将模型公式传递给另一个函数时找不到对象错误 [英] Object not found error when passing model formula to another function

问题描述

我对R有一个奇怪的问题，我似乎无法解决.

I have a weird problem with R that I can't seem to work out.

我试图编写一个函数，以对R中的分步过程选择的模型执行K折交叉验证.(我知道分步过程的问题，这纯粹是出于比较目的):)

I've tried to write a function that performs K-fold cross validation for a model chosen by the stepwise procedure in R. (I'm aware of the issues with stepwise procedures, it's purely for comparison purposes) :)

现在的问题是，如果我定义函数参数(linmod，k，direction)并运行该函数的内容，则它可以正常工作.但是，如果我将其作为函数运行，则会收到一条错误消息，提示找不到datas.train对象.

Now the issue is, that if I define the function parameters (linmod,k,direction) and run the contents of the function, it works flawlessly. BUT, if I run it as a function, I get an error saying the datas.train object can't be found.

我尝试使用debug()逐步执行该函数，并且该对象显然存在，但是R表示在我实际运行该函数时不存在.如果我只是使用lm()拟合模型，那么它就可以正常工作，因此我认为这是函数内部循环中step函数的问题. (尝试注释掉step命令，然后将预测设置为来自普通线性模型的预测.)

I've tried stepping through the function with debug() and the object clearly exists, but R says it doesn't when I actually run the function. If I just fit a model using lm() it works fine, so I believe it's a problem with the step function in the loop, while inside a function. (try commenting out the step command, and set the predictions to those from the ordinary linear model.)

#CREATE A LINEAR MODEL TO TEST FUNCTION
lm.cars <- lm(mpg~.,data=mtcars,x=TRUE,y=TRUE)


#THE FUNCTION
cv.step <- function(linmod,k=10,direction="both"){
  response <- linmod$y
  dmatrix <- linmod$x
  n <- length(response)
  datas <- linmod$model
  form <- formula(linmod$call)

  # generate indices for cross validation
  rar <- n/k
  xval.idx <- list()
  s <- sample(1:n, n) # permutation of 1:n
  for (i in 1:k) {
    xval.idx[[i]] <- s[(ceiling(rar*(i-1))+1):(ceiling(rar*i))]
  }

  #error calculation
  errors <- R2 <- 0

  for (j in 1:k){
     datas.test <- datas[xval.idx[[j]],]
       datas.train <- datas[-xval.idx[[j]],]
       test.idx <- xval.idx[[j]]

       #THE MODELS+
       lm.1 <- lm(form,data= datas.train)
       lm.step <- step(lm.1,direction=direction,trace=0)

      step.pred <- predict(lm.step,newdata= datas.test)
        step.error <- sum((step.pred-response[test.idx])^2)
        errors[j] <- step.error/length(response[test.idx])

        SS.tot <- sum((response[test.idx] - mean(response[test.idx]))^2)
        R2[j] <- 1 - step.error/SS.tot
  }

  CVerror <- sum(errors)/k
  CV.R2 <-  sum(R2)/k

  res <- list()
  res$CV.error <- CVerror
  res$CV.R2 <- CV.R2

return(res)
}


#TESTING OUT THE FUNCTION
cv.step(lm.cars)

有什么想法吗?

推荐答案

在创建公式lm.cars时，已为其分配了自己的环境.除非您明确更改，否则此环境将与公式保持一致.因此，当您使用formula函数提取公式时，将包含模型的原始环​​境.

When you created your formula, lm.cars, in was assigned its own environment. This environment stays with the formula unless you explicitly change it. So when you extract the formula with the formula function, the original environment of the model is included.

我不知道这里是否使用了正确的术语，但是我认为您需要显式更改函数内部公式的环境:

I don't know if I'm using the correct terminology here, but I think you need to explicitly change the environment for the formula inside your function:

cv.step <- function(linmod,k=10,direction="both"){
  response <- linmod$y
  dmatrix <- linmod$x
  n <- length(response)
  datas <- linmod$model
  .env <- environment() ## identify the environment of cv.step

  ## extract the formula in the environment of cv.step
  form <- as.formula(linmod$call, env = .env) 

  ## The rest of your function follows

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