将包含 BigInt 的 RDD 转换为 Spark Dataframe

Question

嗨，我在 Spark 1.6.3 中工作。我有一个rdd，里面有一些BigInt scala类型。如何将其转换为火花数据帧？是否可以在创建数据帧之前强制转换类型？

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))

我已经厌倦了创建一个架构对象;

  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)则会收到此错误

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则会出现以下错误;

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

任何帮助不胜感激

Answer 1

Schema 只能与 RDD[Row] 一起使用。这里使用反射：

sqlContext.createDataFrame(rdd)

您还可以将BigInt更改为受支持的类型之一（BigDecimal？）或对此字段使用二进制编码器。