Question

我遇到了麻烦＆＃34;离开加入＆＃34;在斯卡拉。例如，

val order: Array[Array[String]] = Array(Array("o2","ad1"), Array("o1","ad2"))

val line: Array[Array[String]] = Array(Array("l1","o1","x"),Array("l2","o2","y"))

期望的输出：

Array(Array("l1","o1","x","ad2"),Array("l2","o2","y","ad1"))

Answer 1

数据的建模质量有问题：为什么要将所有内容表示为String数组？但是，好的，你走了：

    val order: Array[Array[String]] = Array(
  Array("o2","ad1"), 
  Array("o1","ad2"),
  Array("o7","ad5")
)
val line: Array[Array[String]] = Array(
  Array("l1","o1","x"),
  Array("l2","o2","y"),
  Array("l3","o13","z")
)

val orderAsMap = order.map{x => (x(0), x(1))}.toMap
val leftOuterJoin = for (x <- line) yield (x :+ orderAsMap.getOrElse(x(1), "<NULL>"))

// print to see the result
for (row <- leftOuterJoin) println(row.mkString(", "))

// output:
// 
// l1, o1, x, ad2
// l2, o2, y, ad1
// l3, o13, z, <NULL>

中间toMap是必要的，这样您就不必每次都遍历整个数组。实际加入是使用:+运算符完成的。

如果您必须定期处理此类问题，请考虑使用适当的库。

编辑：感谢@igorpcholkin提醒我＆＃34;离开加入＆＃34;是＆＃34;左外连接＆＃34;的同义词在通常的命名法中。第一个版本更像是内部联接。我会把它留在这里以防万一你想要一个内部联接：

val order: Array[Array[String]] = Array(Array("o2","ad1"), Array("o1","ad2"))
val line: Array[Array[String]] = Array(
  Array("l1","o1","x"),
  Array("l2","o2","y"),
  Array("l3","o13","z")
)

val orderAsMap = order.map{x => (x(0), x(1))}.toMap
val lj = for {
  x <- line 
  if orderAsMap.contains(x(1))
} yield (x :+ orderAsMap(x(1)))

// print to see the result
for (row <- lj) println(row.mkString(", "))

// output:
// 
// l1, o1, x, ad2
// l2, o2, y, ad1

我如何＆＃34;离开加入＆＃34;在Scala数组[Array [String]]？

1 个答案: