I have a simple function in R that runs summary() via lapply() on many CSVs from one directory that I specify. The function is shown below: 

#   id -- the file name (i.e. 001.csv) so ID == 001. 
#   directory -- location of the CSV files (not my working directory)
#   summarize -- boolean val if summary of the CSV to be output to console. 
getMonitor <- function(id, dir, summarize = FALSE) 
{
    fl <- list.files(dir, pattern = "*.csv", full.names = FALSE)

    fdl <- lapply(fl, read.csv)

    dataSummary <- lapply(fdl, summary)

    if(summarize == TRUE)
    { dataSummary[[id]] }
}

When I try to specify the directory and then pass it as a parameter to the function like so: 

dir <- "C:\\Users\\ST\\My Documents\\R\\specdata"
funcVar <-  getMonitor("001", dir, FALSE)

I receive the error:

Error in file(file, "rt") : cannot open the connection. In addition: Warning message: In file(file, "rt") : cannot open file '001.csv': No such file or directory

Yet when I run the code below on its own: 

fl <- list.files("C:\\Users\\ST\\My Documents\\R\\specdata", 
                  pattern = "*.csv", 
                  full.names = FALSE)
fl[1]

It find the directory I'm pointing to and fl[1] correctly outputs [1] "001.csv" which is the first file listed.

My question is what am I doing wrong when trying to pass this path variable as a parameter to my function. Is R incapable of handling a parameter this way? Is there something I'm just completely missing? I've tried searching around and am familiar with other programming languages so, frankly, I feel kind of stupid/defeated for getting stuck on this right now.

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