Ruby 到 Scala 代码转换 - 在 Scala 中排序 [英] Ruby to Scala code translation - Sorting in Scala

问题描述

我正在将一些代码从 Ruby 转换为 Scala.问题是我这辈子从来没有给 Ruby 编程过.它进展顺利，但现在我遇到了一个我不知道的行，因为我是 Scala 的新手，我不了解排序机制.所以我想把这个 ruby​​ 行翻译成 Scala:

I'm converting some code from Ruby to Scala. Problem is that I never programmed Ruby in my life. It's going well, but now I reached a line that I don't know because I'm new in Scala and I don't understand the sorting mechanism. So I want to translate this ruby line to scala:

fronts[last_front].sort! {|x,y| crowded_comparison_operator(x,y)}

fronts 是 Vector[Vector[Map[String, Any]]]

last_front 是一个 Int

crowded_comparison_operator(x,y) 返回 -1、0 或 1，x 和 y 是 Map[String, Any]

crowded_comparison_operator(x,y) returns -1, 0 or 1, x and y are Map[String, Any]

推荐答案

标准 Scala 集合有两种可能性:

You have two possibilities with standard Scala collections:

• 将 crowded_comparison_operator 的 -1, 0, 1 输出转换为布尔值，告诉您第一个元素是否小于第二个元素，然后使用 sortWith.
• 定义一个新的Ordering，将其显式传递给sorted 方法.

• Convert the -1, 0, 1 output of crowded_comparison_operator into a boolean that tells you whether the first element is less than the second element, then use sortWith.
• define a new Ordering, pass it explicitly to the sorted method.

sortWith 方法

The sortWith method

第一个元素小于第二个元素当且仅当 crowded_comparison_operator 返回 -1，所以你可以这样做:

The first element is less than the second element if and only if crowded_comparison_operator returns -1, so you could do this:

fronts(last_front).sortWith{ (x, y) => crowded_comparison_operator(x, y) < 0 }

<小时>

为sorted

sorted 方法采用一个隐式的 Ordering 参数.您可以定义自己的自定义排序，并明确传递它:

The sorted method takes an implicit Ordering parameter. You can define your own custom ordering, and pass it explicitly:

import scala.math.Ordering

fronts(last_front).sorted(new Ordering[Vector[Map[String, Any]]] {
  def compare(
    x: Vector[Map[String, Any]], 
    y: Vector[Map[String, Any]]
  ): Int = crowded_comparison_operator(x, y)
})

或更短，Scala 版本支持 SAM(从 2.11.5 开始，如果我没记错的话):

or shorter, with scala versions supporting SAM (since 2.11.5, if I remember correctly):

fronts(last_front).sorted(
  (x: Vector[Map[String, Any], y: Vector[Map[String, Any]]) => 
    crowded_comparison_operator(x, y)
)

请注意，正如@mikej 所指出的，Ruby 的 sort! 就地对数组进行排序.这不适用于不可变向量，因此您必须相应地调整代码.

Note that, as @mikej has pointed out, Ruby's sort! sorts the array in-place. This cannot work for an immutable vector, so you have to adjust your code accordingly.

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