Spark unionAll 多个数据帧

问题描述

对于一组数据框

val df1 = sc.parallelize(1 to 4).map(i => (i,i*10)).toDF("id","x")
val df2 = sc.parallelize(1 to 4).map(i => (i,i*100)).toDF("id","y")
val df3 = sc.parallelize(1 to 4).map(i => (i,i*1000)).toDF("id","z")

把我所有的人都结合起来

to union all of them I do

df1.unionAll(df2).unionAll(df3)

对于任意数量的数据帧，是否有更优雅和可扩展的方式来执行此操作，例如来自

Is there a more elegant and scalable way of doing this for any number of dataframes, for example from

Seq(df1, df2, df3)

推荐答案

最简单的解决方案是 reduce 与 union (unionAll in Spark

The simplest solution is to reduce with union (unionAll in Spark < 2.0):

val dfs = Seq(df1, df2, df3)
dfs.reduce(_ union _)

这是相对简洁的，不应该从堆外存储中移动数据需要非线性时间来执行计划分析.如果您尝试合并大量 DataFrames，可能会出现什么问题.

This is relatively concise and shouldn't move data from off-heap storage requires non-linear time to perform plan analysis. what can be a problem if you try to merge large number of DataFrames.

您也可以转换为 RDDs 并使用 SparkContext.union:

You can also convert to RDDs and use SparkContext.union:

dfs match {
  case h :: Nil => Some(h)
  case h :: _   => Some(h.sqlContext.createDataFrame(
                     h.sqlContext.sparkContext.union(dfs.map(_.rdd)),
                     h.schema
                   ))
  case Nil  => None
}

它使 的分析成本保持在较低的水平，但与直接合并 DataFrames 相比效率较低.

It keeps analysis cost low but otherwise it is less efficient than merging DataFrames directly.

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