How do I get the filename without the extension from a path in Python?
"/path/to/some/file.txt" → "file" 28 Answers
Getting the name of the file without the extension:
import os print(os.path.splitext("/path/to/some/file.txt")[0]) Prints:
/path/to/some/file Documentation for os.path.splitext.
Important Note: If the filename has multiple dots, only the extension after the last one is removed. For example:
import os print(os.path.splitext("/path/to/some/file.txt.zip.asc")[0]) Prints:
/path/to/some/file.txt.zip See other answers below if you need to handle that case.
12Use .stem from pathlib in Python 3.4+
from pathlib import Path Path('/root/dir/sub/file.ext').stem will return
'file' Note that if your file has multiple extensions .stem will only remove the last extension. For example, Path('file.tar.gz').stem will return 'file.tar'.
You can make your own with:
>>> import os >>> base=os.path.basename('/root/dir/sub/file.ext') >>> base 'file.ext' >>> os.path.splitext(base) ('file', '.ext') >>> os.path.splitext(base)[0] 'file' Important note: If there is more than one . in the filename, only the last one is removed. For example:
/root/dir/sub/file.ext.zip -> file.ext /root/dir/sub/file.ext.tar.gz -> file.ext.tar See below for other answers that address that.
0>>> print(os.path.splitext(os.path.basename("/path/to/file/hemanth.txt"))[0]) hemanth In Python 3.4+ you can use the pathlib solution
from pathlib import Path print(Path(your_path).resolve().stem) In python 3 pathlib "The pathlib module offers high-level path objects." so,
>>> from pathlib import Path >>> p = Path("/a/b/c.txt") >>> p.with_suffix('') WindowsPath('/a/b/c') >>> p.stem 'c' os.path.splitext() won't work if there are multiple dots in the extension.
For example, images.tar.gz
>>> import os >>> file_path = '/home/dc/images.tar.gz' >>> file_name = os.path.basename(file_path) >>> print os.path.splitext(file_name)[0] images.tar You can just find the index of the first dot in the basename and then slice the basename to get just the filename without extension.
>>> import os >>> file_path = '/home/dc/images.tar.gz' >>> file_name = os.path.basename(file_path) >>> index_of_dot = file_name.index('.') >>> file_name_without_extension = file_name[:index_of_dot] >>> print file_name_without_extension images If you want to keep the path to the file and just remove the extension
>>> file = '/root/dir/ >>> print ('.').join(file.split('.')[:-1]) /root/dir/ As noted by @IceAdor in a comment to @user2902201's solution, rsplit is the simplest solution robust to multiple periods (by limiting the number of splits to maxsplit of just 1 (from the end of the string)).
Here it is spelt out:
file = 'my.report.txt' print file.rsplit('.', maxsplit=1)[0] my.report
Thought I would throw in a variation to the use of the os.path.splitext without the need to use array indexing.
The function always returns a (root, ext) pair so it is safe to use:
root, ext = os.path.splitext(path)
Example:
>>> import os >>> path = 'my_text_file.txt' >>> root, ext = os.path.splitext(path) >>> root 'my_text_file' >>> ext '.txt' But even when I import os, I am not able to call it path.basename. Is it possible to call it as directly as basename?
import os, and then use os.path.basename
importing os doesn't mean you can use os.foo without referring to os.
import os filename, file_extension =os.path.splitext(os.path.basename('/d1/d2/example.cs')) filename is 'example'
file_extension is '.cs'
'
1The other methods don't remove multiple extensions. Some also have problems with filenames that don't have extensions. This snippet deals with both instances and works in both Python 2 and 3. It grabs the basename from the path, splits the value on dots, and returns the first one which is the initial part of the filename.
import os def get_filename_without_extension(file_path): file_basename = os.path.basename(file_path) filename_without_extension = file_basename.split('.')[0] return filename_without_extension Here's a set of examples to run:
example_paths = [ "FileName", "./FileName", "../../FileName", "FileName.txt", "./FileName.txt.zip.asc", "/path/to/some/FileName", "/path/to/some/FileName.txt", "/path/to/some/FileName.txt.zip.asc" ] for example_path in example_paths: print(get_filename_without_extension(example_path)) In every case, the value printed is:
FileName Answers using Pathlib for Several Scenarios
Using Pathlib, it is trivial to get the filename when there is just one extension (or none), but it can be awkward to handle the general case of multiple extensions.
Zero or One extension
from pathlib import Path pth = Path('./thefile.tar') fn = pth.stem print(fn) # thefile # Explanation: # the `stem` attribute returns only the base filename, stripping # any leading path if present, and strips the extension after # the last `.`, if present. # Further tests eg_paths = ['thefile', 'thefile.tar', './thefile', './thefile.tar', '../../thefile.tar', '.././thefile.tar', 'rel/ '/abs/path/to/thefile.tar'] for p in eg_paths: print(Path(p).stem) # prints thefile every time Two or fewer extensions
from pathlib import Path pth = Path('./thefile.tar.gz') fn = pth.with_suffix('').stem print(fn) # thefile # Explanation: # Using the `.with_suffix('')` trick returns a Path object after # stripping one extension, and then we can simply use `.stem`. # Further tests eg_paths += ['./thefile.tar.gz', '/abs/ for p in eg_paths: print(Path(p).with_suffix('').stem) # prints thefile every time Any number of extensions (0, 1, or more)
from pathlib import Path pth = Path('./thefile.tar.gz.bz.7zip') fn = pth.name if len(pth.suffixes) > 0: s = pth.suffixes[0] fn = fn.rsplit(s)[0] # or, equivalently fn = pth.name for s in pth.suffixes: fn = fn.rsplit(s)[0] break # or simply run the full loop fn = pth.name for _ in pth.suffixes: fn = fn.rsplit('.')[0] # In any case: print(fn) # thefile # Explanation # # pth.name -> 'thefile.tar.gz.bz.7zip' # pth.suffixes -> ['.tar', '.gz', '.bz', '.7zip'] # # If there may be more than two extensions, we can test for # that case with an if statement, or simply attempt the loop # and break after rsplitting on the first extension instance. # Alternatively, we may even run the full loop and strip one # extension with every pass. # Further tests eg_paths += ['./thefile.tar.gz.bz.7zip', '/abs/ for p in eg_paths: pth = Path(p) fn = pth.name for s in pth.suffixes: fn = fn.rsplit(s)[0] break print(fn) # prints thefile every time Special case in which the first extension is known
For instance, if the extension could be .tar, .tar.gz, .tar.gz.bz, etc; you can simply rsplit the known extension and take the first element:
pth = Path('foo/bar/) fn = pth.name.rsplit('.tar')[0] print(fn) # thefile A multiple extension aware procedure. Works for str and unicode paths. Works in Python 2 and 3.
import os def file_base_name(file_name): if '.' in file_name: separator_index = file_name.index('.') base_name = file_name[:separator_index] return base_name else: return file_name def path_base_name(path): file_name = os.path.basename(path) return file_base_name(file_name) Behavior:
>>> path_base_name('file') 'file' >>> path_base_name(u'file') u'file' >>> path_base_name('file.txt') 'file' >>> path_base_name(u'file.txt') u'file' >>> path_base_name('file.tar.gz') 'file' >>> path_base_name('file.a.b.c.d.e.f.g') 'file' >>> path_base_name('relative/path/file.ext') 'file' >>> path_base_name('/absolute/path/file.ext') 'file' >>> path_base_name('Relative\\Windows\\Path\\file.txt') 'file' >>> path_base_name('C:\\Absolute\\Windows\\Path\\file.txt') 'file' >>> path_base_name('/path with spaces/file.ext') 'file' >>> path_base_name('C:\\Windows Path With Spaces\\file.txt') 'file' >>> path_base_name('some/path/file name with spaces.tar.gz.zip.rar.7z') 'file name with spaces' import os
filename = C:\\Users\\Public\\Videos\\Sample Videos\\wildlife.wmv This returns the filename without the extension(C:\Users\Public\Videos\Sample Videos\wildlife)
temp = os.path.splitext(filename)[0] Now you can get just the filename from the temp with
os.path.basename(temp) #this returns just the filename (wildlife) Very very very simpely no other modules !!!
import os p = r"C:\Users\bilal\Documents\face Recognition python\imgs\northon.jpg" # Get the filename only from the initial file path. filename = os.path.basename(p) # Use splitext() to get filename and extension separately. (file, ext) = os.path.splitext(filename) # Print outcome. print("Filename without extension =", file) print("Extension =", ext) import os path = "a/b/c/abc.txt" print os.path.splitext(os.path.basename(path))[0] On Windows system I used drivername prefix as well, like:
>>> s = 'c:\\temp\\akarmi.txt' >>> print(os.path.splitext(s)[0]) c:\temp\akarmi So because I do not need drive letter or directory name, I use:
>>> print(os.path.splitext(os.path.basename(s))[0]) akarmi Improving upon @spinup answer:
fn = pth.name for s in pth.suffixes: fn = fn.rsplit(s)[0] break print(fn) # thefile This works for filenames without extension also
I've read the answers, and I notice that there are many good solutions. So, for those who are looking to get either (name or extension), here goes another solution, using the os module, both methods support files with multiple extensions.
import os def get_file_name(path): if not os.path.isdir(path): return os.path.splitext(os.path.basename(path))[0].split(".")[0] def get_file_extension(path): extensions = [] copy_path = path while True: copy_path, result = os.path.splitext(copy_path) if result != '': extensions.append(result) else: break extensions.reverse() return "".join(extensions) Note: this solution on windows does not support file names with the "\" character
We could do some simple split / pop magic as seen here (), to extract the filename (respecting the windows and POSIX differences).
def getFileNameWithoutExtension(path): return path.split('\\').pop().split('/').pop().rsplit('.', 1)[0] getFileNameWithoutExtension('/path/to/file-0.0.1.ext') # => file-0.0.1 getFileNameWithoutExtension('\\path\\to\\file-0.0.1.ext') # => file-0.0.1 For convenience, a simple function wrapping the two methods from os.path :
def filename(path): """Return file name without extension from path. See """ import os.path b = os.path.split(path)[1] # path, *filename* f = os.path.splitext(b)[0] # *file*, ext #print(path, b, f) return f Tested with Python 3.5.
import os list = [] def getFileName( path ): for file in os.listdir(path): #print file try: base=os.path.basename(file) splitbase=os.path.splitext(base) ext = os.path.splitext(base)[1] if(ext): list.append(base) else: newpath = path+"/"+file #print path getFileName(newpath) except: pass return list getFileName("/home/weexcel-java3/Desktop/backup") print list the easiest way to resolve this is to
import ntpath print('Base name is ',ntpath.basename('/path/to/the/file/')) this saves you time and computation cost.
I didn't look very hard but I didn't see anyone who used regex for this problem.
I interpreted the question as "given a path, return the basename without the extension."
e.g.
"path/to/file.json" => "file"
"path/to/my.file.json" => "my.file"
In Python 2.7, where we still live without pathlib...
def get_file_name_prefix(file_path): basename = os.path.basename(file_path) file_name_prefix_match = re.compile(r"^(?P<file_name_pre fix>.*)\..*$").match(basename) if file_name_prefix_match is None: return file_name else: return file_name_prefix_match.group("file_name_prefix") get_file_name_prefix("path/to/file.json") >> file get_file_name_prefix("path/to/my.file.json") >> my.file get_file_name_prefix("path/to/no_extension") >> no_extension What about the following?
import pathlib filename = '/path/to/dir/stem.ext.tar.gz' pathlib.Path(filename).name[:-len(''.join(pathlib.Path(filename).suffixes))] # -> 'stem' or this equivalent?
pathlib.Path(filename).name[:-sum(map(len, pathlib.Path(filename).suffixes))] # use pathlib. the below works with compound filetypes and normal ones source_file = 'spaces.tar.gz.zip.rar.7z' source_path = pathlib.Path(source_file) source_path.name.replace(''.join(source_path.suffixes), '') >>> 'spaces' despite the many working implementations described above I added this ^ as it uses pathlib only and works for compound filetypes and normal ones
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