variable scope & resolution in R function

Question

I want to loop through the vars in a dataframe, calling lm() on each one, and so I wrote this:

findvars <- function(x = samsungData, dv = 'activity', id = 'subject') {
  # Loops through the possible predictor vars, does an lm() predicting the dv
  # from each, and returns a data.frame of coefficients, one row per IV.
  r <- data.frame()
  # All varnames apart from the dependent var, and the case identifier
  ivs <- setdiff(names(x), c(dv, id))
  for (iv in ivs) {
    print(paste("trying", iv))
    m <- lm(dv ~ iv, data = x, na.rm = TRUE)
    # Take the absolute value of the coefficient, then transpose.
    c <- t(as.data.frame(sapply(m$coefficients, abs)))
    c$iv <- iv # which IV produced this row?
    r <- c(r, c)
  }
  return(r)
}

This doesn't work, I believe b/c the formula in the lm() call consists of function-local variables that hold strings naming vars in the passed-in dataframe (e.g., "my_dependant_var" and "this_iv") as opposed to pointers to the actual variable objects.

I tried wrapping that formula in eval(parse(text = )), but could not get that to work.

If I'm right about the problem, can someone explain to me how to get R to resolve the contents of those vars iv & dv into the pointers I need? Or if I'm wrong, can someone explain what else is going on?

Many thanks!

Here is some repro code:

library(datasets)
data(USJudgeRatings)
findvars(x = USJudgeRatings, dv = 'CONT', id = 'DILG')

In R you should forget about "pointers to the objects". Values are passed ... as values. And the "variables" in formulas are not really "strings". Names and symbols are objects of super-class "language". Character vectors are not. — IRTFM
– IRTFM, Commented Dec 4, 2013 at 5:45

joran · Accepted Answer · 2013-12-04 04:29:58Z

3

So there's enough bad stuff happening in your function besides your trouble with the formula, that I think someone should walk you through it all. Here are some annotations, followed by a better version:

  #For small examples, "growing" objects isn't a huge deal,
  # but you will regret it very, very quickly. It's a bad
  # habit. Learn to ditch it now. So don't inititalize
  # empty lists and data frames.
  r <- data.frame()

  ivs <- setdiff(names(x), c(dv, id))
  for (iv in ivs) {
    print(paste("trying", iv))
    #There is no na.rm argument to lm, only na.action
    m <- lm(dv ~ iv, data = x, na.rm = TRUE)
    #Best not to name variables c, its a common function, see two lines from now!
    # Also, use the coef() extractor functions, not $. That way, if/when
    # authors change the object structure your code won't break.
    #Finally, abs is vectorized, no need for sapply
    c <- t(as.data.frame(sapply(m$coefficients, abs)))
    #This is probably best stored in the name
    c$iv <- iv # which IV produced this row?
    #Growing objects == bad! Also, are you sure you know what happens when
    # you concatenate two data frames?
    r <- c(r, c)
  }
  return(r)
}

Try something like this instead:

findvars <- function(x,dv,id){
  ivs <- setdiff(names(x),c(dv,id))
  #initialize result list of the appropriate length
  result <- setNames(vector("list",length(ivs)),ivs)
  for (i in seq_along(ivs)){
    result[[i]] <- abs(coef(lm(paste(dv,ivs[i],sep = "~"),data = x,na.action = na.omit)))
  }
  result
}

answered Dec 4, 2013 at 4:29

joran

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1 Comment

Roy Pardee Over a year ago

Holy cow--thank you so much for breaking all that that down! That of course works, and I have some inkling why it works--at least I think so. I'm comforted to know that lm() will take a string for the formula argument--I guess part of my problem was wrapping my paste() call in eval()? Or something. At any rate--thanks for being so generous w/your time!

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