileStream method for retrieving it (either from disk or memory).
The easiest to accomplish what you want would be to use a ByteArrayInputStream and copy the data to it. I haven't tested it, but the code below should solve your issue. For brevity, it uses Apache IO Commons to copy the streams, but you could just as easily do it natively.
val data = {
val os = new ByteArrayOutputStream()
IOUtils.copy(fileupload.fileStream, os)
os.toByteArray
}
val attach = CSVFile(data)
BTW, you say you are uploading a Word (DOCX) file and expecting it to automatically be CSV when the extension is changed? You will just get a DOCX file with a csv extension unless you actually do some conversion.