火花高阶函数变换输出结构

Question

我如何transform 一个结构数组再次使用 spark 高阶函数的结构？

数据集：

case class Foo(thing1:String, thing2:String, thing3:String)
case class Baz(foo:Foo, other:String)
case class Bar(id:Int, bazes:Seq[Baz])
import spark.implicits._
val df = Seq(Bar(1, Seq(Baz(Foo("first", "second", "third"), "other"), Baz(Foo("1", "2", "3"), "else")))).toDF
df.printSchema
df.show(false)

我想连接所有thing1, thign2, thing3，但保留每个bar 的other 属性。

一个简单的：

scala> df.withColumn("cleaned", expr("transform(bazes, x -> x)")).printSchema
root
 |-- id: integer (nullable = false)
 |-- bazes: array (nullable = true)
 |    |-- element: struct (containsNull = true)
 |    |    |-- foo: struct (nullable = true)
 |    |    |    |-- thing1: string (nullable = true)
 |    |    |    |-- thing2: string (nullable = true)
 |    |    |    |-- thing3: string (nullable = true)
 |    |    |-- other: string (nullable = true)
 |-- cleaned: array (nullable = true)
 |    |-- element: struct (containsNull = true)
 |    |    |-- foo: struct (nullable = true)
 |    |    |    |-- thing1: string (nullable = true)
 |    |    |    |-- thing2: string (nullable = true)
 |    |    |    |-- thing3: string (nullable = true)
 |    |    |-- other: string (nullable = true)

只会把东西复制过来。

所需的连接操作：

 df.withColumn("cleaned", expr("transform(bazes, x -> concat(x.foo.thing1, '::', x.foo.thing2, '::', x.foo.thing3))")).printSchema

很遗憾，将删除 other 列中的所有值：

 +---+----------------------------------------------------+-------------------------------+
|id |bazes                                               |cleaned                        |
+---+----------------------------------------------------+-------------------------------+
|1  |[[[first, second, third], other], [[1, 2, 3], else]]|[first::second::third, 1::2::3]|
+---+----------------------------------------------------+-------------------------------+

如何保留这些？ 试图保留元组：

df.withColumn("cleaned", expr("transform(bazes, x -> (concat(x.foo.thing1, '::', x.foo.thing2, '::', x.foo.thing3), x.other))")).printSchema

失败：

.AnalysisException: cannot resolve 'named_struct('col1', concat(namedlambdavariable().`foo`.`thing1`, '::', namedlambdavariable().`foo`.`thing2`, '::', namedlambdavariable().`foo`.`thing3`), NamePlaceholder(), namedlambdavariable().`other`)' due to data type mismatch: Only foldable string expressions are allowed to appear at odd position, got: NamePlaceholder; line 1 pos 22;

编辑

想要的输出：

• 一个包含内容的新列：

  [[first::second::third, other], [1::2::3,else]

保留列other

Answer 1

通过这种方式，您可以实现您想要的输出。您不能直接访问其他值 bcoz foo 并且其他值共享相同的层次结构。所以你需要单独访问其他。

scala>  df.withColumn("cleaned", expr("transform(bazes, x -> struct(concat(x.foo.thing1, '::', x.foo.thing2, '::', x.foo.thing3),cast(x.other as string)))")).show(false)
+---+----------------------------------------------------+------------------------------------------------+
|id |bazes                                               |cleaned                                         |
+---+----------------------------------------------------+------------------------------------------------+

打印架构

scala>  df.withColumn("cleaned", expr("transform(bazes, x -> struct(concat(x.foo.thing1, '::', x.foo.thing2, '::', x.foo.thing3),cast(x.other as string)))")).printSchema
root
 |-- id: integer (nullable = false)
 |-- bazes: array (nullable = true)
 |    |-- element: struct (containsNull = true)
 |    |    |-- foo: struct (nullable = true)
 |    |    |    |-- thing1: string (nullable = true)
 |    |    |    |-- thing2: string (nullable = true)
 |    |    |    |-- thing3: string (nullable = true)
 |    |    |-- other: string (nullable = true)
 |-- cleaned: array (nullable = true)
 |    |-- element: struct (containsNull = false)
 |    |    |-- col1: string (nullable = true)
 |    |    |-- col2: string (nullable = true)

如果您还有任何与此相关的问题，请告诉我。