Scala 的 groupBy 身份如何工作? [英] How does Scala's groupBy identity work?

问题描述

我在浏览时发现了一个关于按字符对 String 进行分组的问题，例如:

I was browsing around and found a question about grouping a String by it's characters, such as this:

输入:

"aaabbbccccdd"

将产生以下输出:

"aaa"
"bbb"
"cccc"
"ddd"

我发现了这个建议:

val str = "aaabbbccccdd"[
val list = str.groupBy(identity).toList.sortBy(_._1).map(_._2)

这个identity 的家伙让我很好奇.我发现它是在 PreDef 像这样:

And this identity fellow got me curious. I found out it is defined in PreDef like this:

identity[A](x: A): A

所以基本上它会返回给定的任何东西，对吗?但这如何适用于对 groupBy 的调用?

So basically it returns whatever it is given, right? but how does that apply in the call to groupBy?

如果这是一个基本问题，我很抱歉，只是函数式编程仍然让我有点纠结.请让我知道我是否可以提供任何信息以使这个问题更清楚

I'm sorry if this is a basic question, is just that functional programming is still tangling my brains a little. Please let me know if there's any information I can give to make this question clearer

推荐答案

要理解这一点，只需使用 -Xprint:typer 选项调用 scala repl:

To understand this just call scala repl with -Xprint:typer option:

val res2: immutable.Map[Char,String] = augmentString(str).groupBy[Char]({
   ((x: Char) => identity[Char](x))
});

Scalac 将一个简单的 String 转换为 StringOps，它是 TraversableLike 的子类，它有一个 groupBy 方法:

Scalac converts a simple String into StringOps with is a subclass of TraversableLike which has a groupBy method:

def groupBy[K](f: A => K): immutable.Map[K, Repr] = {
    val m = mutable.Map.empty[K, Builder[A, Repr]]
    for (elem <- this) {
      val key = f(elem)
      val bldr = m.getOrElseUpdate(key, newBuilder)
      bldr += elem
    }
    val b = immutable.Map.newBuilder[K, Repr]
    for ((k, v) <- m)
      b += ((k, v.result))

    b.result
  }

因此 groupBy 包含一个映射，其中插入由标识函数返回的字符.

So groupBy contains a map into which inserts chars return by identity function.

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