Extract file extension from file path
This is the sort of thing that easily found with R basic tools. E.g.: ??path.
Anyway, load the tools
package and read ?file_ext
.
Extracting extension from filename in Python
Use os.path.splitext
:
>>> import os
>>> filename, file_extension = os.path.splitext('/path/to/somefile.ext')
>>> filename
'/path/to/somefile'
>>> file_extension
'.ext'
Unlike most manual string-splitting attempts, os.path.splitext
will correctly treat /a/b.c/d
as having no extension instead of having extension .c/d
, and it will treat .bashrc
as having no extension instead of having extension .bashrc
:
>>> os.path.splitext('/a/b.c/d')
('/a/b.c/d', '')
>>> os.path.splitext('.bashrc')
('.bashrc', '')
How to get the File Extension from a string Path
You can use the extension
function in the path
package to get the extension from a file path:
import 'package:path/path.dart' as p;
final path = '/some/path/to/file/file.dart';
final extension = p.extension(path); // '.dart'
If your file has multiple extensions, like file.dart.js
, you can specify the optional level
parameter:
final extension = p.extension('file.dart.js', 2); // '.dart.js'
How can I get file extensions with JavaScript?
Newer Edit: Lots of things have changed since this question was initially posted - there's a lot of really good information in wallacer's revised answer as well as VisioN's excellent breakdown
Edit: Just because this is the accepted answer; wallacer's answer is indeed much better:
return filename.split('.').pop();
My old answer:
return /[^.]+$/.exec(filename);
Should do it.
Edit: In response to PhiLho's comment, use something like:
return (/[.]/.exec(filename)) ? /[^.]+$/.exec(filename) : undefined;
Extract filename and extension in Bash
First, get file name without the path:
filename=$(basename -- "$fullfile")
extension="${filename##*.}"
filename="${filename%.*}"
Alternatively, you can focus on the last '/' of the path instead of the '.' which should work even if you have unpredictable file extensions:
filename="${fullfile##*/}"
You may want to check the documentation :
- On the web at section "3.5.3 Shell Parameter Expansion"
- In the bash manpage at section called "Parameter Expansion"
How to find the extension of a file in C#?
Path.GetExtension
string myFilePath = @"C:\MyFile.txt";
string ext = Path.GetExtension(myFilePath);
// ext would be ".txt"
How do I get the file extension of a file in Java?
In this case, use FilenameUtils.getExtension from Apache Commons IO
Here is an example of how to use it (you may specify either full path or just file name):
import org.apache.commons.io.FilenameUtils;
// ...
String ext1 = FilenameUtils.getExtension("/path/to/file/foo.txt"); // returns "txt"
String ext2 = FilenameUtils.getExtension("bar.exe"); // returns "exe"
Maven dependency:
<dependency>
<groupId>commons-io</groupId>
<artifactId>commons-io</artifactId>
<version>2.6</version>
</dependency>
Gradle Groovy DSL
implementation 'commons-io:commons-io:2.6'
Gradle Kotlin DSL
implementation("commons-io:commons-io:2.6")
Others https://search.maven.org/artifact/commons-io/commons-io/2.6/jar
Remove file extension from a file name string
I used the below, less code
string fileName = "C:\file.docx";
MessageBox.Show(Path.Combine(Path.GetDirectoryName(fileName),Path.GetFileNameWithoutExtension(fileName)));
Output will be
C:\file
Extracting a file extension from a given path in Rust idiomatically
In idiomatic Rust the return type of a function that can fail should be an Option
or a Result
. In general, functions should also accept slices instead of String
s and only create a new String
where necessary. This reduces excessive copying and heap allocations.
You can use the provided extension()
method and then convert the resulting OsStr
to a &str
:
use std::path::Path;
use std::ffi::OsStr;
fn get_extension_from_filename(filename: &str) -> Option<&str> {
Path::new(filename)
.extension()
.and_then(OsStr::to_str)
}
assert_eq!(get_extension_from_filename("abc.gz"), Some("gz"));
Using and_then
is convenient here because it means you don't have to unwrap the Option<&OsStr>
returned by extension()
and deal with the possibility of it being None
before calling to_str
. I also could have used a lambda |s| s.to_str()
instead of OsStr::to_str
- it might be a matter of preference or opinion as to which is more idiomatic.
Notice that both the argument &str
and the return value are references to the original string slice created for the assertion. The returned slice cannot outlive the original slice that it is referencing, so you may need to create an owned String
from this result if you need it to last longer.
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