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Apache Spark 如何 append 新列从列表/数组到 Spark dataframe

[英]Apache Spark how to append new column from list/array to Spark dataframe

I am using Apache Spark 2.0 Dataframe/Dataset API I want to add a new column to my dataframe from List of values.我正在使用 Apache Spark 2.0 数据帧/数据集 API 我想从值列表中向我的 dataframe 添加一个新列。 My list has same number of values like given dataframe.我的列表具有与给定 dataframe 相同数量的值。

val list = List(4,5,10,7,2)
val df   = List("a","b","c","d","e").toDF("row1")

I would like to do something like:我想做类似的事情:

val appendedDF = df.withColumn("row2",somefunc(list))
df.show()
// +----+------+
// |row1 |row2 |
// +----+------+
// |a    |4    |
// |b    |5    |
// |c    |10   |
// |d    |7    |
// |e    |2    |
// +----+------+

For any ideas I would be greatful, my dataframe in reality contains more columns.对于任何想法,我都会很高兴,我的 dataframe 实际上包含更多列。

You could do it like this:你可以这样做:

import org.apache.spark.sql.Row
import org.apache.spark.sql.types._    

// create rdd from the list
val rdd = sc.parallelize(List(4,5,10,7,2))
// rdd: org.apache.spark.rdd.RDD[Int] = ParallelCollectionRDD[31] at parallelize at <console>:28

// zip the data frame with rdd
val rdd_new = df.rdd.zip(rdd).map(r => Row.fromSeq(r._1.toSeq ++ Seq(r._2)))
// rdd_new: org.apache.spark.rdd.RDD[org.apache.spark.sql.Row] = MapPartitionsRDD[33] at map at <console>:32

// create a new data frame from the rdd_new with modified schema
spark.createDataFrame(rdd_new, df.schema.add("new_col", IntegerType)).show
+----+-------+
|row1|new_col|
+----+-------+
|   a|      4|
|   b|      5|
|   c|     10|
|   d|      7|
|   e|      2|
+----+-------+

Adding for completeness: the fact that the input list (which exists in driver memory) has the same size as the DataFrame suggests that this is a small DataFrame to begin with - so you might consider collect() -ing it, zipping with list , and converting back into a DataFrame if needed:添加完整性:输入list (存在于驱动程序内存中)与DataFrame具有相同大小的DataFrame表明这是一个小的 DataFrame 开始 - 所以你可以考虑collect() -ing 它,用list压缩,并在需要时转换回DataFrame

df.collect()
  .map(_.getAs[String]("row1"))
  .zip(list).toList
  .toDF("row1", "row2")

That won't be faster, but if the data is really small it might be negligible and the code is (arguably) clearer.这不会更快,但如果数据真的很小,它可能可以忽略不计,并且代码(可以说)更清晰。

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