Question

此功能使用每个指定的数据框架Cars93。我试图弄清楚如何通过返回结尾处的变量'pass'以及称为'required'的资源数量来返回迭代次数。

factory.function <- function (cars.output=1, trucks.output=1) {
factory <- matrix(c(40,1,60,3),nrow=2,
dimnames=list(c("labor","steel"),c("cars","trucks")))
available <- c(1600,70); names(available) <- rownames(factory)
slack <- c(8,1); names(slack) <- rownames(factory)
output <- c(cars.output, trucks.output); names(output) <- colnames(factory)

passes <- 0 # How many times have we  been around the loop?
repeat {
 passes <- passes + 1
 needed <- factory %*% output # What do we need for that output level?
 # If we're not using too much, and are within the slack, we're done
 if (all(needed <= available) &&
     all((available - needed) <= slack)) {
   break()
 }
 # If we're using too much of everything, cut back by 10%
 if (all(needed > available)) {
   output <- output * 0.9
   next()
 }
 # If we're using too little of everything, increase by 10%
 if (all(needed < available)) {
   output <- output * 1.1
   next()
 }
 # If we're using too much of some resources but not others, randomly
 # tweak the plan by up to 10%
  # runif == Random number, UNIFormly distributed, not "run if"
 output <- output * (1+runif(length(output),min=-0.1,max=0.1))
 }

return(output)
}
factory.function()

Answer 1

你可以使用

return(list(output = output,passes = passes, needed = needed))

您将从包含传递和所需

的函数中返回一个列表对象

如何让此函数返回其他数量？

1 个答案: