基于集合的多作业 [英] Multi-Assignment based on Collection

问题描述

编辑

最初的问题是Collection to Tuple，因为我假设我需要一个元组才能进行可变的多重赋值。事实证明，可以直接对集合进行可变的多重赋值。

有一个简单的Seq [String]派生自一个正则表达式，我想转换为

我目前有：

这是最直接的方法是什么？

  val（clazz，date）= captures match {
 case x：Seq [String] =& （x（0），x（1））
} 
  

我的路由层有一堆regex匹配的路由，我将做val（a，b，c）多重赋值on（捕获组总是知道，因为如果regex不匹配路由不处理）。比较匹配{case .. => ..}会更好一个精简的解决方案

在Scala中将集合转换为元组最短的1-liner是什么？ p>

解决方案

这不是问题的答案，但可能以不同的方式解决问题。

你知道你可以匹配 xs：List [String] ，如下所示：

  val a :: b :: c :: _ = xs 
  

这将列表的前三个元素分配给 a，b，c ？您可以在 val 的声明中匹配 Seq 的其他内容，就像在 语句。请务必注意匹配的错误：

Scala模式匹配和try / catch

Edit

originally the question was "Collection to Tuple" as I assumed I needed a tuple in order to do variable multi-assignment. It turns out that one can do variable multi-assignment directly on collections. Retitled the question accordingly.

Original Have a simple Seq[String] derived from a regex that I would like to convert to a Tuple.

What's the most direct way to do so?

I currently have:

val(clazz, date) = captures match {
  case x: Seq[String] => (x(0), x(1))
}

Which is ok, but my routing layer has a bunch of regex matched routes that I'll be doing val(a,b,c) multi-assignment on (the capture group is always known since the route is not processed if regex does not match). Would be nice to have a leaner solution than match { case.. => ..}

What's the shortest 1-liner to convert collections to tuples in Scala?

解决方案

This is not an answer to the question but might solve the problem in a different way.

You know you can match a xs: List[String] like so:

val a :: b :: c :: _ = xs 

This assigns the first three elements of the list to a,b,c? You can match other things like Seq in the declaration of a val just like inside a case statement. Be sure you take care of matching errors:

Scala pattern matching and try/catch

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