r - 更改 R6 类的子类中的初始化方法

Question

假设我有一个 R6 类Person：

library(R6)

Person <- R6Class("Person",
  public = list(name = NA, hair = NA,
                initialize = function(name, hair) {
                  self$name <- name
                  self$hair <- hair
                  self$greet()
                },
                greet = function() {
                  cat("Hello, my name is ", self$name, ".\n", sep = "")
                })
)

如果我想创建一个子类，它的initialize方法应该是相同的，除了添加一个变量，self我该怎么做？
我尝试了以下方法：

PersonWithSurname <- R6Class("PersonWithSurname",
  inherit = Person,
  public = list(surname = NA,
                initialize = function(name, surname, hair) {
                  Person$new(name, hair)
                  self$surname <- surname
                })
)

但是，当我创建类的新实例时PersonWithSurname，字段name和hair是NA，即类的默认值Person。

PersonWithSurname$new("John", "Doe", "brown")
Hello, my name is John.
<PersonWithSurname>
   Inherits from: <Person>
   Public:
     clone: function (deep = FALSE) 
     greet: function () 
     hair: NA
     initialize: function (name, surname, hair) 
     name: NA
     surname: Doe

在Python我会做以下事情：

class Person(object):
  def __init__(self, name, hair):
    self.name = name
    self.hair = hair
    self.greet()

  def greet(self):
    print "Hello, my name is " + self.name

class PersonWithSurname(Person):
  def __init__(self, name, surname, hair):
    Person.__init__(self, name, hair)
    self.surname = surname

Answer 1

在这方面，R6 的工作方式与 Python 非常相似。也就是说，您只需调用initialize对象super：

PersonWithSurname <- R6Class("PersonWithSurname",
  inherit = Person,
  public = list(surname = NA,
                initialize = function(name, surname, hair) {
                  super$initialize(name, hair)
                  self$surname <- surname
                })
)