What is the fastest way to read an entire file into a String in Scala?

Question

I am currently using the following to do this.

val string = scala.io.Source.fromFile(filePath).mkString

However, I've noticed that it is quite slow. Is there any better (in terms of speed) method to read the whole file into a string?

According to [this answer](http://stackoverflow.com/questions/1284423/read-entire-file-in-scala) you might want to try something like `source.getLines mkString "\n"` — evan.oman, Jun 29 '16 at 16:11
I did new String(Files.readAllBytes(Paths.get(filePath))). Works like a breeze! — pythonic, Jun 29 '16 at 17:04

score 2 · Accepted Answer · answered Jun 29 '16 at 17:06

2

I used the following. This is much faster than my previous approach.

import java.nio.file.Files
import java.nio.file.Paths

val string = new String(Files.readAllBytes(Paths.get(filePath)))

answered Jun 29 '16 at 17:06

pythonic

1

You might want to specify the encoding too. – Gábor Bakos Jun 29 '16 at 18:33

1 Answers1