Help using mapply to run multiple models

Thu, Dec 19, 2013 1:51 PM

Add the line
   call$modelType <- NULL # omit modelType argument
to your function.  Otherwise
   f("glm", ...)
makes the call
   glm(modelType="glm", ...)
where you want it to make the call
   glm(...)

Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com

-----Original Message-----
From: Simon Kiss [mailto:sjkiss at gmail.com]
Sent: Thursday, December 19, 2013 1:49 PM
To: William Dunlap
Cc: Dennis Murphy; r-help at r-project.org
Subject: Re: [R] Help using mapply to run multiple models

Hi there: Just to tie this altogether.

Here is the final function

f<- function (modelType, responseName, predictorNames, data, ..., envir =
parent.frame())
{
  call <- match.call()
  call$formula <- formula(envir = envir, paste(responseName, sep = " ~ ",
                                               paste0("`", predictorNames, "`", collapse = " + ")))
  call[[1]] <- as.name(modelType)
  call$responseName <- NULL # omit responseName=
  call$predictorNames <- NULL # omit 'predictorNames='
  eval(call, envir = envir)
}

Here I call the function to a list of predictor variables and one dependent variable. Note
"glm" and not glm.
z <- lapply(list(c("hp","drat"), c("cyl"), c("am","gear")), FUN=function(preds)f("glm",
"carb", preds, data=mtcars, family='binomial'))

I do get this error:
Error in glm.control(modelType = "glm") :
  unused argument(s) (modelType = "glm")

But
lapply(z, summary)

does seem to return a list of model summaries. It looks like it worked.

I also tried.
z <- lapply(list(c("hp","drat"), c("cyl"), c("am","gear")), FUN=function(preds)f("lm", "mpg",
preds, data=mtcars))

Here, I get:
1: In lm.fit(x, y, offset = offset, singular.ok = singular.ok, ...) :
  extra argument 'modelType' is disregarded.
2: In lm.fit(x, y, offset = offset, singular.ok = singular.ok, ...) :
  extra argument 'modelType' is disregarded.
3: In lm.fit(x, y, offset = offset, singular.ok = singular.ok, ...) :
  extra argument 'modelType' is disregarded.

But again, it actually looks like it worked.
So, thank you very much!
Yours, Simon Kiss

On 2013-12-19, at 1:55 PM, Simon Kiss <sjkiss at gmail.com> wrote:

Hello Bill, that is fantastic and it's quite a bit above what I could write. Is there a way to

make the model type an argument to the function so that you can specify whether one is
running glm, lm and such?

I tried to modify it by inserting an argument modelType below, but that doesn't work.
Yours, simon Kiss

f <- function (modelType, responseName, predictorNames, data, ..., envir =

parent.frame())

  {
      call <- match.call()
      call$formula <- formula(envir = envir, paste(responseName, sep = " ~ ",
          paste0("`", predictorNames, "`", collapse = " + ")))
              call[[1]] <- quote(modelType) # '
      call$responseName <- NULL # omit responseName=
      call$predictorNames <- NULL # omit 'predictorNames='
              eval(call, envir = envir)
  }

On 2013-12-18, at 3:07 PM, William Dunlap <wdunlap at tibco.com> wrote:

f <- function (responseName, predictorNames, data, ..., envir = parent.frame())
  {
      call <- match.call()
      call$formula <- formula(envir = envir, paste(responseName, sep = " ~ ",
          paste0("`", predictorNames, "`", collapse = " + ")))
              call[[1]] <- quote(glm) # 'f' -> 'glm'
      call$responseName <- NULL # omit responseName=
      call$predictorNames <- NULL # omit 'predictorNames='
              eval(call, envir = envir)
  }
as in
  z <- lapply(list(c("hp","drat"), c("cyl"), c("am","gear")), FUN=function(preds)f("carb",

preds, data=mtcars, family=poisson))

  lapply(z, summary)

*********************************
Simon J. Kiss, PhD
Assistant Professor, Wilfrid Laurier University
73 George Street
Brantford, Ontario, Canada
N3T 2C9
Cell: +1 905 746 7606

*********************************
Simon J. Kiss, PhD
Assistant Professor, Wilfrid Laurier University
73 George Street
Brantford, Ontario, Canada
N3T 2C9
Cell: +1 905 746 7606

Help using mapply to run multiple models

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