CODEWITHSUNDEEP
pythonfilepathdirectoryextract
Genspec
Genspec

Reputation: 2419

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

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?

Upvotes: 217

Views: 351464

Answers (7)

bauderr
bauderr

Reputation: 87

I use this to change the current working directory to a folder;

from os import chdir
from os.path import realpath
from os.path import dirname

chdir(realpath(dirname(argv[0])))
  • chdir changes the working directory. I doubt you'll need this.
  • realpath follows symlinks.
  • dirname returns just the path
  • argv is the command line used to execute the program

Upvotes: 0

SethMMorton
SethMMorton

Reputation: 48815

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.

Upvotes: 139

Zipper1365
Zipper1365

Reputation: 461

Anyone trying to do this in the ESRI GIS Table field calculator interface can do this with the Python parser:

PathToContainingFolder = 

"\\".join(!FullFilePathWithFileName!.split("\\")[0:-1])

so that

\Users\me\Desktop\New folder\file.txt

becomes

\Users\me\Desktop\New folder

Upvotes: 0

Sukrit Kalra
Sukrit Kalra

Reputation: 34533

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')

Upvotes: 229

Timothy C. Quinn
Timothy C. Quinn

Reputation: 4515

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

Upvotes: 2

Dan Allan
Dan Allan

Reputation: 35265

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')

Upvotes: 75

Victor Lellis
Victor Lellis

Reputation: 1324

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

Upvotes: 5

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