将结构数组分解为Spark中的列

问题描述

我想将结构数组分解为列（由struct字段定义）。例如，

I'd like to explode an array of structs to columns (as defined by the struct fields). E.g.

root
 |-- arr: array (nullable = true)
 |    |-- element: struct (containsNull = true)
 |    |    |-- id: long (nullable = false)
 |    |    |-- name: string (nullable = true)

应转换为

root
 |-- id: long (nullable = true)
 |-- name: string (nullable = true)

我可以做到

df
  .select(explode($"arr").as("tmp"))
  .select($"tmp.*")

如何在单个选择语句中做到这一点？

How can I do that in a single select statement?

I认为这样做可以，但是不幸的是，它不起作用：

I thought this could work, unfortunately it does not:

df.select(explode($"arr")(".*"))

推荐答案

单步解决方案仅适用于 MapType 列：

Single step solution is available only for MapType columns:

val df = Seq(Tuple1(Map((1L, "bar"), (2L, "foo")))).toDF

df.select(explode($"_1") as Seq("foo", "bar")).show

+---+---+
|foo|bar|
+---+---+
|  1|bar|
|  2|foo|
+---+---+

对于数组，您可以使用 flatMap ：

With arrays you can use flatMap:

val df = Seq(Tuple1(Array((1L, "bar"), (2L, "foo")))).toDF
df.as[Seq[(Long, String)]].flatMap(identity)

单个 SELECT 语句可以用SQL编写：

A single SELECT statement can written in SQL:

 df.createOrReplaceTempView("df")

spark.sql("SELECT x._1, x._2 FROM df LATERAL VIEW explode(_1) t AS x")

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