Scala for 同时循环遍历两个列表 [英] Scala for loop over two lists simultaneously

问题描述

我有一个 List[Message] 和一个 List[Author]，它们具有相同数量的项目，并且应该排序，以便在每个索引处，Message 来自 Author.

I have a List[Message] and a List[Author] which have the same number of items, and should be ordered so that at each index, the Message is from the Author.

我还有一个类，我们在这里称之为 SmartMessage，它的构造函数带有 2 个参数:一个 Message 和相应的 Author.

I also have class that we'll call here SmartMessage, with a constructor taking 2 arguments: a Message and the corresponding Author.

我想要做的是创建一个List[SmartMessage]，结合2个简单列表的数据.

What I want to do, is to create a List[SmartMessage], combining the data of the 2 simple lists.

额外问题:List 在 Scala 中是否保留插入顺序?只是为了确保我创建了具有相同顺序的 List[Message] 和一个 List[Author].

Extra question: does List preserve insertion order in Scala? Just to make sure I create List[Message] and a List[Author] with same ordering.

推荐答案

你可以使用 zip:

val ms: List[Message] = ???
val as: List[Author] = ???

var sms = for ( (m, a) <- (ms zip as)) yield new SmartMessage(m, a)

如果你不喜欢for-comprehensions，你可以使用map:

If you don't like for-comprehensions you could use map:

var sms = (ms zip as).map{ case (m, a) => new SmartMessage(m, a)}

方法 zip 创建对的集合.在这种情况下 List[(Message, Author)].

Method zip creates collection of pairs. In this case List[(Message, Author)].

您也可以在 Tuple2(以及 Tuple3)上使用 zipped 方法:

You could also use zipped method on Tuple2 (and on Tuple3):

var sms = (ms, as).zipped.map{ (m, a) => new SmartMessage(m, a)}

如您所见，在这种情况下，您不需要 map 中的模式匹配.

As you can see you don't need pattern matching in map in this case.

额外

List 是 Seq 并且 Seq 保留顺序.请参阅 scala 集合概览.

List is Seq and Seq preserves order. See scala collections overview.

集合有 3 个主要分支:Seq、Set 和 地图.

There are 3 main branches of collections: Seq, Set and Map.

• Seq 保留元素的顺序.
• Set 不包含重复元素.
• Map 包含从键到值.

• Seq preserves order of elements.
• Set contains no duplicate elements.
• Map contains mappings from keys to values.

List 是 linked list，所以你应该在前面加上元素，而不是追加.请参阅 性能特征 scala 集合.

List in scala is linked list, so you should prepend elements to it, not append. See Performance Characteristics of scala collections.

这篇关于Scala for 同时循环遍历两个列表的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！