[英]scala turn List of Strings to a key/value map
I have a single ordered array of strings imported from an external system in the form of: 我有一个从外部系统导入的单个有序字符串数组,形式如下:
val l = List("key1", "val1", "key2", "val2", etc...)
what's the scala way to turn this into a map where I can get iterate over the keys and get the associated vals? 什么是scala方式将其转换为一个地图,我可以迭代键并获得相关的val?
thx 谢谢
My answer is similar to Daniel's but I would use collect
instead of map
: 我的回答类似于丹尼尔,但我会使用
collect
而不是map
:
l.grouped(2).collect { case List(k, v) => k -> v }.toMap
If you have a list with an unmatched key this won't throw an exception: 如果您有一个包含不匹配键的列表,则不会抛出异常:
scala> val l = List("key1", "val1", "key2", "val2", "key3")
l: List[String] = List(key1, val1, key2, val2, key3)
scala> l.grouped(2).collect { case List(k, v) => k -> v }.toMap
res22: scala.collection.immutable.Map[String,String] = Map(key1 -> val1, key2 -> val2)
scala> l.grouped(2).map { case List(k, v) => k -> v }.toMap
scala.MatchError: List(key3) (of class scala.collection.immutable.$colon$colon)
Some context, grouped
returns a List[List[String]]
where every inner list has two elements: 一些上下文,
grouped
返回List[List[String]]
,其中每个内部列表都有两个元素:
scala> l.grouped(2).toList // the toList is to force the iterator to evaluate.
res26: List[List[String]] = List(List(key1, val1), List(key2, val2))
then with collect
you match on the inner lists and create a tuple, at the end toMap
transforms the list to a Map
. 然后使用
collect
匹配内部列表并创建一个元组,最后toMap
将列表转换为Map
。
l.grouped(2).map { case List(k, v) => k -> v }.toMap
You'll want to get them into pairs then provide them to a map 您需要将它们成对,然后将它们提供给地图
val lst: List[String] = List("key1", "val1", "key2", "val2")
val pairs = lst zip lst.tail
val m: Map[String,String] = pairs.toMap
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