[R] Specifying formula inside a function

[Mark Seeto]

Bill and Erik, thank you very much for your help. In addition to
solving my problem, both solutions contain other good things I didn't
know about.

Regards,
Mark Seeto

On Thu, Jun 10, 2010 at 2:44 PM, Erik Iverson <eriki at ccbr.umn.edu> wrote:
> Hello,
>
>> How does one specify a formula to lm inside a function (with variable
>> names not known in advance) and have the formula appear explicitly in
>> the output?
>>
>> For example,
>>
>> f <- function(d) {
>>  in.model <- sample(c(0,1), ncol(d)-1, replace=T)
>>  current.model <- lm(paste(names(d)[1], "~",
>> paste(names(d[2:ncol(d)])[which(in.model == 1)], collapse= "+")),
>> data=d)  #***
>>  return(current.model)
>> }
>> x1 <- rnorm(50,0,1)
>> x2 <- rnorm(50,0,1)
>> x3 <- rnorm(50,0,1)
>> x4 <- rnorm(50,0,1)
>> y <- rnorm(50,0,1)
>> d <- data.frame(y, x1, x2, x3, x4)
>> f(d)
>>
>> Call:
>> lm(formula = paste(names(d)[1], "~",
>> paste(names(d[2:ncol(d)])[which(in.model ==     1)], collapse = "+")),
>> data = d)
>>
>> Coefficients:
>> (Intercept)           x3           x4
>>    -0.1087       0.2830       0.1024
>>
>> How can I specify the formula in the line marked *** so that the
>> output will show "formula = y ~ x3 + x4" instead of "formula =
>> paste..."?
>>
>
> Well, there could very well be some tricks you can pull with ?substitute, or
> others, but I can't seem to figure it out.  Instead of tricks, it might be
> easier/clearer to assign a class to your object, say 'test', and write a
> print method for that based on print.lm.  There very well may be drawbacks
> to this that I am not realizing :).
>
> f <- function(d) {
>  in.model <- sample(c(0,1), ncol(d)-1, replace=TRUE)
>  current.model <- lm(paste(names(d)[1], "~",
>                            paste(names(d[2:ncol(d)])[which(in.model == 1)],
> collapse= "+")),
>                      data=d)
>  class(current.model) <- c("test", "lm")
>  current.model
> }
>
> # slight modification to print.lm
> print.test <- function (x, digits = max(3, getOption("digits") - 3), ...)
> {
>    cat("\nCall:\n", deparse(x$terms), "\n\n", sep = "")
>    if (length(coef(x))) {
>        cat("Coefficients:\n")
>        print.default(format(coef(x), digits = digits), print.gap = 2,
>            quote = FALSE)
>    }
>    else cat("No coefficients\n")
>    cat("\n")
>    invisible(x)
> }
>
> # should give you what you want...
> f(d)
>
On Thu, Jun 10, 2010 at 2:43 PM,  <Bill.Venables at csiro.au> wrote:
> Here is one way.
>
> f <- function(d) {
>  y <- names(d)[1]
>  xs <- names(d)[-1]
>  nx <- length(xs)
>  xs <- sort(sample(xs, sample(1:nx, 1)))
>  form <- as.formula(paste(y, "~", paste(xs, collapse="+")))
>  Call <- substitute(lm(FORM, data = d), list(FORM = form))
>  eval(Call)
> }
> d <-  within(data.frame(y = rnorm(50)), {
>  x1 <- rnorm(50)
>  x2 <- rnorm(50)
>  x3 <- rnorm(50)
>  x4 <- rnorm(50)
> })
>
> f(d)
>