Question

1）我无法弄清楚错误：

的位置

Error in args[[1]] : subscript out of bounds

来自，如果我尝试运行以下代码：

Part.initialize<-function(...){
  args<-list(...)

    .self$var1 <- if(is.null(args[["var1"]])) vector() else args[["var1"]]
    .self$var2 <- if(is.null(args[["var2"]])) character() else as.character(args[["var2"]])
    .self
}

Part<-setRefClass(Class = "Part"
                 ,fields = c(var1 = "ANY", var2 = "character")
                 ,methods = list(initialize=Part.initialize))

A.initialize<-function(...){
  args<-list(...)

  .self$var1 <- if(is.null(args[["var1"]])) vector() else args[["var1"]]
  .self$var2 <- if(is.null(args[["var2"]])) character() else as.character(args[["var2"]])
  .self$var3 <- if(is.null(args[["var3"]])) list() else as.list(args[["var3"]])
  .self  
}

A<-setRefClass(Class = "A"
              ,contains = "Part"
              ,fields = list(var3 = "list")
              ,methods = list(initialize=A.initialize))

2）是否可以以这种方式组织代码，我必须只添加一行

.self$var3 <- if(is.null(args[["var3"]])) list() else as.list(args[["var3"]])

用于A类中的initialize函数，其余来自父类Part？

Answer 1

使用命名参数而不是解析...;确保默认构造函数在没有调整的情况下调用时工作

Part.initialize<-function(..., var1=vector(), var2=character()){
    callSuper(..., var1=var1, var2=as.character(var2))
}

Part<-setRefClass(Class = "Part"
                 ,fields = c(var1 = "ANY", var2 = "character")
                 ,methods = list(initialize=Part.initialize))

仅解释所考虑的类的参数

A.initialize<-function(..., var3=list()){
    callSuper(..., var3=as.list(var3))
}

A<-setRefClass(Class = "A"
               ,contains = "Part"
               ,fields = list(var3 = "list")
               ,methods = list(initialize=A.initialize))

简单的测试用例

Part()
A()
A(var3=list(a=1))

继承引用类

1 个答案: