【发布时间】:2017-04-21 07:00:12
【问题描述】:
我有一个地图列表:
val input = List( Map("id" -> "abc", "val" -> 100),
Map("id" -> "abc", "val" -> 109),
Map("id" -> "bca", "val" -> 115),
Map("id" -> "acb", "val" -> 100),
Map("id" -> "cba", "val" -> 105),
Map("id" -> "cba", "val" -> 110),
Map("id" -> "cba", "val" -> 116) )
我需要从中过滤掉所有共享键“id”的公共值的映射。换句话说,我必须只保留 key id 的值是唯一的映射。
val output = List( Map("id" -> "bca", "val" -> 115),
Map("id" -> "acb", "val" -> 100) )
【问题讨论】:
标签: scala list dictionary filter