在 Hive 中加载 SparkR 数据帧

loading SparkR data frame in Hive

我需要加载在 SparkR 中创建的 DataFrame,以便在 Hive 中加载。

#created a dataframe df_test
df_test <- createDataFrame(sqlContext, data.frame(mon = c(1,2,3,4,5), year = c(2011,2012,2013,2014,2015)))

#initialized the Hive context
>sc <- sparkR.init()
>hiveContext <- sparkRHive.init(sc)

#used the saveAsTable fn to save dataframe "df_test" in hive table named "table_hive"
>saveAsTable(df_test, "table_hive")

16/08/24 23:08:36 ERROR RBackendHandler: saveAsTable on 13 failed Error in invokeJava(isStatic = FALSE, objId$id, methodName, ...) : java.lang.RuntimeException: Tables created with SQLContext must be TEMPORARY. Use a HiveContext instead. at scala.sys.package$.error(package.scala:27) at org.apache.spark.sql.execution.SparkStrategies$DDLStrategy$.apply(SparkStrategies.scala:392) at org.apache.spark.sql.catalyst.planning.QueryPlanner$$anonfun.apply(QueryPlanner.scala:58) at org.apache.spark.sql.catalyst.planning.QueryPlanner$$anonfun.apply(QueryPlanner.scala:58) at scala.collection.Iterator$$anon.hasNext(Iterator.scala:371) at org.apache.spark.sql.catalyst.planning.QueryPlanner.plan(QueryPlanner.scala:59) at org.apache.spark.sql.execution.QueryExecution.sparkPlan$lzycompute(QueryExecution.scala:47) at org.apache.spark.sql.execution.QueryExecution.sparkPlan(QueryExecution.scala:45) at org.apache.spark.sql.execution.QueryExecution.executedPlan$lzycompute(QueryExecution.scala:52) at org.apache.spark.sql.execution.QueryExecution.executedPlan(QueryExecution.scala:52) at org.apache.spark.sql.execution

抛出上述错误。请帮忙。

范围内有 HiveContext 是不够的。每个数据框都绑定到一个特定的 SQLContext / SparkSession 实例,并且 df_test 显然是使用与 hiveContext

不同的上下文创建的

让我们举例说明:

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 Spark context is available as sc, SQL context is available as sqlContext
> library(magrittr)
> createDataFrame(sqlContext, mtcars) %>% saveAsTable("foo")
16/08/24 20:22:13 ERROR RBackendHandler: saveAsTable on 22 failed
Error in invokeJava(isStatic = FALSE, objId$id, methodName, ...) : 
  java.lang.RuntimeException: Tables created with SQLContext must be TEMPORARY. Use a HiveContext instead.
    at scala.sys.package$.error(package.scala:27)
    at org.apache.spark.sql.execution.SparkStrategies$DDLStrategy$.apply(SparkStrategies.scala:392)
    at org.apache.spark.sql.catalyst.planning.QueryPlanner$$anonfun.apply(QueryPlanner.scala:58)
    at org.apache.spark.sql.catalyst.planning.QueryPlanner$$anonfun.apply(QueryPlanner.scala:58)
    at scala.collection.Iterator$$anon.hasNext(Iterator.scala:396)
    at org.apache.spark.sql.catalyst.planning.QueryPlanner.plan(QueryPlanner.scala:59)
    at org.apache.spark.sql.execution.QueryExecution.sparkPlan$lzycompute(QueryExecution.scala:47)
    at org.apache.spark.sql.execution.QueryExecution.sparkPlan(QueryExecution.scala:45)
    at org.apache.spark.sql.execution.QueryExecution.executedPlan$lzycompute(QueryExecution.scala:52)
    at org.apache.spark.sql.execution.QueryExecution.executedPlan(QueryExecution.scala:52)
    at org.apache.spark.sql.execu
>
> hiveContext <- sparkRHive.init(sc)
> createDataFrame(hiveContext, mtcars) %>% saveAsTable("foo")
NULL