在Spark/Scala中将RDD转换为数据帧

Question

RDD是以数组[数组[字符串]的格式创建的，具有以下值：

val rdd : Array[Array[String]] = Array(
Array("4580056797", "0", "2015-07-29 10:38:42", "0", "1", "1"), 
Array("4580056797", "0", "2015-07-29 10:38:43", "0", "1", "1"))

我想用模式创建一个数据帧：

val schemaString = "callId oCallId callTime duration calltype swId"

接下来的步骤：

scala> val rowRDD = rdd.map(p => Array(p(0), p(1), p(2),p(3),p(4),p(5).trim))
rowRDD: org.apache.spark.rdd.RDD[Array[String]] = MapPartitionsRDD[14] at map at <console>:39
scala> val calDF = sqlContext.createDataFrame(rowRDD, schema)

给出以下错误：

console:45: error: overloaded method value createDataFrame with alternatives:
     (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>
    (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[Array[String]],   
    org.apache.spark.sql.types.StructType)
       val calDF = sqlContext.createDataFrame(rowRDD, schema)

Answer 1

使用spark 1.6.1和scala 2.10

我得到了相同的错误：重载方法值createDataFrame，带有可选项：

对我来说，gotcha是createDataFrame中的签名，我试图使用val-rdd:List[Row]，但它失败了，因为java。util。列出[org.apache.spark.sql.Row]和scala。收集不变的List[org.apache.spark.sql.Row]不一样。

我找到的工作解决方案是通过列表[Array[String]将val-rdd:Array[Array[String]]转换为rdd[Row]。我发现这与文档中的内容最为接近

import org.apache.spark.sql.Row
import org.apache.spark.sql.types.{StructType,StructField,StringType};
val sqlContext = new org.apache.spark.sql.SQLContext(sc)

val rdd_original : Array[Array[String]] = Array(
    Array("4580056797", "0", "2015-07-29 10:38:42", "0", "1", "1"), 
    Array("4580056797", "0", "2015-07-29 10:38:42", "0", "1", "1"))

val rdd : List[Array[String]] = rdd_original.toList

val schemaString = "callId oCallId callTime duration calltype swId"

// Generate the schema based on the string of schema
val schema =
  StructType(
    schemaString.split(" ").map(fieldName => StructField(fieldName, StringType, true)))

// Convert records of the RDD to Rows.
val rowRDD = rdd.map(p => Row(p: _*)) // using splat is easier
// val rowRDD = rdd.map(p => Row(p(0), p(1), p(2), p(3), p(4), p(5))) // this also works

val df = sqlContext.createDataFrame(sc.parallelize(rowRDD:List[Row]), schema)
df.show

Answer 2

您需要首先将数组转换为行，然后定义模式。我假设您的大多数字段都是长的

    val rdd: RDD[Array[String]] = ???
    val rows: RDD[Row] = rdd map {
      case Array(callId, oCallId, callTime, duration, swId) =>
        Row(callId.toLong, oCallId.toLong, callTime, duration.toLong, swId.toLong)
    }

    object schema {
      val callId = StructField("callId", LongType)
      val oCallId = StructField("oCallId", StringType)
      val callTime = StructField("callTime", StringType)
      val duration = StructField("duration", LongType)
      val swId = StructField("swId", LongType)

      val struct = StructType(Array(callId, oCallId, callTime, duration, swId))
    }

    sqlContext.createDataFrame(rows, schema.struct)

Answer 3

只需粘贴到火花壳中：

val a = 
  Array(
    Array("4580056797", "0", "2015-07-29 10:38:42", "0", "1", "1"), 
    Array("4580056797", "0", "2015-07-29 10:38:42", "0", "1", "1"))

val rdd = sc.makeRDD(a)

case class X(callId: String, oCallId: String, 
  callTime: String, duration: String, calltype: String, swId: String)

然后在RDD上映射（）以创建case类的实例，然后使用toDF（）创建数据帧：

scala> val df = rdd.map { 
  case Array(s0, s1, s2, s3, s4, s5) => X(s0, s1, s2, s3, s4, s5) }.toDF()
df: org.apache.spark.sql.DataFrame = 
  [callId: string, oCallId: string, callTime: string, 
    duration: string, calltype: string, swId: string]

这从case类推断出模式。

然后您可以继续：

scala> df.printSchema()
root
 |-- callId: string (nullable = true)
 |-- oCallId: string (nullable = true)
 |-- callTime: string (nullable = true)
 |-- duration: string (nullable = true)
 |-- calltype: string (nullable = true)
 |-- swId: string (nullable = true)

scala> df.show()
+----------+-------+-------------------+--------+--------+----+
|    callId|oCallId|           callTime|duration|calltype|swId|
+----------+-------+-------------------+--------+--------+----+
|4580056797|      0|2015-07-29 10:38:42|       0|       1|   1|
|4580056797|      0|2015-07-29 10:38:42|       0|       1|   1|
+----------+-------+-------------------+--------+--------+----+

如果您想在普通程序中（不在火花壳中）使用toDF（），请确保（此处引用）：

• 导入sqlContext。隐含_ 使用toDF（）在方法外定义case类