我正在尝试将spark数据帧 org.apache.spark.sql.DataFrame 转换为sparklyr表 tbl_spark . 我尝试使用 sdf_register ，但失败并出现以下错误 .
在这里，df是spark数据帧 .
sdf_register(df, name = "my_tbl")
错误是，
Error: org.apache.spark.sql.AnalysisException: Table not found: my_tbl; line 2 pos 17
at org.apache.spark.sql.catalyst.analysis.package$AnalysisErrorAt.failAnalysis(package.scala:42)
at org.apache.spark.sql.catalyst.analysis.Analyzer$ResolveRelations$.getTable(Analyzer.scala:306)
at org.apache.spark.sql.catalyst.analysis.Analyzer$ResolveRelations$$anonfun$apply$9.applyOrElse(Analyzer.scala:315)
at org.apache.spark.sql.catalyst.analysis.Analyzer$ResolveRelations$$anonfun$apply$9.applyOrElse(Analyzer.scala:310)
at org.apache.spark.sql.catalyst.plans.logical.LogicalPlan$$anonfun$resolveOperators$1.apply(LogicalPlan.scala:57)
at org.apache.spark.sql.catalyst.plans.logical.LogicalPlan$$anonfun$resolveOperators$1.apply(LogicalPlan.scala:57)
at org.apache.spark.sql.catalyst.trees.CurrentOrigin$.withOrigin(TreeNode.scala:69)
at org.apache.spark.sql.catalyst.plans.logical.LogicalPlan.resolveOperators(LogicalPlan.scala:56)
at org.apache.spark.sql.catalyst.plans.logical.LogicalPlan$$anonfun$1.apply(LogicalPlan.scala:54)
at org.apache.spark.sql.catalyst.plans.logical.LogicalPlan$$anonfun$1.apply(LogicalPlan.scala:54)
at org.apache.spark.sql.catalyst.trees.TreeNode$$anonfun$4.apply(TreeNode.scala:281)
at scala.collection.Iterator$$anon$11.next(Iterator.scala:328)
at scala.collection.Iterator$class.foreach(Iterator.scala:727)
我错过了什么吗？或者有没有更好的方法将其转换为 tbl_spark ？
谢谢！
1 回答
使用
sdf_copy_to()或dplyr::copy_to()，例如my_tbl <- sdf_copy_to(sc, df, "my_tbl")