scala Spark 的 Column.isin 函数不带 List
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Spark's Column.isin function does not take List
提问by Jake Fund
I am trying to filter out rows from my Spark Dataframe.
我正在尝试从 Spark 数据框中过滤掉行。
val sequence = Seq(1,2,3,4,5)
df.filter(df("column").isin(sequence))
Unfortunately, I get an unsupported literal type error
不幸的是，我收到了一个不受支持的文字类型错误
java.lang.RuntimeException: Unsupported literal type class scala.collection.immutable.$colon$colon List(1,2,3,4,5)
according to the documentationit takes a scala.collection.Seq list
根据文档，它需要一个 scala.collection.Seq 列表
I guess I don't want a literal? Then what can I take in, some sort of wrapper class?
我想我不想要文字？那么我可以接受什么，某种包装类？
回答by eliasah
@JustinPihony's answer is correct but it's incomplete. The isinfunction takes a repeated parameterfor argument, so you'll need to pass it as so :
@JustinPihony 的回答是正确的，但不完整。该isin函数采用重复的参数作为参数，因此您需要将其传递如下：
scala> val df = sc.parallelize(Seq(1,2,3,4,5,6,7,8,9)).toDF("column")
// df: org.apache.spark.sql.DataFrame = [column: int]
scala> val sequence = Seq(1,2,3,4,5)
// sequence: Seq[Int] = List(1, 2, 3, 4, 5)
scala> val result = df.filter(df("column").isin(sequence : _*))
// result: org.apache.spark.sql.DataFrame = [column: int]
scala> result.show
// +------+
// |column|
// +------+
// | 1|
// | 2|
// | 3|
// | 4|
// | 5|
// +------+
回答by Justin Pihony
This is happening because the underlying Scala implementation uses varargs, so the documentation in Java is not quite correct. It is using the @varargsannotation, so you can just pass in an array.
发生这种情况是因为底层Scala 实现使用varargs，因此 Java 中的文档不太正确。它正在使用@varargs注释，因此您只需传入一个数组即可。
