Name of a list not the contents

Question

I have a list containing 18 elements called bx2.

I want to use bx2 in a function,

XlsMaker  <- function(x) {
  library("XLConnect")
a  <- length(x)
b  <- paste0(x,".xlsx")
for (i in 1:a){
writeWorksheetToFile(b, data = x[[i]], sheet = names(x[i]))
}
}

but when put I bx2 into the function it pulls in all the elements of the list rather than just the name of the list.

Is it possible to re-write the function so that b becomes bx2.xlsx?

score 1 · Answer 1 · answered Dec 18 '14 at 22:09

I think you just want to deparse the parameter name

XlsMaker  <- function(x) {
    varname  <- deparse(substitute(x))
    library("XLConnect")
    a  <- length(x)
    b  <- paste0(varname ,".xlsx")
    for (i in 1:a){
        writeWorksheetToFile(b, data = x[[i]], sheet = names(x[i]))
    }
}
bx2 <-list(1:3, 4:6)
XlsMaker(bx2)

score 1 · Accepted Answer · answered Dec 18 '14 at 22:13

The line b <- paste0(x,".xlsx") is wrong. That calls paste0 on the object x itself which is not at all what you want to do. You want to call it on the name of the object.

This in general opens a can of worms because objects can have two different names in two different places. Consider: the object named bx2 in your global environment is now named x within the function's scope. If you only want to call this function from the top level (e.g. for interactive use), you can safely get the name of the object from the parent environment (the environment you called the function from) by replacing that line with:

x_name <- deparse(substitute(x))
b <- paste0(x_name, ".xlsx")

The substitute function gets the name of x in the parent environment, as a special name object. The deparse function converts this name into a character vector of length one.

The reason I said this is only safe to use at the top level is that substitute can return surprising or unintended results if not used carefully. Hadley Wickham goes into detail on this point in his book.

Name of a list not the contents

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