scala - 将结构数组分解为 Spark 中的列

Question

我想将一个结构数组分解为列(由结构字段定义)。例如。

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

应该转化为

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

我可以用

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

如何在单个 select 语句中做到这一点？

我认为这可以工作，不幸的是它没有:

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

Exception in thread "main" org.apache.spark.sql.AnalysisException: No such struct field .* in col;

Answer 1

单步解决方案仅适用于 MapType列:

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 :

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

单 SELECT语句可以用 SQL 编写:

 df.createOrReplaceTempView("df")

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