变化的R6类的子类初始化方法

Question

比方说，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，其initialize方法应该是一样的，除了加入多一个变量来self我将如何做到这一点？
我试过如下：

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PersonWithSurname类中的字段name和hair是NA的新实例，即Person类的默认值。

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我会做到以下几点：

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工作方式非常像Python在这方面;也就是说，你只需要调用initialize的super对象：

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