In our app, we are in need to compute file hash, so we can compare if the file was updated later.
The way I am doing it right now is with this little method:
protected[services] def computeMigrationHash(toVersion: Int): String = {
val migrationClassName = MigrationClassNameFormat.format(toVersion, toVersion)
val migrationClass = Class.forName(migrationClassName)
val fileName = migrationClass.getName.replace('.', '/') + ".class"
val resource = getClass.getClassLoader.getResource(fileName)
logger.debug("Migration file - " + resource.getFile)
val file = new File(resource.getFile)
val hc = Files.hash(file, Hashing.md5())
logger.debug("Calculated migration file hash - " + hc.toString)
hc.toString
}
It all works perfectly, until the code get's deployed into different environment and file file is located in a different absolute path. I guess, the hashing take the path into account as well.
What is the best way to calculate some sort of reliable hash of a file content that well produce the same result for as log as the content of a file stays the same?
Thanks,
Having perused the source code https://github.com/google/guava/blob/master/guava/src/com/google/common/io/Files.java - only the file contents are hashed - the path does not come into play.
public static HashCode hash(File file, HashFunction hashFunction) throws IOException {
return asByteSource(file).hash(hashFunction);
}
Therefore you need not worry about locality of the file. Now why you end up with a different hash on a different fs .. maybe you should compare the size/contents to ensure eg no compound eol's were introduced.
Related
I write some tests and to get absolute path from relative path i use this function
private def getAbsolutePath(filePath: String): String = {
getClass.getResource(filePath).getFile
}
and then i do:
println(getAbsolutePath("/parquetIncrementalProcessor/withPartitioning/"))
println(getAbsolutePath("/parquetIncrementalProcessor/withPartitioning/own_loading_id=1/partition_column=test/"))
i get:
/Users/19658296/csp-fp-snaphot/library/target/scala-2.11/test-classes/parquetIncrementalProcessor/withPartitioning/
/Users/19658296/csp-fp-snaphot/library/target/scala-2.11/test-classes/parquetIncrementalProcessor/withPartitioning/own_loading_id%3d1/partition_column%3dtest/
As you can see, instead of =, I get some strange symbol. At the same time, when I try to read these files with a park, he can read the path without %3d, and with %3d he gets the error "Path does not exist".
How can I fix this?
Seems like its URL encoded, maybe because using stuff from files and resources are designed to work with Universal Resource Locators. You can URLDecode it like so:
import java.net.URLDecoder
def getAbsolutePath(filePath: String): String = {
val path = getClass.getResource(filePath).getFile
URLDecoder.decode(path, "UTF-8")
}
Currently I have a configuration file like this:
project {
inputs {
baseFile {
paths = ["project/src/test/resources/inputs/parquet1/date=2020-11-01/"]
type = parquet
applyConversions = false
}
}
}
And I want to change the date "2020-11-01" to another one during run time. I read I need a new config object since it's immutable, I'm trying this but I'm not quite sure how to edit paths since it's a list and not a String and it definitely needs to be a list or else it's going to say I haven't configured a path for the parquet.
val newConfig = config.withValue("project.inputs.baseFile.paths"(0),
ConfigValueFactory.fromAnyRef("project/src/test/resources/inputs/parquet1/date=2020-10-01/"))
But I'm getting a:
Error com.typesafe.config.ConfigException$BadPath: path parameter: Invalid path 'project.inputs.baseFile.': path has a leading, trailing, or two adjacent period '.' (use quoted "" empty string if you want an empty element)
What's the correct way to set the new path?
One option you have, is to override the entire array:
import scala.collection.JavaConverters._
val mergedConfig = config.withValue("project.inputs.baseFile.paths",
ConfigValueFactory.fromAnyRef(Seq("project/src/test/resources/inputs/parquet1/date=2020-10-01/").asJava))
But a more elegant way to do this (IMHO), is to create a new config, and to use the existing as a fallback.
For example, we can create a new config:
val newJsonString = """project {
|inputs {
|baseFile {
| paths = ["project/src/test/resources/inputs/parquet1/date=2020-10-01/"]
|}}}""".stripMargin
val newConfig = ConfigFactory.parseString(newJsonString)
And now to merge them:
val mergedConfig = newConfig.withFallback(config)
The output of:
println(mergedConfig.getList("project.inputs.baseFile.paths"))
println(mergedConfig.getString("project.inputs.baseFile.type"))
is:
SimpleConfigList(["project/src/test/resources/inputs/parquet1/date=2020-10-01/"])
parquet
As expected.
You can read more about Merging config trees. Code run at Scastie.
I didn't find any way to replace one element of the array with withValue.
I'm trying to read some files from my Scala project, and if I use: java.io.File(".").getCanonicalPath() I find that my current directory is far away from them (exactly where I have installed Scala Eclipse). So how can I change the current directory to the root of my project, or get the path to my project? I really don't want to have an absolute path to my input files.
val PATH = raw"E:\lang\scala\progfun\src\examples\"
def printFileContents(filename: String) {
try {
println("\n" + PATH + filename)
io.Source.fromFile(PATH + filename).getLines.foreach(println)
} catch {
case _:Throwable => println("filename " + filename + " not found")
}
}
val filenames = List("random.txt", "a.txt", "b.txt", "c.txt")
filenames foreach printFileContents
Add your files to src/main/resources/<packageName> where <packageName> is your class package.
Change the line val PATH = getClass.getResource("").getPath
new File(".").getCanonicalPath
will give you the base-path you need
Another workaround is to put the path you need in an user environmental variable, and call it with sys.env (returns exception if failure) or System.getenv (returns null if failure), for example val PATH = sys.env("ScalaProjectPath") but the problem is that if you move the project you have to update the variable, which I didn't want.
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).
I'm working on an SBT plugin where I'd like to zip up a directory. This is possible due to the following method in IO:
def zip(sources: Traversable[(File,String)], outputZip: File): Unit
After tinkering with this method, it seems that simply passing it a directory and expecting the resulting zip file to have the same file & folder structure is wrong.. Passing a directory (empty or otherwise) results in the following:
[error]...:zipper: java.util.zip.ZipException: ZIP file must have at least one entry
Therefore, it appears that the way to get use the zip method is by stepping through the directory and adding each file individually to the Traversable object.
Assuming my understanding is correct, this strikes me as very odd - vey rarely do users need to cherry-pick what is to be added to an archive.
Any thoughts on this?
It seems like you can use this to compose a zip with files from multiple places. I can see the use of that in a build system.
A bit late to the party, but this should do what you need:
val parentFolder: File = ???
val folderName: String = ???
val src: File = parentFolder / folderName
val tgt: File = parentFolder / s"$folderName.zip"
IO.zip(allSubpaths(src), tgt)
Here is some code for zipping directories using sbt's IO class:
IO.withTemporaryDirectory(base => {
val dirToZip = new File(base, "lib")
IO.createDirectory(dirToZip)
IO.write(dirToZip / "test1", "test")
IO.write(dirToZip / "test2", "test")
val zip: File = base / ("test.zip")
IO.zip(allSubpaths(dirToZip), zip)
val out: File = base / "out"
IO.createDirectory(out)
IO.unzip(zip,out) mustEqual(Set(out /"test1", out / "test2"))
IO.delete((out ** "*").get)
//Create a zip containing this lib directory but under a different directory in the zip
val finder: PathFinder = dirToZip ** "*" --- dirToZip //Remove dirToZip as you can't rebase a directory to itself
IO.zip(finder x rebase(dirToZip, "newlib"), base / "rebased.zip")
IO.createDirectory(out)
IO.unzip(base / "rebased.zip",out) mustEqual(Set(out /"newlib"/"test1", out / "newlib"/ "test2"))
})
See the docs
http://www.scala-sbt.org/0.12.2/docs/Detailed-Topics/Mapping-Files.html
http://www.scala-sbt.org/0.12.3/docs/Detailed-Topics/Paths.html
for tips on creating the Traversable object to pass to IO.zip