Scala：将地图大小n拆分为列表（地图最大大小3）

Stepanov Andrey 发表于 Dev

斯蒂芬诺夫·安德烈（Stepanov Andrey）

在：

Map("k1" -> "v1", "k2" -> "v2", "k3" -> "v3", "k4" -> "v4", "k5" -> "v5", "k6" -> "v6", "k7" -> "v7", "k8" -> "v8", "k9" -> "v9", "k0" -> "v0")

出：

List(Map("k1" -> "v1", "k2" -> "v2", "k3" -> "v3), Map("k4" -> "v4", "k5" -> "v5", "k6" -> "v6), Map("k7" -> "v7", "k8" -> "v8", "k9" -> "v9), Map("k0" -> "v0"))

宁静

val a = Map("k1" -> "v1", "k2" -> "v2", "k3" -> "v3", "k4" -> "v4", "k5" -> "v5", "k6" -> "v6", "k7" -> "v7", "k8" -> "v8", "k9" -> "v9", "k0" -> "v0")
a.grouped(3).toList

这给您：

res2: List[scala.collection.immutable.Map[String,String]] = List(Map(k2 -> v2, k0 -> v0, k5 -> v5), Map(k9 -> v9, k6 -> v6, k7 -> v7), Map(k1 -> v1, k4 -> v4, k3 -> v3), Map(k8 -> v8))

唯一没有订购的东西

为了保持顺序，您可以执行以下操作：

a.toList.sortBy(_._1).grouped(3).toList.map(_.toMap)

这给你：

res6: List[scala.collection.immutable.Map[String,String]] = List(Map(k0 -> v0, k1 -> v1, k2 -> v2), Map(k3 -> v3, k4 -> v4, k5 -> v5), Map(k6 -> v6, k7 -> v7, k8 -> v8), Map(k9 -> v9))

请注意，您的姓名首字母Map未正确排序（最后一个元素为“ k0”，但它应该是第一个）。但是，如果您要保留插入顺序并将地图列表按3分组，则该列表应该可以工作：

val b = scala.collection.mutable.LinkedHashMap("k1" -> "v1", "k2" -> "v2", "k3" -> "v3", "k4" -> "v4", "k5" -> "v5", "k6" -> "v6", "k7" -> "v7", "k8" -> "v8", "k9" -> "v9", "k0" -> "v0")
b.toList.grouped(3).toList.map(_.toMap)

结果是：

res8: List[scala.collection.immutable.Map[String,String]] = List(Map(k1 -> v1, k2 -> v2, k3 -> v3), Map(k4 -> v4, k5 -> v5, k6 -> v6), Map(k7 -> v7, k8 -> v8, k9 -> v9), Map(k0 -> v0))

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