Read entire file in Scala?

Question

What's a simple and canonical way to read an entire file into memory in Scala? (Ideally, with control over character encoding.)

The best I can come up with is:

scala.io.Source.fromPath("file.txt").getLines.reduceLeft(_+_)

or am I supposed to use one of Java's god-awful idioms, the best of which (without using an external library) seems to be:

import java.util.Scanner
import java.io.File
new Scanner(new File("file.txt")).useDelimiter("\\Z").next()

From reading mailing list discussions, it's not clear to me that scala.io.Source is even supposed to be the canonical I/O library. I don't understand what its intended purpose is, exactly.

... I'd like something dead-simple and easy to remember. For example, in these languages it's very hard to forget the idiom ...

Ruby    open("file.txt").read
Ruby    File.read("file.txt")
Python  open("file.txt").read()

Answer 1

val lines = scala.io.Source.fromFile("file.txt").mkString

By the way, "scala." isn't really necessary, as it's always in scope anyway, and you can, of course, import io's contents, fully or partially, and avoid having to prepend "io." too.

The above leaves the file open, however. To avoid problems, you should close it like this:

val source = scala.io.Source.fromFile("file.txt")
val lines = try source.mkString finally source.close()

Another problem with the code above is that it is horribly slow due to its implementation. For larger files one should use:

source.getLines mkString "\n"

Answer 2

Just to expand on Daniel's solution, you can shorten things up tremendously by inserting the following import into any file which requires file manipulation:

import scala.io.Source._

With this, you can now do:

val lines = fromFile("file.txt").getLines

I would be wary of reading an entire file into a single String. It's a very bad habit, one which will bite you sooner and harder than you think. The getLines method returns a value of type Iterator[String]. It's effectively a lazy cursor into the file, allowing you to examine just the data you need without risking memory glut.

Oh, and to answer your implied question about Source: yes, it is the canonical I/O library. Most code ends up using java.io due to its lower-level interface and better compatibility with existing frameworks, but any code which has a choice should be using Source, particularly for simple file manipulation.

Answer 3

Java 8+

import java.nio.charset.StandardCharsets
import java.nio.file.{Files, Paths}

val path = Paths.get("file.txt")
new String(Files.readAllBytes(path), StandardCharsets.UTF_8)

Java 11+

import java.nio.charset.StandardCharsets
import java.nio.file.{Files, Path}

val path = Path.of("file.txt")
Files.readString(path, StandardCharsets.UTF_8)

These offer control over character encoding, and no resources to clean up. It's also faster than other patterns (e.g. getLines().mkString("\n")) due to more efficient allocation patterns.

Answer 4

// for file with utf-8 encoding
val lines = scala.io.Source.fromFile("file.txt", "utf-8").getLines.mkString

Answer 5

(EDIT: This does not work in scala 2.9 and maybe not 2.8 either)

Use trunk:

scala> io.File("/etc/passwd").slurp
res0: String = 
##
# User Database
# 
... etc

Answer 6

I've been told that Source.fromFile is problematic. Personally, I have had problems opening large files with Source.fromFile and have had to resort to Java InputStreams.

Another interesting solution is using scalax. Here's an example of some well commented code that opens a log file using ManagedResource to open a file with scalax helpers: http://pastie.org/pastes/420714

Answer 7

Using getLines() on scala.io.Source discards what characters were used for line terminators (\n, \r, \r\n, etc.)

The following should preserve it character-for-character, and doesn't do excessive string concatenation (performance problems):

def fileToString(file: File, encoding: String) = {
  val inStream = new FileInputStream(file)
  val outStream = new ByteArrayOutputStream
  try {
    var reading = true
    while ( reading ) {
      inStream.read() match {
        case -1 => reading = false
        case c => outStream.write(c)
      }
    }
    outStream.flush()
  }
  finally {
    inStream.close()
  }
  new String(outStream.toByteArray(), encoding)
}

Answer 8

If you don't mind a third-party dependency, you should consider using my OS-Lib library. This makes reading/writing files and working with the filesystem very convenient:

// Make sure working directory exists and is empty
val wd = os.pwd/"out"/"splash"
os.remove.all(wd)
os.makeDir.all(wd)

// Read/write files
os.write(wd/"file.txt", "hello")
os.read(wd/"file.txt") ==> "hello"

// Perform filesystem operations
os.copy(wd/"file.txt", wd/"copied.txt")
os.list(wd) ==> Seq(wd/"copied.txt", wd/"file.txt")

with one-line helpers for reading bytes, reading chunks, reading lines, and many other useful/common operations

Answer 9

Just like in Java, using CommonsIO library:

FileUtils.readFileToString(file, StandardCharsets.UTF_8)

Also, many answers here forget Charset. It's better to always provide it explicitly, or it will hit one day.

Answer 10

One more: https://github.com/pathikrit/better-files#streams-and-codecs

Various ways to slurp a file without loading the contents into memory:

val bytes  : Iterator[Byte]            = file.bytes
val chars  : Iterator[Char]            = file.chars
val lines  : Iterator[String]          = file.lines
val source : scala.io.BufferedSource   = file.content 

You can supply your own codec too for anything that does a read/write (it assumes scala.io.Codec.default if you don't provide one):

val content: String = file.contentAsString  // default codec
// custom codec:
import scala.io.Codec
file.contentAsString(Codec.ISO8859)
//or
import scala.io.Codec.string2codec
file.write("hello world")(codec = "US-ASCII")

Answer 11

For emulating Ruby syntax (and convey the semantics) of opening and reading a file, consider this implicit class (Scala 2.10 and upper),

import java.io.File

def open(filename: String) = new File(filename)

implicit class RichFile(val file: File) extends AnyVal {
  def read = io.Source.fromFile(file).getLines.mkString("\n")
}

In this way,

open("file.txt").read

Answer 12

You do not need to parse every single line and then concatenate them again...

Source.fromFile(path)(Codec.UTF8).mkString

I prefer to use this:

import scala.io.{BufferedSource, Codec, Source}
import scala.util.Try

def readFileUtf8(path: String): Try[String] = Try {
  val source: BufferedSource = Source.fromFile(path)(Codec.UTF8)
  val content = source.mkString
  source.close()
  content
}

Answer 13

The obvious question being "why do you want to read in the entire file?" This is obviously not a scalable solution if your files get very large. The scala.io.Source gives you back an Iterator[String] from the getLines method, which is very useful and concise.

It's not much of a job to come up with an implicit conversion using the underlying java IO utilities to convert a File, a Reader or an InputStream to a String. I think that the lack of scalability means that they are correct not to add this to the standard API.

Answer 14

you can also use Path from scala io to read and process files.

import scalax.file.Path

Now you can get file path using this:-

val filePath = Path("path_of_file_to_b_read", '/')
val lines = file.lines(includeTerminator = true)

You can also Include terminators but by default it is set to false..

Answer 15

as a few people mentioned scala.io.Source is best to be avoided due to connection leaks.

Probably scalax and pure java libs like commons-io are the best options until the new incubator project (ie scala-io) gets merged.

Answer 16

For faster overall reading / uploading a (large) file, consider increasing the size of bufferSize (Source.DefaultBufSize set to 2048), for instance as follows,

val file = new java.io.File("myFilename")
io.Source.fromFile(file, bufferSize = Source.DefaultBufSize * 2)

Note Source.scala. For further discussion see Scala fast text file read and upload to memory.

Answer 17

print every line, like use Java BufferedReader read ervery line, and print it:

scala.io.Source.fromFile("test.txt" ).foreach{  print  }

equivalent:

scala.io.Source.fromFile("test.txt" ).foreach( x => print(x))

Answer 18

import scala.io.source
object ReadLine{
def main(args:Array[String]){
if (args.length>0){
for (line <- Source.fromLine(args(0)).getLine())
println(line)
}
}

in arguments you can give file path and it will return all lines

Answer 19

You can use

Source.fromFile(fileName).getLines().mkString

however it should be noticed that getLines() removes all new line characters. If you want save formatting you should use

Source.fromFile(fileName).iter.mkString