Scala SortedMap:获取大于给定键的所有键 [英] Scala SortedMap : Get all keys greater than a given key

问题描述

给定一个 Scala collection.SortedMap 和一个键 k，获取所有键的最有效方法是什么(甚至更好，所有键值对) 大于 k 存储在排序映射中.返回的一组键应保留键的顺序.当然，我想避免仔细阅读整个数据结构(即使用 filterKeys)，并利用地图已排序的事实.

Given a Scala collection.SortedMap and a key k, what is the most efficient way of getting all keys (or even better, all key-value pairs) greater than k stored in the sorted map. The returned set of keys should preserve the order of keys. Of course, I would like to avoid to peruse the whole data structure (i.e. using filterKeys), and take advantage of the fact that the map is sorted.

我想做类似的事情:

val m = collection.SortedMap((1,1) -> "somevalue", (1,2) -> "somevalue", 
  (1,3) -> "somevalue", (2,1) -> "somevalue", (3,1) -> "somevalue")
m.getKeysGreaterThan((2,1))
// res0: scala.collection.SortedSet[(Int, Int)] = TreeSet((2,1), (3,1))

如果你能想到更合适的类似地图的数据结构，请提出.

If you can think of a more appropriate map-like data structure, please suggest it.

推荐答案

从 API 文档中试试这个:

m.from((2,1))

注意，结果包含键值.

我刚刚检查了 Scala 2.10，TreeMap.from 在 RedBlackTree 上调用了 from，这似乎是一个有效的实现(通常的 O(log n) 用于基于树的数据结构.

I just checked in Scala 2.10, TreeMap.from calls from on RedBlackTree, which appears to be an efficient implementation (the usual O(log n) for tree-based data-structures).

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