在 Scala 中读取整个文件? [英] Read entire file in Scala?

问题描述

在 Scala 中将整个文件读入内存的简单而规范的方法是什么?(理想情况下，可以控制字符编码.)

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

我能想到的最好的是:

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

或者我应该使用 Java 的可怕成语之一，其中最好的(不使用外部库)似乎是:

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()

通过阅读邮件列表讨论，我不清楚 scala.io.Source 甚至应该是规范的 I/O 库.我不明白它的预期目的究竟是什么.

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.

...我想要一些简单易记的东西.例如，在这些语言中，很难忘记习语......

... 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()

推荐答案

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

顺便说一句，scala."并不是真正必要的，因为它总是在范围内，当然，您可以完全或部分导入 io 的内容，而不必加上io".

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.

然而，上面的代码使文件保持打开状态.为避免出现问题，您应该像这样关闭它:

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()

上面代码的另一个问题是，由于其实现性质，它的速度非常慢.对于较大的文件，应使用:

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

source.getLines mkString "\n"

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