将包含BigInt的RDD转换为Spark Dataframe

时间:2017-10-12 15:37:49

标签: scala apache-spark apache-spark-sql spark-dataframe scala-collections

您好我在Spark 1.6.3工作。我有一个rdd,里面有一些BigInt scala类型。我如何将其转换为spark数据帧? 是否可以在创建数据框之前转换类型?

我的rdd:

Array[(BigInt, String, String, BigInt, BigInt, BigInt, BigInt, List[String])] = Array((14183197,Browse,3393626f-98e3-4973-8d38-6b2fb17454b5_27331247X28X6839X1506087469573,80161702,8702170626376335,59,527780275219,List(NavigationLevel, Session)), (14183197,Browse,3393626f-98e3-4973-8d38-6b2fb17454b5_27331247X28X6839X1506087469573,80161356,8702171157207449,72,527780278061,List(StartPlay, Action, Session)))
打印出来:

(14183197,Browse,3393626f-98e3-4973-8d38-6b2fb17454b5_27331247X28X6839X1506087469573,80161356,8702171157207449,72,527780278061,List(StartPlay, Action, Session))
(14183197,Browse,3393626f-98e3-4973-8d38-6b2fb17454b5_27331247X28X6839X1506087469573,80161702,8702170626376335,59,527780275219,List(NavigationLevel, Session))

我已经厌倦了创建架构对象;

  val schema = StructType(Array(
    StructField("trackId", LongType, true),
    StructField("location", StringType, true),
    StructField("listId", StringType, true),
    StructField("videoId", LongType, true),
    StructField("id", LongType, true),
    StructField("sequence", LongType, true),
    StructField("time", LongType, true),
    StructField("type", ArrayType(StringType), true)
  ))

如果我尝试val df = sqlContext.createDataFrame(rdd, schema)我收到此错误

error: overloaded method value createDataFrame with alternatives:
  (data: java.util.List[_],beanClass: Class[_])org.apache.spark.sql.DataFrame <and>
  (rdd: org.apache.spark.api.java.JavaRDD[_],beanClass: Class[_])org.apache.spark.sql.DataFrame <and>
  (rdd: org.apache.spark.rdd.RDD[_],beanClass: Class[_])org.apache.spark.sql.DataFrame <and>
  (rows: java.util.List[org.apache.spark.sql.Row],schema: org.apache.spark.sql.types.StructType)org.apache.spark.sql.DataFrame <and>
  (rowRDD: org.apache.spark.api.java.JavaRDD[org.apache.spark.sql.Row],schema: org.apache.spark.sql.types.StructType)org.apache.spark.sql.DataFrame <and>
  (rowRDD: org.apache.spark.rdd.RDD[org.apache.spark.sql.Row],schema: org.apache.spark.sql.types.StructType)org.apache.spark.sql.DataFrame
 cannot be applied to (org.apache.spark.rdd.RDD[(BigInt, String, String, BigInt, BigInt, BigInt, BigInt, scala.collection.immutable.List[String])], org.apache.spark.sql.types.StructType)

或者,如果我尝试val df = sc.parallelize(rdd.toSeq).toDF,我会收到以下错误;

error: value toSeq is not a member of org.apache.spark.rdd.RDD[(BigInt, String, String, BigInt, BigInt, BigInt, BigInt, List[String])]

感谢任何帮助

1 个答案:

答案 0 :(得分:0)

架构只能与RDD[Row]一起使用。这里使用反射:

sqlContext.createDataFrame(rdd)

您还将BigInt更改为one of the supported typesBigDecimal?)或use binary encoder for this field