Question

当我在函数定义中使用可选参数时，省略号有问题。为了澄清，我定义了以下功能：

func1 <- function (x) (x-2)^2

func3 <- function (fun, arg.curve.user){
  arg.curve.user$expr <-  substitute(func1)
  arg.curve.default <- list(col = "blue", n = 1000, main = "This is a test")
  arg.curve <- modifyList (arg.curve.default, arg.curve.user)
  do.call("curve", arg.curve)
}

# optimizes func1 and call func2 to plot func1
func2 <- function (lb, ub, n.restarts = 5, n.sim = 10, ...){
  arg.curve.user <- as.list(substitute(list(...)))
  output <- gosolnp(fun = func1, LB = lb, UB = ub,  n.restarts =  n.restarts, 
  n.sim =  n.sim)$par
  func3(fun = func1, arg.curve.user = arg.curve.user)
   return(output)
}

通过调用func2，func1已经过优化，并且还通过func3调用绘制（需要包 Rsolnp ）。

func2 ( lb = 0, ub = 8, n.restarts = 5, n.sim = 10, n = 200, from = 0, to = 8)

但假设用户拼错了n.restarts并写了nrestarts：

func2 ( lb = 0, ub = 8, nrestarts = 5, n.sim = 10, n = 200, from = 0, to = 8)

在这种情况下，我希望R实施以下计划来处理n.restarts的缺席：

将默认值（即5）指定为n.restarts作为可选参数
在结尾处声明警告：“nrestarts”不是图形参数

但这不会发生， R会将n（200）的值分配给n.restarts 而不是!!

有人可以帮我解决这个问题吗？

非常感谢

Answer 1

当n参数未被用户提供时，n.restarts参数与n部分匹配。相反，与@Andrie（当然会有效）的建议相反，有一种机制允许您以参数n.restarts和参数...的方式继续。诀窍是在 func2 <- function (lb, ub, ..., n.restarts = 5, n.sim = 10){ writeLines(paste("Value of `n.restarts` is", n.restarts)) arg.curve.user <- as.list(substitute(list(...))) output <- gosolnp(fun = func1, LB = lb, UB = ub, n.restarts = n.restarts, n.sim = n.sim)$par func3(fun = func1, arg.curve.user = arg.curve.user) output }之后放置想要匹配完全 的参数。

> func2 (lb = 0, ub = 8, n.restarts = 2, n.sim = 10, n = 200, + from = 0, to = 8) Value of `n.restarts` is 2 <---- Here! Iter: 1 fn: 6.926e-15 Pars: 2.00000 Iter: 2 fn: 2.501e-15 Pars: 2.00000 solnp--> Completed in 2 iterations Iter: 1 fn: 8.336e-16 Pars: 2.00000 Iter: 2 fn: 8.336e-16 Pars: 2.00000 solnp--> Completed in 2 iterations [1] 2 > func2 (lb = 0, ub = 8, nrestarts = 2, n.sim = 10, n = 200, + from = 0, to = 8) Value of `n.restarts` is 5 <---- Here! Default Iter: 1 fn: 2.83e-15 Pars: 2.00000 Iter: 2 fn: 2.5e-15 Pars: 2.00000 solnp--> Completed in 2 iterations Iter: 1 fn: 2.037e-15 Pars: 2.00000 Iter: 2 fn: 2.037e-15 Pars: 2.00000 solnp--> Completed in 2 iterations Iter: 1 fn: 1.087e-15 Pars: 2.00000 Iter: 2 fn: 1.087e-15 Pars: 2.00000 solnp--> Completed in 2 iterations Iter: 1 fn: 8.558e-16 Pars: 2.00000 Iter: 2 fn: 8.558e-16 Pars: 2.00000 solnp--> Completed in 2 iterations Iter: 1 fn: 7.147e-16 Pars: 2.00000 Iter: 2 fn: 7.147e-16 Pars: 2.00000 solnp--> Completed in 2 iterations [1] 2 Warning messages: 1: In plot.window(...) : "nrestarts" is not a graphical parameter 2: In plot.xy(xy, type, ...) : "nrestarts" is not a graphical parameter 3: In axis(side = side, at = at, labels = labels, ...) : "nrestarts" is not a graphical parameter 4: In axis(side = side, at = at, labels = labels, ...) : "nrestarts" is not a graphical parameter 5: In box(...) : "nrestarts" is not a graphical parameter 6: In title(...) : "nrestarts" is not a graphical parameter

在使用中，这给出了：

{{1}}

Answer 2

如果您坚持使用常规参数评估，您更有可能获得警告。同样，您不需要使用特殊评估来获得所需的行为。使用非标准评估是一个坏主意，因为您不太可能完全重现R的默认行为，从而为您和您的用户带来微妙的错误。

library(Rsolnp)

func1 <- function (x) (x-2)^2

func3 <- function (fun, col = "blue", n = 1000, main = "This is a test", ...){
  curve(func1, ..., n = n, col = col, main = main)
}

# optimizes func1 and call func2 to plot func1
func2 <- function (lb, ub, n.restarts = 5, n.sim = 10, ...){
  output <- gosolnp(fun = func1, LB = lb, UB = ub, n.restarts = n.restarts, 
  n.sim =  n.sim)$par
  func3(fun = func1, ...)
  return(output)
}

然后当你跑：

func2 ( lb = 0, ub = 8, nrestarts = 5, n.sim = 10, n = 200, from = 0, to = 8)

你会收到警告

Warning messages:
1: In plot.window(...) : "nrestarts" is not a graphical parameter
2: In plot.xy(xy, type, ...) : "nrestarts" is not a graphical parameter
3: In axis(side = side, at = at, labels = labels, ...) :
  "nrestarts" is not a graphical parameter
4: In axis(side = side, at = at, labels = labels, ...) :
  "nrestarts" is not a graphical parameter
5: In box(...) : "nrestarts" is not a graphical parameter
6: In title(...) : "nrestarts" is not a graphical parameter

如何在存在可选参数的情况下处理省略号（...）？

2 个答案: