Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
634 views
in Technique[技术] by (71.8m points)

r - Object not found error when passing model formula to another function

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

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) :)

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.

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)

Any thoughts?

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

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.

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

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...