如何从镶木地板文件中读取和写入自定义类

Question

我正在尝试使用 DataFrame/datasets 为某个类类型编写 parquet 读/写类

类架构：

class A {
  long count;
  List<B> listOfValues;
}
class B {
  String id;
  long count;
}

代码：

  String path = "some path";
  List<A> entries = somerandomAentries();
  JavaRDD<A> rdd = sc.parallelize(entries, 1);
  DataFrame df = sqlContext.createDataFrame(rdd, A.class);

  df.write().parquet(path);
  DataFrame newDataDF = sqlContext.read().parquet(path);
  newDataDF.show();

当我尝试运行它时，它会引发错误。我在这里想念什么？创建数据框时是否需要为整个班级提供架构 错误：

    Caused by: scala.MatchError: B(Id=abc, count=0) (of class B)
    at org.apache.spark.sql.catalyst.CatalystTypeConverters$StructConverter.toCatalystImpl(CatalystTypeConverters.scala:255)
    at org.apache.spark.sql.catalyst.CatalystTypeConverters$StructConverter.toCatalystImpl(CatalystTypeConverters.scala:250)
    at org.apache.spark.sql.catalyst.CatalystTypeConverters$CatalystTypeConverter.toCatalyst(CatalystTypeConverters.scala:102)
    at org.apache.spark.sql.catalyst.CatalystTypeConverters$ArrayConverter.toCatalystImpl(CatalystTypeConverters.scala:169)
    at org.apache.spark.sql.catalyst.CatalystTypeConverters$ArrayConverter.toCatalystImpl(CatalystTypeConverters.scala:153)
    at org.apache.spark.sql.catalyst.CatalystTypeConverters$CatalystTypeConverter.toCatalyst(CatalystTypeConverters.scala:102)
    at org.apache.spark.sql.catalyst.CatalystTypeConverters$$anonfun$createToCatalystConverter$2.apply(CatalystTypeConverters.scala:401)
    at org.apache.spark.sql.SQLContext$$anonfun$org$apache$spark$sql$SQLContext$$beansToRows$1$$anonfun$apply$1.apply(SQLContext.scala:1358)
    at org.apache.spark.sql.SQLContext$$anonfun$org$apache$spark$sql$SQLContext$$beansToRows$1$$anonfun$apply$1.apply(SQLContext.scala:1358)
    at scala.collection.TraversableLike$$anonfun$map$1.apply(TraversableLike.scala:244)
    at scala.collection.TraversableLike$$anonfun$map$1.apply(TraversableLike.scala:244)
    at scala.collection.IndexedSeqOptimized$class.foreach(IndexedSeqOptimized.scala:33)
    at scala.collection.mutable.ArrayOps$ofRef.foreach(ArrayOps.scala:108)
    at scala.collection.TraversableLike$class.map(TraversableLike.scala:244)
    at scala.collection.mutable.ArrayOps$ofRef.map(ArrayOps.scala:108)
    at org.apache.spark.sql.SQLContext$$anonfun$org$apache$spark$sql$SQLContext$$beansToRows$1.apply(SQLContext.scala:1358)
    at org.apache.spark.sql.SQLContext$$anonfun$org$apache$spark$sql$SQLContext$$beansToRows$1.apply(SQLContext.scala:1356)
    at scala.collection.Iterator$$anon$11.next(Iterator.scala:328)
    at org.apache.spark.sql.execution.datasources.DefaultWriterContainer.writeRows(WriterContainer.scala:263)
    ... 8 more

Answer 1

您收到错误是因为 Spark 1.6 版本不支持嵌套的 JavaBean。请看https://spark.apache.org/docs/1.6.0/sql-programming-guide.html#inferring-the-schema-using-reflection

目前，Spark SQL 不支持包含嵌套或复杂类型（如列表或数组）的 JavaBean。