Java.lang.IllegalArgumentException：要求失败：在 Double 中找不到列

Question

我在 Spark 中工作，我有许多包含行的 csv 文件，一行看起来像这样：

2017,16,16,51,1,1,4,-79.6,-101.90,-98.900

它可以包含更多或更少的字段，取决于 csv 文件

每个文件对应一个 cassandra 表，我需要在其中插入文件包含的所有行，所以我基本上所做的就是获取该行，拆分其元素并将它们放入 List[Double]

sc.stop
import com.datastax.spark.connector._, org.apache.spark.SparkContext, org.apache.spark.SparkContext._, org.apache.spark.SparkConf


val conf = new SparkConf(true).set("spark.cassandra.connection.host", "localhost")
val sc = new SparkContext(conf)
val nameTable = "artport"
val ligne = "20171,16,165481,51,1,1,4,-79.6000,-101.7000,-98.9000"
val linetoinsert : List[String] = ligne.split(",").toList
var ainserer : Array[Double] = new Array[Double](linetoinsert.length)
for (l <- 0 to linetoinsert.length)yield {ainserer(l) = linetoinsert(l).toDouble}
val liste = ainserer.toList
val rdd = sc.parallelize(liste)
rdd.saveToCassandra("db", nameTable) //db is the name of my keyspace in cassandra

当我运行代码时，出现此错误

java.lang.IllegalArgumentException: requirement failed: Columns not found in Double: [collecttime, sbnid, enodebid, rackid, shelfid, slotid, channelid, c373910000, c373910001, c373910002]
  at scala.Predef$.require(Predef.scala:224)
  at com.datastax.spark.connector.mapper.DefaultColumnMapper.columnMapForWriting(DefaultColumnMapper.scala:108)
  at com.datastax.spark.connector.writer.MappedToGettableDataConverter$$anon$1.<init>(MappedToGettableDataConverter.scala:37)
  at com.datastax.spark.connector.writer.MappedToGettableDataConverter$.apply(MappedToGettableDataConverter.scala:28)
  at com.datastax.spark.connector.writer.DefaultRowWriter.<init>(DefaultRowWriter.scala:17)
  at com.datastax.spark.connector.writer.DefaultRowWriter$$anon$1.rowWriter(DefaultRowWriter.scala:31)
  at com.datastax.spark.connector.writer.DefaultRowWriter$$anon$1.rowWriter(DefaultRowWriter.scala:29)
  at com.datastax.spark.connector.writer.TableWriter$.apply(TableWriter.scala:382)
  at com.datastax.spark.connector.RDDFunctions.saveToCassandra(RDDFunctions.scala:35)
  ... 60 elided

我发现如果我的 RDD 类型为 :

，插入就可以工作

rdd: org.apache.spark.rdd.RDD[(Double, Double, Double, Double, Double, Double, Double, Double, Double, Double)]

但是我从我正在做的事情中得到的是 RDD

org.apache.spark.rdd.RDD[Double]

例如，我无法使用 scala Tuple9，因为我不知道执行前列表将包含的元素数量，这个解决方案也不适合我的问题，因为有时我的 csv 中有超过 100 列，元组停在 Tuple22

感谢您的帮助

Answer 1

正如 @SergGr 提到的，Cassandra 表具有包含已知列的架构。因此，在保存到 Cassandra 数据库之前，您需要将

Array

映射到

Cassandra schema

。您可以使用

Case Class

来实现此目的。尝试以下代码，我假设

Cassandra

表中的每一列都是

Double

类型。

//create a case class equivalent to your Cassandra table
case class Schema(collecttime: Double,
                  sbnid: Double,
                  enodebid: Double,
                  rackid: Double,
                  shelfid: Double,
                  slotid: Double,
                  channelid: Double,
                  c373910000: Double,
                  c373910001: Double,
                  c373910002: Double)
object test {

  import com.datastax.spark.connector._, org.apache.spark.SparkContext, org.apache.spark.SparkContext._, org.apache.spark.SparkConf

  def main(args: Array[String]): Unit = {
    val conf = new SparkConf(true).set("spark.cassandra.connection.host", "localhost")
    val sc = new SparkContext(conf)
    val nameTable = "artport"
    val ligne = "20171,16,165481,51,1,1,4,-79.6000,-101.7000,-98.9000"
    //parse ligne string Schema case class
    val schema = parseString(ligne)
    //get RDD[Schema]
    val rdd = sc.parallelize(Seq(schema))
    //now you can save this RDD to cassandra
    rdd.saveToCassandra("db", nameTable)
    }


    //function to parse string to Schema case class
    def parseString(s: String): Schema = {
       //get each field from string array
       val Array(collecttime, sbnid, enodebid, rackid, shelfid, slotid,
       channelid, c373910000, c373910001, c373910002, _*) = s.split(",").map(_.toDouble)

       //map those fields to Schema class
       Schema(collecttime,
         sbnid,
         enodebid,
         rackid,
         shelfid,
         slotid,
         channelid,
         c373910000,
         c373910001,
         c373910002)
     }
}

Answer 2

当您在连接列中看到此情况时 - 有时您需要将连接值包装在

Tuple1

中

例如

map({case a=> Tuple1(a.p:BigInt) })
  .joinWithCassandraTable("keyspacename", "tableName", joinColumns = SomeColumns("columnName"))