如何按字典顺序比较scala元组? [英] How to lexicographically compare scala tuples?

问题描述

给定两个相同数量的元组，我如何按字典顺序比较它们?看起来这应该像下面的代码片段一样简单，但事实并非如此.有什么简单的例子来说明如何做吗?

Given two tuples of the same arity, how can I lexicographically compare them? It seems like this should be as simple as in the following snippet, but it isn't. Any simple example of how to do it?

var x = (1,2,3) < (1,2,4)

如果它们是列表，我可以定义一个递归函数来比较列表的头部，直到找到差异或列表的结尾，但我认为我不能对元组这样做.

Were they lists, I could define a recursive function that would compare the lists' head until a difference or the end of a list was found, but I don't think I could do that for tuples.

推荐答案

不简单，因为while

It's not simple because while

var x = (1,2,3) < (1,2)

看起来很简单，

var x = (1,false,3) < (1,2)

不是.你如何处理非有序类型?如何处理同一个元组位置的不同类型?

is not. How do you deal with non-ordered types? How do you deal with different types in the same tuple position?

您是否要求所有类型都相同?在这种情况下，您没有元组.元组的全部意义在于它的元数是固定的(你静态地知道它有多大)并且每个元素可以是不同的类型.

Do you mandate all types to be the same? In that case, you do not have a tuple. The whole point of a tuple is that its arity is fixed (you statically know how big it is) and each element can be of a different type.

如果我发现自己遇到了这个问题——而且我会非常努力地避免——我会抓住 Shapeless，将元组转换成 HLists 之类的东西，然后尝试进行比较.

If I found myself with that problem -- and I'd try very hard not to -- I'd grab Shapeless, convert the tuples into something like HLists, and then try to compare on that.

编辑

啊，现在容易多了:

import scala.math.Ordering.Implicits._
var x = (1,2,3) < (1,2,4)

这些额外的隐式不会自动可用，因为它们在某些情况下会导致隐式发散.

These extra implicits are not automatically available because they can result in diverging implicits under some circumstances.

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