如何将这个具有不可变成员的非功能scala代码转换为优雅的解决方案？

如何避免可变索引并使其更加优雅？我知道必须使用Option更改Null，我只是对答案感到好奇

class Person(val name: String, val department: String)

var people = Array(new Person(“Jones”, “Marketing”), new Person(“Smith”, “Engineering”))
var engineer: Person = null
var index = 0

while (index < people.length) {

  if (people(index).department == “Engineering”) 
    engineer = people(index)

  index = index + 1


}
println(engineer.name + “ is an engineer”)

class-Person（val-name:String，val-department:String）
var people=Array（新人（“琼斯”、“营销”）、新人（“史密斯”、“工程”））
变量工程师：Person=null
var指数=0
while（指数<人长）{
if（人员（索引）。部门==“工程”）
工程师=人员（索引）
索引=索引+1
}
println（engineer.name+“是工程师”）

以下是我重构的方法：

// Refactored into a case class, since it's a simple data container
case class Person(name: String, department: String)

// Using the case class convenience apply method to drop `new` 
val people = Array(Person(“Jones”, “Marketing”), Person(“Smith”, “Engineering”))

// Selects all the engineers. You could add `.headOption` to get the first. 
val engineers = people.filter(_.department == "Engineering")

// Functional way of iterating the collection of engineers
// Also, using string interpolation to print
for (engineer <- engineers) println(s"${engineer.name} is an engineer.")

最后一个提示，如果你的部门是固定的选项的有限集合，你也应该考虑把它变成一个类型：

sealed trait Department
case object Engineering extends Department
case object Marketing extends Department
// ... for each valid department

---

然后你可以根据身份而不是价值来匹配。这使您可以依赖类型系统，而不必经常验证字符串（有些人称为）。最佳做法是尽早将数据验证为类型，将其作为类型化数据处理，然后仅转换回字符串，以便将数据从系统中导出（例如，打印到屏幕、记录、通过API提供服务）。

您可以使用

查找

首先查找：

people
  .find { _.department == "Engineering" }
  .foreach { engineer => println(engineer.name + " is an engineer") }

或

过滤器

查找所有：

people
  .filter { _.department == "Engineering" }
  .foreach { engineer => println(engineer.name + " is an engineer") }

顺便说一下，只需将增量操作移到

if

块之外，就可以修复代码：

if (people(index).department == "Engineering") {
  engineer = people(index)
  // index = index + 1
}
index = index + 1

之后，您应该检查

engineer

中的

null

，因为您的数组可能不包含符合您的条件的

人员

因此，您似乎希望找到最后一个
人
，因此您可以使用：

people
  .foldLeft(None: Option[Person])((r, p) =>
    if (p.department == "Engineering") Some(p) else r)
  .foreach { engineer => println(engineer.name + " is an engineer") }

另外，在避免所有
var
之后，您还可以将
数组
（可变结构）更改为
列表
（默认为scala.collection.immutable.List）
如果您想找到数组中的最后一个工程师，您可能会使用：

case class Person(val name: String, val department: String)

val people = Array(Person(“Jones”, “Marketing”), Person(“Smith”, “Engineering”))

def findLastEngineer(l: Seq[Person]) : Option[Person] =  
  people.foldLeft(None) { 
    case (previousOpt, eng) => if (eng.department == "Engineering") Some(eng) else previousOpt
  }
}

println(findLastEngineer(people).map(_.name).getOrElse("Not found"))

我会这样做：

case class Person(name: String, department: String)

val people = List(Person("Jones", "Marketing"), Person("Smith", "Engineering"))

val engineers = people.filter { person:Person => person.department == "Engineering" }

engineers.map { engineer: Person => println(engineer.name + " is an engineer") }

尝试使用函数转换其他类型中的类型。通常我们使用map/reduce/filter函数来实现这一点。
此代码包含bug（无限循环）。你想在功能代码中得到这个bug吗？如果你能用文字描述代码片段实际应该做什么，也许会有帮助。所以你想打印列表名称上的最后一个工程师？或者任何？或者首先？这不是一个真正的问题：这是从StackOverflowCareers上的招聘广告中剪下来粘贴的：也许OP应该遵循他们的建议：“如果你必须用谷歌搜索任何东西来做这件事，那么现在可能不是你申请的合适时机。”谁在乎申请呢？谢谢你，这很有启发性和解释性。顺便说一句，工程设计后缺少双引号。对，函数map/reduce/filter已经处理集合中的所有元素，因此我们不必麻烦使用可变索引，只需
people.filter（Person.department.equals（“工程”）。last
case class Person(val name: String, val department: String)

val people = Array(Person(“Jones”, “Marketing”), Person(“Smith”, “Engineering”))

def findLastEngineer(l: Seq[Person]) : Option[Person] =  
  people.foldLeft(None) { 
    case (previousOpt, eng) => if (eng.department == "Engineering") Some(eng) else previousOpt
  }
}

println(findLastEngineer(people).map(_.name).getOrElse("Not found"))

case class Person(name: String, department: String)

val people = List(Person("Jones", "Marketing"), Person("Smith", "Engineering"))

val engineers = people.filter { person:Person => person.department == "Engineering" }

engineers.map { engineer: Person => println(engineer.name + " is an engineer") }