如何使用Scala从Spark中的列表或数组创建行 [英] How to create a Row from a List or Array in Spark using Scala

问题描述

我正在尝试根据用户输入创建行(org.apache.spark.sql.catalyst.expressions.Row).我无法随机创建行".

I'm trying to create a Row (org.apache.spark.sql.catalyst.expressions.Row) based on the user input. I'm not able to create a Row randomly.

是否有任何功能可以从List或Array创建行.

Is there any functionality to create a Row from List or Array.

例如，如果我有一个具有以下格式的.csv文件，

For eg., If I have a .csv file with the following format,

"91xxxxxxxxxx,21.31,15,0,0"

如果用户输入[1, 2]，那么我只需要第二列和第三列以及第一列customer_id

If the user input [1, 2] then I need to take only 2nd column and 3rd column along with the customer_id which is the first column

我尝试用代码解析它:

val l3 = sc.textFile("/SparkTest/abc.csv").map(_.split(" ")).map(r => (foo(input,r(0)))) `

其中foo被定义为

def f(n: List[Int], s: String) : Row = {
    val n = input.length
    var out = new Array[Any](n+1)
    var r = s.split(",")
    out(0) = r(0)
    for (i <- 1 to n)
        out(i) = r(input(i-1)).toDouble
    Row(out)
}

并且输入的是列表内容

val input = List(1,2)

执行此代码，我得到的l3为:

Executing this code I get l3 as:

Array[org.apache.spark.sql.Row] = Array([[Ljava.lang.Object;@234d2916])

但是我想要的是:

Array[org.apache.spark.sql.catalyst.expressions.Row] = Array([9xxxxxxxxxx,21.31,15])`

必须通过此操作才能在Spark SQL中创建架构

This has to be passed to create a schema in Spark SQL

推荐答案

类似以下的方法应该起作用:

Something like the following should work:

import org.apache.spark.sql._

def f(n: List[Int], s: String) : Row =
  Row.fromSeq(s.split(",").zipWithIndex.collect{case (a,b) if n.contains(b) => a}.toSeq)

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