如何将 Column.isin 与列表一起使用? [英] How to use Column.isin with list?

问题描述

val items = List("a", "b", "c")

sqlContext.sql("select c1 from table")
          .filter($"c1".isin(items))
          .collect
          .foreach(println)

上面的代码抛出以下异常.

The code above throws the following exception.

Exception in thread "main" java.lang.RuntimeException: Unsupported literal type class scala.collection.immutable.$colon$colon List(a, b, c) 
at org.apache.spark.sql.catalyst.expressions.Literal$.apply(literals.scala:49)
at org.apache.spark.sql.functions$.lit(functions.scala:89)
at org.apache.spark.sql.Column$$anonfun$isin$1.apply(Column.scala:642)
at org.apache.spark.sql.Column$$anonfun$isin$1.apply(Column.scala:642)
at scala.collection.TraversableLike$$anonfun$map$1.apply(TraversableLike.scala:245)
at scala.collection.TraversableLike$$anonfun$map$1.apply(TraversableLike.scala:245)
at scala.collection.IndexedSeqOptimized$class.foreach(IndexedSeqOptimized.scala:33)
at scala.collection.mutable.WrappedArray.foreach(WrappedArray.scala:35)
at scala.collection.TraversableLike$class.map(TraversableLike.scala:245)
at scala.collection.AbstractTraversable.map(Traversable.scala:104)
at org.apache.spark.sql.Column.isin(Column.scala:642)

下面是我修复它的尝试.它编译并运行但不返回任何匹配项.不知道为什么.

Below is my attempt to fix it. It compiles and runs but doesn't return any match. Not sure why.

val items = List("a", "b", "c").mkString("\"","\",\"","\"")

sqlContext.sql("select c1 from table")
          .filter($"c1".isin(items))
          .collect
          .foreach(println)

推荐答案

根据文档，isin 需要一个可变参数，而不是一个列表.List在这里实际上是一个令人困惑的名字.您可以尝试将您的 List 转换为 vararg ，如下所示:

According to documentation, isin takes a vararg, not a list. List is actually a confusing name here. You can try converting your List to vararg like this:

val items = List("a", "b", "c")

sqlContext.sql("select c1 from table")
          .filter($"c1".isin(items:_*))
          .collect
          .foreach(println)

您使用 mkString 的变体可以编译，因为单个 String 也是可变参数(参数数量等于 1)，但这可能不是您想要实现的.

Your variant with mkString compiles, because one single String is also a vararg (with number of arguments equal to 1), but it is proably not what you want to achieve.

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