Edit Question - KnowledgeHub

Title Body

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.

Tags (comma-separated) Cancel