CODEWITHSUNDEEP
scalaapache-sparkapache-spark-sql
Steve YN
Steve YN

Reputation: 69

How to create a scala case class instance with a Map instance

I want to create a scala case class whose fields come form a map . And , here is the case class 

 case class UserFeature(uid: String = null,
                     age: String = null,
                     marriageStatus: String = null,
                     consumptionAbility: String = null,
                     LBS: String = null,
                     interest1: String = null,
                     interest2: String = null,
                     interest3: String = null,
                     interest4: String = null,
                     interest5: String = null,
                     kw1: String = null,
                     kw2: String = null,
                     kw3: String = null,
                     topic1: String = null,
                     topic2: String = null,
                     topic3: String = null,
                     appIdInstall: String = null,
                     appIdAction: String = null,
                     ct: String = null,
                     os: String = null,
                     carrier: String = null,
                     house: String = null
                          )

suppose the map instance is

Map("uid" -> "4564131",
    "age" -> "5",
    "ct" -> "bk7755")

how can I apply the keys&values of the map to the fields&values of case class?

Upvotes: 3

Views: 2844

Answers (3)

Levi Ramsey
Levi Ramsey

Reputation: 20611

Synthesizing the other two answers, I would convert all the Strings in UserFeature that you're defaulting to null (which you should basically never use in Scala unless interacting with poorly-written Java code requires it, and even then use it as little as possible) to Option[String]. I leave that search-and-replace out of the answer.

Then you can do:

object UserFeature {
  def apply(map: Map[String, String]): UserFeature =
    UserFeature(map.get("uid"), map.get("age") ...)
}

Which lets you use:

val someMap: Map[String, String] = ...
val userFeature = UserFeature(someMap)

With the change to Option[String], there will be some other changes that need to be made in your codebase. https://danielwestheide.com/blog/2012/12/19/the-neophytes-guide-to-scala-part-5-the-option-type.html is a good tutorial for how to deal with Option.

Upvotes: 1

pramesh
pramesh

Reputation: 1964

You can do UserFeature(uid = map_var("uid"), age = map_var("age"), ct = map_var("ct")) assuming the variable holding Map is map_var and the keys are available

Upvotes: 1

Tim
Tim

Reputation: 27421

It is not a good idea to use null to represent missing string values. Use Option[String] instead.

case class UserFeature(uid: Option[String] = None,
                       age: Option[String] = None,
                       marriageStatus: Option[String] = None,
                       ...

Once you have done that, you can use get on the map to retrieve the value.

UserFeature(map.get("uid"), map.get("age"), map.get("marriageStatus") ...)

Values that are present in the map will be Some(value) and missing values will be None. The Option class has lots of useful methods for processing optional values in a safe way.

Upvotes: 1

Related Questions