Python pyspark 1.3.0将数据帧保存到配置单元表中
我正在python上使用spark 1.3.0 我有:Python pyspark 1.3.0将数据帧保存到配置单元表中,python,hadoop,pyspark,Python,Hadoop,Pyspark,我正在python上使用spark 1.3.0 我有: DF.show(3) ID Date Hour TimeInCluster Cluster Xcluster Ycluster 25342438156 2012-11-30 15:00:00 26 T 130270 165620 25342438156 2012-11-30 16:00:00 86 D 136850
DF.show(3)
ID Date Hour TimeInCluster Cluster Xcluster Ycluster
25342438156 2012-11-30 15:00:00 26 T 130270 165620
25342438156 2012-11-30 16:00:00 86 D 136850 177070
25342438156 2012-11-30 17:00:00 35 D 136850 177070
我试图将DF保存到不存在的蜂巢表中
我该怎么做
多谢各位
我将代码更改为:
sqlContext = HiveContext(sc)
FinalDf.write().mode(SaveMode.Overwrite).saveAsTable("myDB.sixuserstablediary")
但我犯了那个错误
py4j.protocol.Py4JJavaError: An error occurred while calling o280.apply.
: org.apache.spark.sql.AnalysisException: Cannot resolve column name "write" among (IMSI, Date, Hour, TimeInCluster, Cluster, Xcluster, Ycluster);
at org.apache.spark.sql.DataFrame$$anonfun$resolve$1.apply(DataFrame.scala:162)
at org.apache.spark.sql.DataFrame$$anonfun$resolve$1.apply(DataFrame.scala:162)
at scala.Option.getOrElse(Option.scala:120)
at org.apache.spark.sql.DataFrame.resolve(DataFrame.scala:161)
at org.apache.spark.sql.DataFrame.col(DataFrame.scala:436)
at org.apache.spark.sql.DataFrame.apply(DataFrame.scala:426)
at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method)
at sun.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:62)
at sun.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:43)
at java.lang.reflect.Method.invoke(Method.java:497)
at py4j.reflection.MethodInvoker.invoke(MethodInvoker.java:231)
at py4j.reflection.ReflectionEngine.invoke(ReflectionEngine.java:379)
at py4j.Gateway.invoke(Gateway.java:259)
at py4j.commands.AbstractCommand.invokeMethod(AbstractCommand.java:133)
at py4j.commands.CallCommand.execute(CallCommand.java:79)
at py4j.GatewayConnection.run(GatewayConnection.java:207)
at java.lang.Thread.run(Thread.java:745)
您需要使用Spark HiveContext 进口Spark HiveContex
from pyspark.sql import HiveContext
sqlContext = HiveContext(sc)
从dataframe创建一个临时表,然后通过从临时表中选择数据插入配置单元表
// Register the dataframe
df.registerTempTable("tbl_tmp")
sqlContext.sql("create table default.tbl_hive_data as select * from tbl_tmp")
请参考下面的链接:-我尝试了,但得到了错误,我的代码:FinalDf.registertEmptableMyTestableSqlContext.sqlcreate table sixuserstablediary as select*from MyTestable错误:sqlContext.sqlcreate table sixuserstablediary as select*from MyTestable;文件/opt/cloudera/parcels/CDH-5.4.7-1.cdh5.4.7.p0.3/lib/spark/pyspark/sql/context.py,第528行,在sql返回数据框架self中,在调用文件/opt/cloudera/parcels/CDH-5.4.7-1.cdh5.4.7.p0.3/lib/spark/python/lib/py4j-0.8.2.1-src.zip/py4j/protocol.py第300行的get_return_值中