[英]Unable to convert an rdd with zipWithIndex to a dataframe in spark
[英]Unable to convert an RDD[Row] to a DataFrame
对于以下代码-将DataFrame转换为RDD[Row]
并通过mapPartitions
追加新列的数据:
// df is a DataFrame
val dfRdd = df.rdd.mapPartitions {
val bfMap = df.rdd.sparkContext.broadcast(factorsMap)
iter =>
val locMap = bfMap.value
iter.map { r =>
val newseq = r.toSeq :+ locMap(r.getAs[String](inColName))
Row(newseq)
}
}
对于另一RDD[Row]
的RDD[Row]
,输出正确:
println("**dfrdd\n" + dfRdd.take(5).mkString("\n"))
**dfrdd
[ArrayBuffer(0021BEC286CC, 4, Series, series, bc514da3e0d534da8207e3aab231d1cb, livetv, 148818)]
[ArrayBuffer(0021BEE7C556, 4, Series, series, bc514da3e0d534da8207e3aab231d1cb, livetv, 26908)]
[ArrayBuffer(8C7F3BFD4B82, 4, Series, series, bc514da3e0d534da8207e3aab231d1cb, livetv, 99942)]
[ArrayBuffer(0021BEC8F8B8, 1, Series, series, 0d2debc63efa3790a444c7959249712b, livetv, 53994)]
[ArrayBuffer(10EA59F10C8B, 1, Series, series, 0d2debc63efa3790a444c7959249712b, livetv, 1427)]
让我们尝试将RDD[Row]
转换回DataFrame:
val newSchema = df.schema.add(StructField("userf",IntegerType))
现在让我们创建更新的DataFrame:
val df2 = df.sqlContext.createDataFrame(dfRdd,newSchema)
新架构看起来正确吗?
newSchema.show()
root
|-- user: string (nullable = true)
|-- score: long (nullable = true)
|-- programType: string (nullable = true)
|-- source: string (nullable = true)
|-- item: string (nullable = true)
|-- playType: string (nullable = true)
|-- userf: integer (nullable = true)
注意,我们确实看到了新的userf
列。
但是,它不起作用:
println("df2: " + df2.take(1))
Job aborted due to stage failure: Task 0 in stage 9.0 failed 1 times,
most recent failure: Lost task 0.0 in stage 9.0 (TID 9, localhost, executor driver): java.lang.RuntimeException: Error while encoding:
java.lang.RuntimeException: scala.collection.mutable.ArrayBuffer is not a
valid external type for schema of string
if (assertnotnull(input[0, org.apache.spark.sql.Row, true], top level row object).isNullAt) null else staticinvoke(class org.apache.spark.unsafe.types.UTF8String, StringType, fromString, validateexternaltype(getexternalrowfield(assertnotnull(input[0, org.apache.spark.sql.Row, true], top level row object), 0, user), StringType), true) AS user#28
+- if (assertnotnull(input[0, org.apache.spark.sql.Row, true], top level row object).isNullAt) null else staticinvoke(class org.apache.spark.unsafe.types.UTF8String, StringType, fromString, validateexternaltype(getexternalrowfield(assertnotnull(input[0, org.apache.spark.sql.Row, true], top level row object), 0, user), StringType), true)
:- assertnotnull(input[0, org.apache.spark.sql.Row, true], top level row object).isNullAt
: :- assertnotnull(input[0, org.apache.spark.sql.Row, true], top level row object)
: : +- input[0, org.apache.spark.sql.Row, true]
: +- 0
:- null
那么:这里缺少什么细节?
注意:我对不同的方法不感兴趣:例如withColumn
或Datasets
..让我们仅考虑以下方法:
调用Row
的构造函数似乎有一个小错误:
val newseq = r.toSeq :+ locMap(r.getAs[String](inColName))
Row(newseq)
此“构造函数”的签名(实际上是apply方法)为:
def apply(values: Any*): Row
当您传递Seq[Any]
,会将其视为Seq[Any]
类型的单个值 。 您要传递此序列的元素 ,因此应使用:
val newseq = r.toSeq :+ locMap(r.getAs[String](inColName))
Row(newseq: _*)
解决此问题后,行将与您构建的架构匹配,您将获得预期的结果。
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