Python extract file path

Get the filename, directory, extension from a path string in Python

In Python, you can get the filename (basename), directory (folder) name, and extension from a path string or join the strings to generate the path string with the os.path module in the standard library.

• Difference in path separator by OS
• Get the filename (basename) from a path: os.path.basename()
  • Filename with extension
  • Filename without extension
  • Notes on when a path string indicates a directory
  • Create a path string with a different extension
  • Get the extension without dot (period)
  • Examples of cases like .tar.gz
  • Create a path string for another file in the same directory
  • Backslash and raw string
  • Examples of getting file name, folder name, extension
  • Get and join a drive letter: os.path.splitdrive()

  Use the following path string as an example.

  import os filepath = './dir/subdir/filename.ext' 

  The sample code below is running on a Mac. Examples for Windows are shown at the end.

  In Python 3.4 or later, you can also get the filename, directory (folder) name, extension, etc., with the pathlib module that treats paths as objects.

  Difference in path separator by OS

  The path separator varies depending on the OS. UNIX (including Mac) uses the slash / , while Windows uses the backslash \ .

  You can get the separator in the OS running Python with os.sep or os.path.sep .

  print(os.sep) # / print(os.sep is os.path.sep) # True 

  Get the filename (basename) from a path: os.path.basename()

  Use os.path.basename() to get the filename (basename) from a path string.

  Filename with extension

  os.path.basename() returns the filename with the extension.

  filepath = './dir/subdir/filename.ext' basename = os.path.basename(filepath) print(basename) # filename.ext print(type(basename)) # 

  Filename without extension

  You can get the filename without the extension with os.path.splitext() described later.

  filepath = './dir/subdir/filename.ext' basename_without_ext = os.path.splitext(os.path.basename(filepath))[0] print(basename_without_ext) # filename 

  os.path.splitext() split at the last (right) dot . . If you want to split by the first (left) dot . , use split() .

  filepath_tar_gz = './dir/subdir/filename.tar.gz' print(os.path.splitext(os.path.basename(filepath_tar_gz))[0]) # filename.tar print(os.path.basename(filepath_tar_gz).split('.', 1)[0]) # filename 

  Get the directory (folder) name from a path: os.path.dirname()

  Use os.path.dirname() to get the directory folder (name) from a path string.

  filepath = './dir/subdir/filename.ext' dirname = os.path.dirname(filepath) print(dirname) # ./dir/subdir print(type(dirname)) # 

  If you want to get only the directory name directly above the file, use os.path.basename() .

  subdirname = os.path.basename(os.path.dirname(filepath)) print(subdirname) # subdir 

  Get the file and directory name pair: os.path.split()

  Use os.path.split() to get both the file and directory (folder) name.

  os.path.split() returns a tuple of filename returned by os.path.basename() and directory name returned by os.path.dirname() .

  filepath = './dir/subdir/filename.ext' base_dir_pair = os.path.split(filepath) print(base_dir_pair) # ('./dir/subdir', 'filename.ext') print(type(base_dir_pair)) # print(os.path.split(filepath)[0] == os.path.dirname(filepath)) # True print(os.path.split(filepath)[1] == os.path.basename(filepath)) # True 

  You can unpack tuple to assign to each variable.

  dirname, basename = os.path.split(filepath) print(dirname) # ./dir/subdir print(basename) # filename.ext 

  Use os.path.join() described later to rejoin the file and directory names.

  Notes on when a path string indicates a directory

  Note that the result will differ for a path string indicating a directory (folder) based on the presence or absence of a separator at the end.

  dirpath_without_sep = './dir/subdir' print(os.path.split(dirpath_without_sep)) # ('./dir', 'subdir') print(os.path.basename(dirpath_without_sep)) # subdir 

  If there is a separator at the end, use os.path.dirname() and os.path.basename() to get the bottom folder name.

  dirpath_with_sep = './dir/subdir/' print(os.path.split(dirpath_with_sep)) # ('./dir/subdir', '') print(os.path.basename(os.path.dirname(dirpath_with_sep))) # subdir 

  Get the extension: os.path.splitext()

  Use os.path.splitext() to get the extension.

  os.path.splitext() splits the extension and others and returns it as a tuple. The extension contains the dot . .

  filepath = './dir/subdir/filename.ext' root_ext_pair = os.path.splitext(filepath) print(root_ext_pair) # ('./dir/subdir/filename', '.ext') print(type(root_ext_pair)) # 

  You can concatenate with the + operator to return the original path string.

  root, ext = os.path.splitext(filepath) print(root) # ./dir/subdir/filename print(ext) # .ext path = root + ext print(path) # ./dir/subdir/filename.ext 

  Create a path string with a different extension

  To create a path string with only the extension changed from the original, concatenate the first element of the tuple returned by os.path.splitext() with any extension.

  filepath = './dir/subdir/filename.ext' other_ext_filepath = os.path.splitext(filepath)[0] + '.jpg' print(other_ext_filepath) # ./dir/subdir/filename.jpg 

  Get the extension without dot (period)

  If you want to get the extension without the dot (period) . , specify the second and subsequent strings with slice [1:] .

  filepath = './dir/subdir/filename.ext' ext_without_dot = os.path.splitext(filepath)[1][1:] print(ext_without_dot) # ext 

  Examples of cases like .tar.gz

  Note that os.path.splitext() splits at the last (right) dot . as demonstrated in the previous example. Be cautious with extensions like .tar.gz .

  filepath_tar_gz = './dir/subdir/filename.tar.gz' print(os.path.splitext(filepath_tar_gz)) # ('./dir/subdir/filename.tar', '.gz') 

  If you want to split by the first (left) dot . in the file name, use the split() method of the string, but it doesn’t work if the directory name also contains the dot . .

  print(filepath_tar_gz.split('.', 1)) # ['', '/dir/subdir/filename.tar.gz'] 

  After splitting with os.path.split() , apply the split() method of the string and join with os.path.join() described later.

  The string returned by split() does not contain a delimiter, so be careful if you want to get an extension with a dot . like os.path.splitext() .

  dirname, basename = os.path.split(filepath_tar_gz) basename_without_ext, ext = basename.split('.', 1) path_without_ext = os.path.join(dirname, basename_without_ext) print(path_without_ext) # ./dir/subdir/filename print(ext) # tar.gz ext_with_dot = '.' + ext print(ext_with_dot) # .tar.gz 

  Create a path string by combining the file and directory names: os.path.join()

  Use os.path.join() to join file and directory names to create a new path string.

  path = os.path.join('dir', 'subdir', 'filename.ext') print(path) # dir/subdir/filename.ext 

  Create a path string for another file in the same directory

  If you want to create a path string for another file in the same directory of one file, use os.path.dirname() and os.path.join() .

  filepath = './dir/subdir/filename.ext' other_filepath = os.path.join(os.path.dirname(filepath), 'other_file.ext') print(other_filepath) # ./dir/subdir/other_file.ext 

  Use different OS formats

  If you want to manipulate path strings in an OS format that is not the OS on which Python is currently running, import and use different modules instead of the os module.

  • UNIX (including current Mac): posixpath
  • Windows: ntpath
  • Macintosh 9 and earlier: macpath

  Since each module has the same interface as os.path , you can change the os.path part of the sample code so far to their module names (such as ntpath ).

  Examples for Windows

  The sample code below is running on Mac using the ntpath module mentioned above. When running on Windows, you can replace ntpath with os.path .

  Backslash and raw string

  The path separator in Windows is the backslash \ .

  To write a backslash in a string, you need to write two backslashes to escape. print() outputs one backslash.

  import ntpath print(ntpath.sep) # \ print('\\') # \ print(ntpath.sep is '\\') # True 

  Using a raw string ( r’xxx’ ) simplifies writing a Windows path, as it allows you to write a backslash directly. Raw strings and normal strings have equal value.

  file_path = 'c:\\dir\\subdir\\filename.ext' file_path_raw = r'c:\dir\subdir\filename.ext' print(file_path == file_path_raw) # True 

  For more information about raw strings, see the following article.

  Examples of getting file name, folder name, extension

  These examples also work on Windows.

  print(ntpath.basename(file_path)) # filename.ext print(ntpath.dirname(file_path)) # c:\dir\subdir print(ntpath.split(file_path)) # ('c:\\dir\\subdir', 'filename.ext') 

  Get and join a drive letter: os.path.splitdrive()

  Use os.path.splitdrive() to get the drive letter. The sample code below uses ntpath.splitdrive() .

  os.path.splitdrive() splits the drive letter including the colon : and others.

  print(ntpath.splitdrive(file_path)) # ('c:', '\\dir\\subdir\\filename.ext') 

  If you want to get only the drive letter, select the first character.

  drive_letter = ntpath.splitdrive(file_path)[0][0] print(drive_letter) # c 

  Be careful when joining drive characters.

  If you pass it to os.path.join() as it is, it will not work.

  print(ntpath.join('c:', 'dir', 'subdir', 'filename.ext')) # c:dir\subdir\filename.ext 

  You can also specify os.sep ( ntpath.sep in the sample code) in the argument of os.path.join() or add a separator to the drive letter.

  print(ntpath.join('c:', ntpath.sep, 'dir', 'subdir', 'filename.ext')) # c:\dir\subdir\filename.ext print(ntpath.join('c:\\', 'dir', 'subdir', 'filename.ext')) # c:\dir\subdir\filename.ext 

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  Get file path without file name

  Don’t assume that the path separator is always / if you are planning to run this on different operating systems.

  Its for an internal project where we could set a standard, but you are right, someone might input a filepath from a different OS where its \ instead of /

  4 Answers 4

  dirname, fname = os.path.split(fullpath) 

  Split the pathname path into a pair, (head, tail) where tail is the last pathname component and head is everything leading up to that. The tail part will never contain a slash; if path ends in a slash, tail will be empty. If there is no slash in path, head will be empty.

  os.path is always the module suitable for the platform that the code is running on.

  fullpath = '/path/to/some/file.jpg' import os os.path.dirname(fullpath) 

  Using pathlib you can get the path without the file name using the .parent attribute:

  from pathlib import Path fullpath = Path("/path/to/some/file.jpg") filepath = str(fullpath.parent) # /path/to/some/ 

  This handles both UNIX and Windows paths correctly.

  fullpath = '/path/to/some/file.jpg' index = fullpath.rfind('/') fullpath[0:index] 

  Based on the users given and then criteria, this should be good. But I agree this would not work in case of different OS.

  It doesn’t work correctly with my first example either. We should not suggest fail-prone methods when there is a documented, platform agnostic built-in method of doing something, such as the os.path.split method suggested by @LevLevitsky. The OP has suggested a similar method themselves but they are looking for a better method because they think it is open to errors.

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