将包含BigInt的RDD转换为Spark Dataframe

Question

嗨，我正在使用spark 1.6.3。我有一个rdd，里面有一些BigInt scala类型。我将如何将其转换为火花数据框？ 在创建数据框之前是否可以转换类型？

我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模式对象;

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)我得到这个错误

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我碰到下面的错误;

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

任何帮助理解

Answer 1

架构只能与RDD[Row]使用。这里使用反射：

sqlContext.createDataFrame(rdd) 

你也有变化BigInt到one of the supported types（BigDecimal？）或use binary encoder for this field。