Question

I want to write a function f that given:

x <- 1  
f(.(y==x))

It returns the string:

"y == 1"

The function should be general enough, that the expression can be arbitrary. For instance, f(.(y==x & z==x)) should return "y == 1 & z == 1".

Any ideas?

The function:

g <- function(e) as.character(unlist(e))

does not work for me, as it returns "y == x". I want the replacement to take place.

UPDATE:

Also, this question is not a duplicate of How do I substitute symbols in a language object? I want to create a function f, that means that somehow I need to get the value of x from the caller s environment. My current attempt is this:

require(plyr)
x <- 1
f <- function(e) do.call(substitute, list(e[[1]], parent.frame(1)))
f(.(y==x & z==x))

But it still does not work. Maybe I should use something else instead of parent.frame?

Many thanks.

Answer 1

 library(plyr)
 g <- function(e) { 
      e[[1]][[3]] <- eval.parent(e[[1]][[3]])
      deparse(e[[1]]) 
 }
 x <- 1
 g(.(y==x))

This works, but is perhaps not as general as you wanted. (I would rather have a solution that substituted values if possible and left the values alone otherwise.)

UPDATE: not working yet, but maybe helpful.

 g <- function(e) {
      e1 <- e[[1]]
      pe <- parent.env(environment())
      vars <- all.vars(e1)
      ss <- setNames(lapply(vars,
                   function(v) if (exists(v,pe)) get(v,pe) else v),
                     vars)
      ## at this point I m stumped -- have tried various
      ## versions of eval, lapply, rapply, substitute ...
      ## afraid I would actually have to make up my own recursive
      ## function since rapply() only works on lists ...
 }

Answer 2

f <- function(e) {
    deparse(do.call("substitute", list(e, as.list(parent.frame()))))
}

x <- 1
f(quote(y == x)) # "y == 1"
f(quote(y==x & z==x)) # "y == 1 & z == 1"

I use do.call to construct a substitute call that has the arguments in args evaluated. In the case of f(quote(y == x)), the call will be substitute(y == x, list(x = 1, f = ...)).
I have to use the list as.list(parent.frame()) instead of the environment parent.frame() because according to ?substitute

If it is an ordinary variable, its value is substituted, unless env is .GlobalEnv in which case the symbol is left unchanged.

Calling parent.frame() inside f result in <environment: R_GlobalEnv>

I am not familiar with plyr::. so I use quote in base instead.

友情链接