py4j.protocol.Py4JJavaError when selecting nested column in dataframe using select statementment

Question

我正在尝试在 spark 数据框 (python) 中执行一个简单的任务，即通过从另一个数据框中选择特定列和嵌套列来创建新的数据框例如：

df.printSchema()
root
 |-- time_stamp: long (nullable = true)
 |-- country: struct (nullable = true)
 |    |-- code: string (nullable = true)
 |    |-- id: long (nullable = true)
 |    |-- time_zone: string (nullable = true)
 |-- event_name: string (nullable = true)
 |-- order: struct (nullable = true)
 |    |-- created_at: string (nullable = true)
 |    |-- creation_type: struct (nullable = true)
 |    |    |-- id: long (nullable = true)
 |    |    |-- name: string (nullable = true)
 |    |-- destination: struct (nullable = true)
 |    |    |-- state: string (nullable = true)
 |    |-- ordering_user: struct (nullable = true)
 |    |    |-- cancellation_score: long (nullable = true)
 |    |    |-- id: long (nullable = true)
 |    |    |-- is_test: boolean (nullable = true)

df2=df.sqlContext.sql("""select a.country_code as country_code,
a.order_destination_state as order_destination_state,
a.order_ordering_user_id as order_ordering_user_id,
a.order_ordering_user_is_test as order_ordering_user_is_test,
a.time_stamp as time_stamp
from
(select
flat_order_creation.order.destination.state as order_destination_state,
flat_order_creation.order.ordering_user.id as order_ordering_user_id,
flat_order_creation.order.ordering_user.is_test as   order_ordering_user_is_test,
flat_order_creation.time_stamp as time_stamp
from flat_order_creation) a""")

我收到以下错误：

Traceback (most recent call last):
  File "/home/hadoop/scripts/orders_all.py", line 180, in <module>
    df2=sqlContext.sql(q)
  File "/usr/lib/spark/python/lib/pyspark.zip/pyspark/sql/context.py", line 552, in sql
  File "/usr/lib/spark/python/lib/py4j-0.8.2.1-src.zip/py4j/java_gateway.py", line 538, in __call__
  File "/usr/lib/spark/python/lib/pyspark.zip/pyspark/sql/utils.py", line 36, in deco
  File "/usr/lib/spark/python/lib/py4j-0.8.2.1-src.zip/py4j/protocol.py", line 300, in get_return_value
py4j.protocol.Py4JJavaError: An error occurred while calling o60.sql.
: java.lang.RuntimeException: [6.21] failure: ``*'' expected but `order' found

flat_order_creation.order.destination.state as order_destination_state,

我在本地模式下使用 spark-submit 和 master 来运行这段代码。重要的是要提到当我连接到 pyspark shell 和运行时代码（逐行）有效，但是在提交时（即使在本地模式下）它失败了。 another thing is important to mention is that when selecting a non nested field it works as well. 我在 EMR（版本 4.2.0）上使用 spark 1.5.2

Answer 1

没有 a Minimal, Complete, and Verifiable example 我只能猜测，但看起来您在交互式 shell 和独立程序中使用了不同的 SparkContext 实现。

只要在 shell 中提供的 Hive 支持 sqlContext 构建了 Spark 二进制文件就是 HiveContext。在其他差异中，它提供了比普通 SQLContext 更复杂的 SQL 解析器。您可以按如下方式轻松重现您的问题：

import org.apache.spark.{SparkConf, SparkContext}
import org.apache.spark.sql.SQLContext
import org.apache.spark.sql.hive.HiveContext

val conf: SparkConf = ???
val sc: SparkContext = ???
val query = "SELECT df.foobar.order FROM df"

val hiveContext: SQLContext = new HiveContext(sc)
val sqlContext: SQLContext = new SQLContext(sc)
val json = sc.parallelize(Seq("""{"foobar": {"order": 1}}"""))

sqlContext.read.json(json).registerTempTable("df")
sqlContext.sql(query).show
// java.lang.RuntimeException: [1.18] failure: ``*'' expected but `order' found
// ...

hiveContext.read.json(json).registerTempTable("df")
hiveContext.sql(query)
// org.apache.spark.sql.DataFrame = [order: bigint]

在独立程序中使用 HiveContext 初始化 sqlContext 应该可以解决问题：

from pyspark.sql import HiveContext

sqlContext = HiveContext(sc) 

df = sqlContext.createDataFrame(...)
df.registerTempTable("flat_order_creation")

sqlContext.sql(...)

请务必注意，问题不在于嵌套本身，而是使用 ORDER 关键字作为列名。因此，如果使用 HiveContext 不是一个选项，只需将字段名称更改为其他名称即可。

py4j.protocol.Py4JJavaError when selecting nested column in dataframe using select statementment

py4j.protocol.Py4JJavaError when selecting nested column in dataframe using select statetment

apache-spark

pyspark

apache-spark-sql

pyspark-sql