使用带有选项字段的案例类将数据框转换为数据集 [英] spark convert dataframe to dataset using case class with option fields
问题描述
我有以下案例类:
case class Person(name: String, lastname: Option[String] = None, age: BigInt) {}
以及以下json:
{ "name": "bemjamin", "age" : 1 }
当我尝试将数据框转换为数据集时:
When I try to transform my dataframe into a dataset:
spark.read.json("example.json")
.as[Person].show()
它显示了以下错误:
线程主要" org.apache.spark.sql.AnalysisException中的异常:给定以下输入列,无法解析"
姓
":[年龄,姓名];
Exception in thread "main" org.apache.spark.sql.AnalysisException: cannot resolve '
lastname
' given input columns: [age, name];
我的问题是:如果我的模式是我的案例类,并且它定义了姓氏是可选的,那么as()是否应该进行转换?
My question is: If my schema is my case class and it defines that the lastname is optional, shouldn't the as() do the conversion?
我可以使用.map轻松修复此问题,但我想知道是否还有其他更清洁的替代方法.
I can easily fix this using a .map but I would like to know if there is another cleaner alternative to this.
推荐答案
我们还有一个解决上述问题的选项.需要2个步骤
We have one more option to solve above issue.There are 2 steps required
-
确保将可能缺失的字段声明为可为空Scala类型(如Option [_]).
Make sure that fields that can be missing are declared as nullable Scala types(like Option[_]).
提供一个模式参数,而不依赖于模式推断.例如,可以使用 Spark SQL Encoder :
Provide a schema argument and not depend on schema inference.You can use for example use Spark SQL Encoder:
import org.apache.spark.sql.Encoders
val schema = Encoders.product[Person].schema
您可以如下更新代码.
val schema = Encoders.product[Person].schema
val df = spark.read
.schema(schema)
.json("/Users/../Desktop/example.json")
.as[Person]
+--------+--------+---+
| name|lastname|age|
+--------+--------+---+
|bemjamin| null| 1|
+--------+--------+---+
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