How can I extract the folder path from file path in Python?

Question

I would like to get just the folder path from the full path to a file.

For example T:\Data\DBDesign\DBDesign_93_v141b.mdb and I would like to get just T:\Data\DBDesign (excluding the \DBDesign_93_v141b.mdb).

I have tried something like this:

existGDBPath = r'T:\Data\DBDesign\DBDesign_93_v141b.mdb'
wkspFldr = str(existGDBPath.split('\\')[0:-1])
print wkspFldr 

but it gave me a result like this:

['T:', 'Data', 'DBDesign']

which is not the result that I require (being T:\Data\DBDesign).

Any ideas on how I can get the path to my file?

Solution 1

You were almost there with your use of the split function. You just needed to join the strings, like follows.

>>> import os
>>> '\\'.join(existGDBPath.split('\\')[0:-1])
'T:\\Data\\DBDesign'

Although, I would recommend using the os.path.dirname function to do this, you just need to pass the string, and it'll do the work for you. Since, you seem to be on windows, consider using the abspath function too. An example:

>>> import os
>>> os.path.dirname(os.path.abspath(existGDBPath))
'T:\\Data\\DBDesign'

If you want both the file name and the directory path after being split, you can use the os.path.split function which returns a tuple, as follows.

>>> import os
>>> os.path.split(os.path.abspath(existGDBPath))
('T:\\Data\\DBDesign', 'DBDesign_93_v141b.mdb')

Solution 2

WITH PATHLIB MODULE (UPDATED ANSWER)

One should consider using pathlib for new development. It is in the stdlib for Python3.4, but available on PyPI for earlier versions. This library provides a more object-orented method to manipulate paths <opinion> and is much easier read and program with </opinion>.

>>> import pathlib
>>> existGDBPath = pathlib.Path(r'T:\Data\DBDesign\DBDesign_93_v141b.mdb')
>>> wkspFldr = existGDBPath.parent
>>> print wkspFldr
Path('T:\Data\DBDesign')

WITH OS MODULE

Use the os.path module:

>>> import os
>>> existGDBPath = r'T:\Data\DBDesign\DBDesign_93_v141b.mdb'
>>> wkspFldr = os.path.dirname(existGDBPath)
>>> print wkspFldr 
'T:\Data\DBDesign'

You can go ahead and assume that if you need to do some sort of filename manipulation it's already been implemented in os.path. If not, you'll still probably need to use this module as the building block.

Solution 3

The built-in submodule os.path has a function for that very task.

import os
os.path.dirname('T:\Data\DBDesign\DBDesign_93_v141b.mdb')

Solution 4

Here is the code:

import os
existGDBPath = r'T:\Data\DBDesign\DBDesign_93_v141b.mdb'
wkspFldr = os.path.dirname(existGDBPath)
print wkspFldr # T:\Data\DBDesign

Solution 5

Here is my little utility helper for splitting paths int file, path tokens:

import os    
# usage: file, path = splitPath(s)
def splitPath(s):
    f = os.path.basename(s)
    p = s[:-(len(f))-1]
    return f, p

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Comments

[Dave Babbitt]

os.sep.join(existGDBPath.split(os.sep)[:-1] looks prettier.

[iedmrc]

Path().parent what I was looking for!

[Sébastien Dawans]

'\\'.join() is not portable, use os.path.join() instead

[Petr Vepřek]

Using str methods split and join is not portable (think / on *nix). os.path has all methods you need.

[Vexen Crabtree]

Marked-up simply because you stuck <opinion /> in tags!