在R6类的子类中更改initialize方法

假设我有一个R6类

人

：

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 = "")
                })
)

但是，当我用urname创建class

persons的新实例时，
name
和
hair
字段是
NA
，即class
Person
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

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
                })
)

PersonWithUrName实际上在Python中应该使用
super（）
，而不是父类的名称。这可以防止错误。这有记录在案的地方吗？我在小插曲中找不到任何关于
super
的内容。@Tommy确实，文档在这方面有点欠缺，但至少提到了
super
，尽管它没有具体解释超类初始化。这种行为可以从R6的（非平凡的）实现中猜测出来。此外，项目包含显示此行为的项目。更新的链接指向。
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

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