将RDD转换为Spark Dataframe(Pyspark)。这很有效。但是给出了新的错误

时间:2018-04-16 06:17:12

标签: python apache-spark pyspark spark-dataframe rdd

我有一个RDD:

rd.take(2)

[Row(id=0, items=['ab', 'nccd], actor='brad'),
 Row(id=1, items=['rd', 'fh'], actor='tony')]

我正在尝试将其转换为spark数据帧:

df = spark.createDataFrame(rd)

这对我有用。

但是现在当我试图运行它时:

df.show()

这给了我错误。这很有效。 请给我一些关于这个的见解

Error:

Py4JJavaError: An error occurred while calling o1264.showString.
: java.lang.IllegalStateException: SparkContext has been shutdown
at org.apache.spark.SparkContext.runJob(SparkContext.scala:2021)
at org.apache.spark.SparkContext.runJob(SparkContext.scala:2050)
at org.apache.spark.SparkContext.runJob(SparkContext.scala:2069)
at org.apache.spark.sql.execution.SparkPlan.executeTake(SparkPlan.scala:336)
at org.apache.spark.sql.execution.CollectLimitExec.executeCollect(limit.scala:38)
at org.apache.spark.sql.Dataset.org$apache$spark$sql$Dataset$$collectFromPlan(Dataset.scala:2861)
at org.apache.spark.sql.Dataset$$anonfun$head$1.apply(Dataset.scala:2150)
at org.apache.spark.sql.Dataset$$anonfun$head$1.apply(Dataset.scala:2150)
at org.apache.spark.sql.Dataset$$anonfun$55.apply(Dataset.scala:2842)
at org.apache.spark.sql.execution.SQLExecution$.withNewExecutionId(SQLExecution.scala:65)
at org.apache.spark.sql.Dataset.withAction(Dataset.scala:2841)
at org.apache.spark.sql.Dataset.head(Dataset.scala:2150)
at org.apache.spark.sql.Dataset.take(Dataset.scala:2363)
at org.apache.spark.sql.Dataset.showString(Dataset.scala:241)
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:498)
at py4j.reflection.MethodInvoker.invoke(MethodInvoker.java:244)
at py4j.reflection.ReflectionEngine.invoke(ReflectionEngine.java:357)
at py4j.Gateway.invoke(Gateway.java:280)
at py4j.commands.AbstractCommand.invokeMethod(AbstractCommand.java:132)
at py4j.commands.CallCommand.execute(CallCommand.java:79)
at py4j.GatewayConnection.run(GatewayConnection.java:214)
at java.lang.Thread.run(Thread.java:748)

2 个答案:

答案 0 :(得分:1)

您可能知道Apache Spark是一个懒惰的评估者。您可以执行操作和转换。调用动作时都会调用转换。因此,当您进行show()或collect()调用时,您之前调用的所有函数都将被处理。所以你对createDataFrame的调用显然不起作用。

请阅读这篇文章,了解如何实现所需的输出:Creating a DataFrame from Row results in 'infer schema issue'

答案 1 :(得分:1)

除了@pissall所说的,以下内容应该有效:

from pyspark.sql.types import *

schema = StructType([StructField('id', IntegerType()), 
                     StructField('items', ArrayType(StringType())), 
                     StructField('actor', StringType())
                    ])
df = spark.createDataFrame(rd, schema)