更改 R6 class 的 subclass 中的初始化方法

Question

假设我有一个 R6 class 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 subclass，其 initialize 方法除了向 self 添加一个变量外应该是相同的，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
                })
)

然而，当我创建 class PersonWithSurname 的新实例时，字段 name 和 hair 是 NA，即默认值 class 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 中，我将执行以下操作：

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

R6 在这方面与 Python 非常相似；也就是说，您只需在 super 对象上调用 initialize：

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

更改 R6 class 的 subclass 中的初始化方法

Change initialize method in subclass of an R6 class

r

r6