Question

例如，student列有一个StructType(("id", "int"), ("name", "string"))

| student     |
| ------------|
| [123,james] |

如何使用Student将一个列值转换为Dataset<Row>.map()类实例？我应该将列的值视为字符串数组并解析它以构造实例吗？

Answer 1

嵌套列值也是Row。因此，我们可以按名称获取属性的值，然后构造一个实例。

这是一些演示代码：

scala> val df = Seq((1, "james"), (2, "tony")).toDF("id", "name")
df: org.apache.spark.sql.DataFrame = [id: int, name: string]

scala> val dd = df.select(struct("*").alias("students"))
dd: org.apache.spark.sql.DataFrame = [students: struct<id: int, name: string>]

scala> dd.show()
+--------------------+
|            students|
+--------------------+
|     [1,james]      |
|     [2,tony]       |
+--------------------+

scala> rows(0).getStruct(0)
res9: org.apache.spark.sql.Row = [1,james]

我们可以看到，rows(0).getStruct(0)返回的单元格值为Row。

Answer 2

使用编码器然后收集。

scala> val df = Seq((1, "james"), (2, "tony")).toDF("id", "name")
df: org.apache.spark.sql.DataFrame = [id: int, name: string]

scala> val dd = df.select(struct("*").alias("students"))
dd: org.apache.spark.sql.DataFrame = [students: struct<id: int, name: string>]

scala> dd.show()
+--------------------+
|            students|
+--------------------+
|     [1,james]      |
|     [2,tony]       |
+--------------------+

scala> case class Student(id: Int, name: String)
defined class Student

scala> dd.select("students.*").as[Student].collectAsList
res6: java.util.List[Student] = [Student(1,james), Student(2,tony)]

如何将具有嵌套StructType的列转换为Spark SQL中的类实例？

2 个答案: