我使用 Apache Spark 2.1.1(使用过 2.1.0,它是相同的,今天切换)。
我有一个数据集:
root
|-- muons: array (nullable = true)
| |-- element: struct (containsNull = true)
| | |-- reco::Candidate: struct (nullable = true)
| | |-- qx3_: integer (nullable = true)
| | |-- pt_: float (nullable = true)
| | |-- eta_: float (nullable = true)
| | |-- phi_: float (nullable = true)
| | |-- mass_: float (nullable = true)
| | |-- vertex_: struct (nullable = true)
| | | |-- fCoordinates: struct (nullable = true)
| | | | |-- fX: float (nullable = true)
| | | | |-- fY: float (nullable = true)
| | | | |-- fZ: float (nullable = true)
| | |-- pdgId_: integer (nullable = true)
| | |-- status_: integer (nullable = true)
| | |-- cachePolarFixed_: struct (nullable = true)
| | |-- cacheCartesianFixed_: struct (nullable = true)
如您所见,此架构中有 3 个空结构。我 100% 知道我可以阅读/操作/做任何事情。但是,当我尝试在镶木地板中写入磁盘时,出现以下异常:
dsReduced.write.format("parquet").save(outputPathName):
java.lang.IllegalStateException: Cannot build an empty group
at org.apache.parquet.Preconditions.checkState(Preconditions.java:91)
at org.apache.parquet.schema.Types$BaseGroupBuilder.build(Types.java:622)
at org.apache.parquet.schema.Types$BaseGroupBuilder.build(Types.java:497)
at org.apache.parquet.schema.Types$Builder.named(Types.java:286)
at org.apache.spark.sql.execution.datasources.parquet.ParquetSchemaConverter.convertField(ParquetSchemaConverter.scala:535)
at org.apache.spark.sql.execution.datasources.parquet.ParquetSchemaConverter.convertField(ParquetSchemaConverter.scala:321)
at org.apache.spark.sql.execution.datasources.parquet.ParquetSchemaConverter$$anonfun$convertField$1.apply(ParquetSchemaConverter.scala:534)
at org.apache.spark.sql.execution.datasources.parquet.ParquetSchemaConverter$$anonfun$convertField$1.apply(ParquetSchemaConverter.scala:533)
所以,基本上我想了解这是一个错误还是一个预期的行为???我还假设它与空结构有关。任何帮助将非常感激!
更新:我很快就创建了精简版本,并且可以正常工作!任何见解都会非常有帮助!
VK
Parquet 不写入空结构:
欲了解更多信息 - 请参阅此处https://issues.apache.org/jira/browse/SPARK-20593 https://issues.apache.org/jira/browse/SPARK-20593
VK
本文内容由网友自发贡献,版权归原作者所有,本站不承担相应法律责任。如您发现有涉嫌抄袭侵权的内容,请联系:hwhale#tublm.com(使用前将#替换为@)