如何在循环中将函数添加到 R 中的 R6Class

Question

我想在 R6Class 上写包装器，但它不起作用，在发现这个问题后我尝试了很多东西 dynamically add function to r6 class instance

所以我试过了，它们都不起作用：

get <- function(x = list()) {
  class <- R6::R6Class(classname = "class")
  for (name in names(x)) {
    class$set("public", name, function() name)
  }
  class
}

x <- get(x = list(a = 10, b = 20))$new()
x$a()
# b
x$b()
# b

这是因为闭包循环不会创建新的作用域。所以我试过这个：

get <- function(x = list()) {
  class <- R6::R6Class(classname = "class")
  lapply(names(x), function(name) {
    print(name)
    class$set("public", name, function() name)
  })
  class
}

x <- get(x = list(a = 10, b = 10))$new()
x$a()

这将抛出名称未定义的错误，因为 R6Class 的这种行为在 eval substitute 中，因此无法创建从其所在位置获取 scope/environment 的新函数被称为。或者有什么办法吗？

我真正的问题是我想创建函数包装器并且我想调用：

fn <- function() {
    x <- 10
    y <- myFunction(public = list(
       foo = function(y) {
          x + y
       }
    })
    z <- y$new()
    z$foo(10)
    ## I want 20 as result
}

有没有办法创建 myFunction 函数来创建 R6Class？我想要这个的原因是因为我有基于 R6Class 的组件系统，并且想删除一些需要添加到每个 class 的样板，这样它更容易使用。我不想创建新的 class 系统，我想使用 R6 classes.

Answer 1

我在添加这个问题后在 GitHub 上提问，他们很快就给出了答案。这是 re-post 的答案：

get <- function(x = list()) {
  class <- R6::R6Class(classname = "class")
  lapply(names(x), function(name) {
    fn <- eval(substitute(function() subst_name, list(subst_name = name)))
    class$set("public", name, fn)
  })
  class
}
x <- get(x = list(a = 10, b = 20))$new()
x$a()

如果你想在打印 x$a 时使用更好的名称，你可以使用以下方法清除名称引用：

attr(fn, "srcref") <- NULL

编辑:

这是添加到 class 的值是函数的示例（这是我改进的代码）：

constructor <- function(public = list(), private = list()) {
  class <- R6::R6Class(classname = "class")
  lapply(names(public), function(name) {
    if (is.function(public[[name]])) {
      env <- environment(public[[name]])
      env$self <- public
      env$private <- private
      fn <- eval(substitute(function(...) fn(...), list(fn = public[[name]])))
      class$set("public", name, fn)
    } else {
      class$set("public", name, public[[name]])
    }
  })
  class
}
test <- function() {
  a <- 10
  class <- constructor(
     public = list(
         a = function() { a + self$b },
         b = 20
     )
  )
  x <- class$new()
  x$a()
}

test()

如果您想访问 super，您需要使用此代码：

EDIT2:

component <- function(public = NULL,
                      private = NULL,
                      static = NULL,
                      ...) {
  class <- R6::R6Class(...)
  r6.class.add(class, public)
  r6.class.add(class, private)
  class$extend <- make.extend(class)
  class
}

#' helper function for adding properties to R6Class
r6.class.add <- function(class, seq) {
  prop.name <- as.character(substitute(seq)) # so we don't need to write name as string
  lapply(names(seq), function(name) {
    if (is.function(seq[[name]])) {
      ## the only way to have scope from when function was create with self and private
      ## eval substitute simply circument R6 encapsulation and use scope from where function
      ## was created (closure) and env.fn patch the env of inner function so it get self
      ## and private as magic names - this is done so component function the same as
      ## when R6Class is created inline - so component is referencial transparent and can
      ## be replaced with R6Class
      fn <- eval(substitute(function(...) {
        ## patch function env
        fn <- fn.expr # fn.expr will be inline function expression
        parent <- parent.env(environment())
        ## we don't overwrite function scope so you can nest one constructor
        ## in another constructor
        env <- new.env(parent = environment(fn))
        env$self <- parent$self
        env$super <- parent$super
        env$private <- parent$private
        environment(fn) <- env
        fn(...)
      }, list(fn.expr = seq[[name]], name = name)))
      class$set(prop.name, name, fn)
    } else {
      class$set(prop.name, name, seq[[name]])
    }
  })
}

您也可以使用 get("self", parent) 而不是 env$self <- parent$self（它将在环境链中搜索变量）。

如何在循环中将函数添加到 R 中的 R6Class

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