Get folder name of the file in Python

You are looking to use dirname. If you only want that one directory, you can use os.path.basename,

When put all together it looks like this:

os.path.basename(os.path.dirname('dir/sub_dir/other_sub_dir/file_name.txt'))

That should get you "other_sub_dir"

The following is not the ideal approach, but I originally proposed,using os.path.split, and simply get the last item. which would look like this:

os.path.split(os.path.dirname('dir/sub_dir/other_sub_dir/file_name.txt'))[-1]

You can use dirname:

os.path.dirname(path)

Return the directory name of pathname path. This is the first element of the pair returned by passing path to the function split().

And given the full path, then you can split normally to get the last portion of the path. For example, by using basename:

os.path.basename(path)

Return the base name of pathname path. This is the second element of the pair returned by passing path to the function split(). Note that the result of this function is different from the Unix basename program; where basename for '/foo/bar/' returns 'bar', the basename() function returns an empty string ('').

---

All together:

>>> import os
>>> path=os.path.dirname("C:/folder1/folder2/filename.xml")
>>> path
'C:/folder1/folder2'
>>> os.path.basename(path)
'folder2'

this is pretty old, but if you are using Python 3.4 or above use PathLib.

# using OS
import os
path=os.path.dirname("C:/folder1/folder2/filename.xml")
print(path)
print(os.path.basename(path))

# using pathlib
import pathlib
path = pathlib.PurePath("C:/folder1/folder2/filename.xml")
print(path.parent)
print(path.parent.name)