[英]Spark Scala case class with array and map datatype
我有这样的数据:
[Michael, 100, Montreal,Toronto, Male,30, DB:80, Product:DeveloperLead]
[Will, 101, Montreal, Male,35, Perl:85, Product:Lead,Test:Lead]
[Steven, 102, New York, Female,27, Python:80, Test:Lead,COE:Architect]
[Lucy, 103, Vancouver, Female,57, Sales:89,HR:94, Sales:Lead]
所以我必须读取这些数据并使用 Spark 定义一个案例类。 我已经编写了以下程序,但是在将案例类转换为数据框时出现错误。 我的代码有什么问题,我该如何纠正?
case class Ayush(name: String,employee_id:String ,work_place: Array[String],sex_age: Map [String,String],skills_score: Map[String,String],depart_title: Map[String,Array[String]])
我在下面的行中收到一个错误(见下图):
val d = df.map(w=> Ayush(w(0),w(1),w(2)._1,w(2)._2,w(3)._1,w(3)._2,w(4)._1,w(4)._2,w(5)._1,w(5)._2._1,w(5)._2._2))).toDF
我已经更改了您的数据。 将工作场所和部门数据用双引号括起来,以便我可以使用逗号分隔值获取数据。 然后添加一个自定义分隔符,以便稍后我可以使用分隔符来分隔数据。 您可以使用自己的分隔符。 图像如下:
数据如下:
迈克尔,100,“蒙特利尔,多伦多”,男,30,DB:80,“产品,开发主管”威尔,101,蒙特利尔,男,35,Perl:85,“产品,主管,测试,主管”史蒂文,102,纽约,女性,27,Python:80,“测试,领导,COE,建筑师” Lucy,103,温哥华,女性,57,销售:89_HR:94,“销售,领导”
以下是我执行的代码更改,对我来说效果很好:
val df = spark.read.csv("CSV PATH HERE")
case class Ayush(name: String,employee_id:String ,work_place: Array[String],sex_age: Map [String,String],skills_score: Map[String,String],depart_title: Map[String,Array[String]])
val resultDF = df.map { x => {
val departTitleData = x(6).toString
val skill_score = x(5).toString
val skill_Map = scala.collection.mutable.Map[String, String]()
// Separate skill by underscore I can get each skill:Num then i will add each one in map
skill_score.split("_").foreach { x => skill_Map += (x.split(":")(0) -> x.split(":")(1)) }
// Putting data into case class
new Ayush(x(0).toString(), x(1).toString, x(2).toString.split(","), Map(x(3).toString -> x(4).toString), skill_Map.toMap, Map(x(6).toString.split(",")(0) -> x(6).toString.split(",")) )
}}
//End Here
上面的代码输出是:
================================================== ==============================
+-------+-----------+--------------------+------------------+--------------------+--------------------+
| name|employee_id| work_place| sex_age| skills_score| depart_title|
+-------+-----------+--------------------+------------------+--------------------+--------------------+
|Michael| 100|[ Montreal, Toronto]| Map( Male -> 30)| Map( DB -> 80)|Map( Product -> W...|
| Will| 101| [ Montreal]| Map( Male -> 35)| Map( Perl -> 85)|Map( Product -> W...|
| Steven| 102| [ New York]|Map( Female -> 27)| Map( Python -> 80)|Map( Test -> Wrap...|
| Lucy| 103| [ Vancouver]|Map( Female -> 57)|Map(HR -> 94, Sa...|Map( Sales -> Wra...|
+-------+-----------+--------------------+------------------+--------------------+--------------------+
@vishal我不知道这个问题是否仍然有效,但这是我不更改源数据的解决方案,公平警告它可能有点令人讨厌:)
def main(args:Array[String]):Unit= {
val conf=new SparkConf().setAppName("first_demo").setMaster("local[*]")
val sc=new SparkContext(conf)
val spark=SparkSession.builder().getOrCreate()
import spark.implicits._
val rdd1=sc.textFile("file:///C:/Users/k.sandeep.varma/Downloads/documents/documents/spark_data/employee_data.txt")
val clean_rdd=rdd1.map(x=>x.replace("[","")).map(x=>x.replace("]",""))
val schema_rdd=clean_rdd.map(x=>x.split(", ")).map(x=>schema(x(0),x(1),x(2).split(","),Map(x(3).split(",")(0)->x(3).split(",")(1)),Map(x(4).split(":")(0)->x(4).split(":")(1)),Map(x(5).split(":")(0)->x(5).split(":"))))
val df1=schema_rdd.toDF()
df1.printSchema()
df1.show(false)
输出:
|name |employee_id|work_place |sex_age |skills_score |depart_title |
+-------+-----------+-------------------+--------------+----------------+---------------------------------------+
|Michael|100 |[Montreal, Toronto]|[Male -> 30] |[DB -> 80] |[Product -> [Product, DeveloperLead]] |
|Will |101 |[Montreal] |[Male -> 35] |[Perl -> 85] |[Product -> [Product, Lead,Test, Lead]]|
|Steven |102 |[New York] |[Female -> 27]|[Python -> 80] |[Test -> [Test, Lead,COE, Architect]] |
|Lucy |103 |[Vancouver] |[Female -> 57]|[Sales -> 89,HR]|[Sales -> [Sales, Lead]] |
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