对于元组的关键元素的Scala图，如何从元组中选择（项目？）特定项目？

Question

假设我有一个具有多属性值的地图，我想选择一个特定的属性。

例如，表示具有姓名，性别，年龄，说明的人员的表格的地图。

在SQL中，我会写“选择的人年龄其中名称=‘任何人’”

我怎么才能在斯卡拉这种效果？

val people = Map(
"Walter White" -> ("male",52,"bad boy"), 
"Skyler White" -> ("female",42,"morally challenged mom") 
) 

// equivalent of select * from people. This works. 
for ((name,(gender,age,desc)) <- people) println(s"$name is a $age year old $gender and is a $desc") 

// what should be the syntax to get "the age of Walter White is 52"? 
// in SQL, it would be "'The age of Walter White is ' || (select age from people where name='Walter White')" 
// what would it be in Scala? 
println("The age of Walter White is " + people("Walter White")(1)) // not this! 

Answer 1

您可以创建一个Person案例类，并创建Person！而非元组新的地图。

case class Person(name: String, gender: String, age: Int, description: String) 

val persons = people map { case (name, (gender, age, descr)) => 
    name -> Person(name, gender, age, descr) 
} 

这种方式，你可以写：

persons("Walter White").age    // Int = 52 
persons.get("Skyler White").map(_.age) // Option[Int] = Some(42) 

你也可以使用原来的代码访问的年龄：

people("Walter White")._2    // Int = 52