Get filename without extension in R

Question

I have a file

ABCD.csv 

The length before the .csv is not fixed and vary to any length.

How can I extract the portion

Answer 1

There's a built in file_path_sans_ext from the standard install tools package that grabs the file without the extension.

tools::file_path_sans_ext("ABCD.csv")
## [1] "ABCD"

Answer 2

You can try this also:

data <- "ABCD.csv"
gsub(pattern = "\\.csv$", "", data)

#[1] "ABCD"

This will be helpful in case of list of files as well, say

data <- list.files(pattern="\\.csv$") , using the code will remove extension of all the files in the list.

Answer 3

Loading the library needed :

> library(stringr)

Extracting all the matches from the regex:

> str_match("ABCD.csv", "(.*)\\..*$")
     [,1]       [,2]  
[1,] "ABCD.csv" "ABCD"

Returning only the second part of the result, which corresponds to the group matching the file name:

> str_match("ABCD.csv", "(.*)\\..*$")[,2]
[1] "ABCD"

EDIT for @U-10-Forward:

It is basically the same principle as the other answer. Just that I found this solution more robust.

Regex wise it means:

• () = group

• .* = any single character except the newline character any number of time

• // is escape notation, thus //. means literally "."

• .* = any characters any number of time again

• $ means should be at the end of the input string

The logic is then that it will return the group preceding a "." followed by a group of characters at the end of the string (which equals the file extension in this case).

Answer 4

If you have filenames with multiple (possible extensions) and you want to strip off only the last extension, you can try the following.

Consider the filename foo.bar.baz.txt this

sub('\\..[^\\.]*$', '', "foo.bar.baz.txt")

will leave you with foo.bar.baz.

Answer 5

You can use sub or substr

sub('\\.csv$', '', str1) 
#[1] "ABCD"

or

substr(str1, 1, nchar(str1)-4)
#[1] "ABCD"

Using the 'file_path' from @JasonV's post

sub('\\..*$', '', basename(filepath))
#[1] "ABCD"

Or

library(stringr)
str_extract(filepath,  perl('(?<=[/])([^/]+)(?=\\.[^.]+)'))
#[1] "ABCD"

data

str1 <- 'ABCD.csv'

Answer 6

basename will also remove the path leading to the file. And with this regex, any extension will be removed.

filepath <- "d:/Some Dir/ABCD.csv"
sub(pattern = "(.*)\\..*$", replacement = "\\1", basename(filepath))

# [1] "ABCD"

Or, using file_path_sans_ext as Tyler Rinker suggested:

file_path_sans_ext(basename(filepath))

# [1] "ABCD"