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I need to extract the name of the parent directory of a certain path. This is what it looks like:
C:\stuff\directory_i_need\subdir\file.jpg
I would like to extract directory_i_need.
import os
## first file in current dir (with full path)
file = os.path.join(os.getcwd(), os.listdir(os.getcwd())[0])
file
os.path.dirname(file) ## directory of file
os.path.dirname(os.path.dirname(file)) ## directory of directory of file
...
And you can continue doing this as many times as necessary...
Edit: from os.path, you can use either os.path.split or os.path.basename:
dir = os.path.dirname(os.path.dirname(file)) ## dir of dir of file
## once you're at the directory level you want, with the desired directory as the final path node:
dirname1 = os.path.basename(dir)
dirname2 = os.path.split(dir)[1] ## if you look at the documentation, this is exactly what os.path.basename does.
For Python 3.4+, try the pathlib module:
>>> from pathlib import Path
>>> p = Path('C:\\Program Files\\Internet Explorer\\iexplore.exe')
>>> str(p.parent)
'C:\\Program Files\\Internet Explorer'
>>> p.name
'iexplore.exe'
>>> p.suffix
'.exe'
>>> p.parts
('C:\\', 'Program Files', 'Internet Explorer', 'iexplore.exe')
>>> p.relative_to('C:\\Program Files')
WindowsPath('Internet Explorer/iexplore.exe')
>>> p.exists()
True
All you need is parent part if you use pathlib.
from pathlib import Path
p = Path(r'C:\Program Files\Internet Explorer\iexplore.exe')
print(p.parent)
Will output:
C:\Program Files\Internet Explorer
Case you need all parts (already covered in other answers) use parts:
p = Path(r'C:\Program Files\Internet Explorer\iexplore.exe')
print(p.parts)
Then you will get a list:
('C:\\', 'Program Files', 'Internet Explorer', 'iexplore.exe')
Saves tone of time.
First, see if you have splitunc() as an available function within os.path. The first item returned should be what you want... but I am on Linux and I do not have this function when I import os and try to use it.
Otherwise, one semi-ugly way that gets the job done is to use:
>>> pathname = "\\C:\\mystuff\\project\\file.py"
>>> pathname
'\\C:\\mystuff\\project\\file.py'
>>> print pathname
\C:\mystuff\project\file.py
>>> "\\".join(pathname.split('\\')[:-2])
'\\C:\\mystuff'
>>> "\\".join(pathname.split('\\')[:-1])
'\\C:\\mystuff\\project'
which shows retrieving the directory just above the file, and the directory just above that.
import os
directory = os.path.abspath('\\') # root directory
print(directory) # e.g. 'C:\'
directory = os.path.abspath('.') # current directory
print(directory) # e.g. 'C:\Users\User\Desktop'
parent_directory, directory_name = os.path.split(directory)
print(directory_name) # e.g. 'Desktop'
parent_parent_directory, parent_directory_name = os.path.split(parent_directory)
print(parent_directory_name) # e.g. 'User'
This should also do the trick.
This is what I did to extract the piece of the directory:
for path in file_list:
directories = path.rsplit('\\')
directories.reverse()
line_replace_add_directory = line_replace+directories[2]
Thank you for your help.
You have to put the entire path as a parameter to os.path.split. See The docs. It doesn't work like string split.
Given a file path
/path/to/some/file.jpg
How would I get
/path/to/some
I'm currently doing
fullpath = '/path/to/some/file.jpg'
filepath = '/'.join(fullpath.split('/')[:-1])
But I think it is open to errors
With os.path.split:
dirname, fname = os.path.split(fullpath)
Per the docs:
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.
Please try this
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.
With String rfind method.
fullpath = '/path/to/some/file.jpg'
index = fullpath.rfind('/')
fullpath[0:index]
The following code in Python gives me the current path.
import os
DIR = os.path.dirname(os.path.dirname(__file__))
How can I now use the variable DIR to go down one more directory? I don't want to change the value of DIR as it is used elsewhere.
I have tried this:
DIR + "../path/"
But it does not seems to work.
Call one more dirname:
os.path.join(os.path.dirname(DIR), 'path')
Try:
import os.path
print(os.path.abspath(os.path.join(DIR, os.pardir)))
When you join a path via '+' you have to add a 'r':
path = r'C:/home/' + r'user/dekstop'
or write double backslashes:
path = 'C://home//' + 'user//dekstop'
Anyway you should never use that!
That's the best way:
import os
path = os.path.join('C:/home/', 'user/dekstop')
Could someone tell me how to get the parent directory of a path in Python in a cross platform way. E.g.
C:\Program Files ---> C:\
and
C:\ ---> C:\
If the directory doesn't have a parent directory, it returns the directory itself. The question might seem simple but I couldn't dig it up through Google.
Python 3.4
Use the pathlib module.
from pathlib import Path
path = Path("/here/your/path/file.txt")
print(path.parent.absolute())
Old answer
Try this:
import os
print os.path.abspath(os.path.join(yourpath, os.pardir))
where yourpath is the path you want the parent for.
Using os.path.dirname:
>>> os.path.dirname(r'C:\Program Files')
'C:\\'
>>> os.path.dirname('C:\\')
'C:\\'
>>>
Caveat: os.path.dirname() gives different results depending on whether a trailing slash is included in the path. This may or may not be the semantics you want. Cf. #kender's answer using os.path.join(yourpath, os.pardir).
The Pathlib method (Python 3.4+)
from pathlib import Path
Path('C:\Program Files').parent
# Returns a Pathlib object
The traditional method
import os.path
os.path.dirname('C:\Program Files')
# Returns a string
Which method should I use?
Use the traditional method if:
You are worried about existing code generating errors if it were to use a Pathlib object. (Since Pathlib objects cannot be concatenated with strings.)
Your Python version is less than 3.4.
You need a string, and you received a string. Say for example you have a string representing a filepath, and you want to get the parent directory so you can put it in a JSON string. It would be kind of silly to convert to a Pathlib object and back again for that.
If none of the above apply, use Pathlib.
What is Pathlib?
If you don't know what Pathlib is, the Pathlib module is a terrific module that makes working with files even easier for you. Most if not all of the built in Python modules that work with files will accept both Pathlib objects and strings. I've highlighted below a couple of examples from the Pathlib documentation that showcase some of the neat things you can do with Pathlib.
Navigating inside a directory tree:
>>> p = Path('/etc')
>>> q = p / 'init.d' / 'reboot'
>>> q
PosixPath('/etc/init.d/reboot')
>>> q.resolve()
PosixPath('/etc/rc.d/init.d/halt')
Querying path properties:
>>> q.exists()
True
>>> q.is_dir()
False
import os
p = os.path.abspath('..')
C:\Program Files ---> C:\\\
C:\ ---> C:\\\
An alternate solution of #kender
import os
os.path.dirname(os.path.normpath(yourpath))
where yourpath is the path you want the parent for.
But this solution is not perfect, since it will not handle the case where yourpath is an empty string, or a dot.
This other solution will handle more nicely this corner case:
import os
os.path.normpath(os.path.join(yourpath, os.pardir))
Here the outputs for every case that can find (Input path is relative):
os.path.dirname(os.path.normpath('a/b/')) => 'a'
os.path.normpath(os.path.join('a/b/', os.pardir)) => 'a'
os.path.dirname(os.path.normpath('a/b')) => 'a'
os.path.normpath(os.path.join('a/b', os.pardir)) => 'a'
os.path.dirname(os.path.normpath('a/')) => ''
os.path.normpath(os.path.join('a/', os.pardir)) => '.'
os.path.dirname(os.path.normpath('a')) => ''
os.path.normpath(os.path.join('a', os.pardir)) => '.'
os.path.dirname(os.path.normpath('.')) => ''
os.path.normpath(os.path.join('.', os.pardir)) => '..'
os.path.dirname(os.path.normpath('')) => ''
os.path.normpath(os.path.join('', os.pardir)) => '..'
os.path.dirname(os.path.normpath('..')) => ''
os.path.normpath(os.path.join('..', os.pardir)) => '../..'
Input path is absolute (Linux path):
os.path.dirname(os.path.normpath('/a/b')) => '/a'
os.path.normpath(os.path.join('/a/b', os.pardir)) => '/a'
os.path.dirname(os.path.normpath('/a')) => '/'
os.path.normpath(os.path.join('/a', os.pardir)) => '/'
os.path.dirname(os.path.normpath('/')) => '/'
os.path.normpath(os.path.join('/', os.pardir)) => '/'
os.path.split(os.path.abspath(mydir))[0]
os.path.abspath(os.path.join(somepath, '..'))
Observe:
import posixpath
import ntpath
print ntpath.abspath(ntpath.join('C:\\', '..'))
print ntpath.abspath(ntpath.join('C:\\foo', '..'))
print posixpath.abspath(posixpath.join('/', '..'))
print posixpath.abspath(posixpath.join('/home', '..'))
import os
print"------------------------------------------------------------"
SITE_ROOT = os.path.dirname(os.path.realpath(__file__))
print("example 1: "+SITE_ROOT)
PARENT_ROOT=os.path.abspath(os.path.join(SITE_ROOT, os.pardir))
print("example 2: "+PARENT_ROOT)
GRANDPAPA_ROOT=os.path.abspath(os.path.join(PARENT_ROOT, os.pardir))
print("example 3: "+GRANDPAPA_ROOT)
print "------------------------------------------------------------"
>>> import os
>>> os.path.basename(os.path.dirname(<your_path>))
For example in Ubuntu:
>>> my_path = '/home/user/documents'
>>> os.path.basename(os.path.dirname(my_path))
# Output: 'user'
For example in Windows:
>>> my_path = 'C:\WINDOWS\system32'
>>> os.path.basename(os.path.dirname(my_path))
# Output: 'WINDOWS'
Both examples tried in Python 2.7
Suppose we have directory structure like
1]
/home/User/P/Q/R
We want to access the path of "P" from the directory R then we can access using
ROOT = os.path.abspath(os.path.join("..", os.pardir));
2]
/home/User/P/Q/R
We want to access the path of "Q" directory from the directory R then we can access using
ROOT = os.path.abspath(os.path.join(".", os.pardir));
If you want only the name of the folder that is the immediate parent of the file provided as an argument and not the absolute path to that file:
os.path.split(os.path.dirname(currentDir))[1]
i.e. with a currentDir value of /home/user/path/to/myfile/file.ext
The above command will return:
myfile
import os
dir_path = os.path.dirname(os.path.realpath(__file__))
parent_path = os.path.abspath(os.path.join(dir_path, os.pardir))
import os.path
os.path.abspath(os.pardir)
Just adding something to the Tung's answer (you need to use rstrip('/') to be more of the safer side if you're on a unix box).
>>> input1 = "../data/replies/"
>>> os.path.dirname(input1.rstrip('/'))
'../data'
>>> input1 = "../data/replies"
>>> os.path.dirname(input1.rstrip('/'))
'../data'
But, if you don't use rstrip('/'), given your input is
>>> input1 = "../data/replies/"
would output,
>>> os.path.dirname(input1)
'../data/replies'
which is probably not what you're looking at as you want both "../data/replies/" and "../data/replies" to behave the same way.
print os.path.abspath(os.path.join(os.getcwd(), os.path.pardir))
You can use this to get the parent directory of the current location of your py file.
GET Parent Directory Path and make New directory (name new_dir)
Get Parent Directory Path
os.path.abspath('..')
os.pardir
Example 1
import os
print os.makedirs(os.path.join(os.path.dirname(__file__), os.pardir, 'new_dir'))
Example 2
import os
print os.makedirs(os.path.join(os.path.dirname(__file__), os.path.abspath('..'), 'new_dir'))
os.path.abspath('D:\Dir1\Dir2\..')
>>> 'D:\Dir1'
So a .. helps
import os
def parent_filedir(n):
return parent_filedir_iter(n, os.path.dirname(__file__))
def parent_filedir_iter(n, path):
n = int(n)
if n <= 1:
return path
return parent_filedir_iter(n - 1, os.path.dirname(path))
test_dir = os.path.abspath(parent_filedir(2))
The answers given above are all perfectly fine for going up one or two directory levels, but they may get a bit cumbersome if one needs to traverse the directory tree by many levels (say, 5 or 10). This can be done concisely by joining a list of N os.pardirs in os.path.join. Example:
import os
# Create list of ".." times 5
upup = [os.pardir]*5
# Extract list as arguments of join()
go_upup = os.path.join(*upup)
# Get abspath for current file
up_dir = os.path.abspath(os.path.join(__file__, go_upup))
To find the parent of the current working directory:
import pathlib
pathlib.Path().resolve().parent
import os
def parent_directory():
# Create a relative path to the parent of the current working directory
relative_parent = os.path.join(os.getcwd(), "..") # .. means parent directory
# Return the absolute path of the parent directory
return os.path.abspath(relative_parent)
print(parent_directory())
I want to change a.txt to b.kml.
Use os.rename:
import os
os.rename('a.txt', 'b.kml')
Usage:
os.rename('from.extension.whatever','to.another.extension')
File may be inside a directory, in that case specify the path:
import os
old_file = os.path.join("directory", "a.txt")
new_file = os.path.join("directory", "b.kml")
os.rename(old_file, new_file)
As of Python 3.4 one can use the pathlib module to solve this.
If you happen to be on an older version, you can use the backported version found here
Let's assume you are not in the root path (just to add a bit of difficulty to it) you want to rename, and have to provide a full path, we can look at this:
some_path = 'a/b/c/the_file.extension'
So, you can take your path and create a Path object out of it:
from pathlib import Path
p = Path(some_path)
Just to provide some information around this object we have now, we can extract things out of it. For example, if for whatever reason we want to rename the file by modifying the filename from the_file to the_file_1, then we can get the filename part:
name_without_extension = p.stem
And still hold the extension in hand as well:
ext = p.suffix
We can perform our modification with a simple string manipulation:
Python 3.6 and greater make use of f-strings!
new_file_name = f"{name_without_extension}_1"
Otherwise:
new_file_name = "{}_{}".format(name_without_extension, 1)
And now we can perform our rename by calling the rename method on the path object we created and appending the ext to complete the proper rename structure we want:
p.rename(Path(p.parent, new_file_name + ext))
More shortly to showcase its simplicity:
Python 3.6+:
from pathlib import Path
p = Path(some_path)
p.rename(Path(p.parent, f"{p.stem}_1_{p.suffix}"))
Versions less than Python 3.6 use the string format method instead:
from pathlib import Path
p = Path(some_path)
p.rename(Path(p.parent, "{}_{}_{}".format(p.stem, 1, p.suffix))
import shutil
shutil.move('a.txt', 'b.kml')
This will work to rename or move a file.
os.rename(old, new)
This is found in the Python docs: http://docs.python.org/library/os.html
As of Python version 3.3 and later, it is generally preferred to use os.replace instead of os.rename so FileExistsError is not raised if the destination file already exists.
assert os.path.isfile('old.txt')
assert os.path.isfile('new.txt')
os.rename('old.txt', 'new.txt')
# Raises FileExistsError
os.replace('old.txt', 'new.txt')
# Does not raise exception
assert not os.path.isfile('old.txt')
assert os.path.isfile('new.txt')
See the documentation.
Use os.rename. But you have to pass full path of both files to the function. If I have a file a.txt on my desktop so I will do and also I have to give full of renamed file too.
os.rename('C:\\Users\\Desktop\\a.txt', 'C:\\Users\\Desktop\\b.kml')
One important point to note here, we should check if any files exists with the new filename.
suppose if b.kml file exists then renaming other file with the same filename leads to deletion of existing b.kml.
import os
if not os.path.exists('b.kml'):
os.rename('a.txt','b.kml')
import os
# Set the path
path = 'a\\b\\c'
# save current working directory
saved_cwd = os.getcwd()
# change your cwd to the directory which contains files
os.chdir(path)
os.rename('a.txt', 'b.klm')
# moving back to the directory you were in
os.chdir(saved_cwd)
Using the Pathlib library's Path.rename instead of os.rename:
import pathlib
original_path = pathlib.Path('a.txt')
new_path = original_path.rename('b.kml')
Here is an example using pathlib only without touching os which changes the names of all files in a directory, based on a string replace operation without using also string concatenation:
from pathlib import Path
path = Path('/talend/studio/plugins/org.talend.designer.components.bigdata_7.3.1.20200214_1052\components/tMongoDB44Connection')
for p in path.glob("tMongoDBConnection*"):
new_name = p.name.replace("tMongoDBConnection", "tMongoDB44Connection")
new_name = p.parent/new_name
p.rename(new_name)
import shutil
import os
files = os.listdir("./pics/")
for key in range(0, len(files)):
print files[key]
shutil.move("./pics/" + files[key],"./pics/img" + str(key) + ".jpeg")
This should do it. python 3+
How to change the first letter of filename in a directory:
import os
path = "/"
for file in os.listdir(path):
os.rename(path + file, path + file.lower().capitalize())
then = os.listdir(path)
print(then)
If you are Using Windows and you want to rename your 1000s of files in a folder then:
You can use the below code. (Python3)
import os
path = os.chdir(input("Enter the path of the Your Image Folder : ")) #Here put the path of your folder where your images are stored
image_name = input("Enter your Image name : ") #Here, enter the name you want your images to have
i = 0
for file in os.listdir(path):
new_file_name = image_name+"_" + str(i) + ".jpg" #here you can change the extention of your renmamed file.
os.rename(file,new_file_name)
i = i + 1
input("Renamed all Images!!")
os.chdir(r"D:\Folder1\Folder2")
os.rename(src,dst)
#src and dst should be inside Folder2
import os
import re
from pathlib import Path
for f in os.listdir(training_data_dir2):
for file in os.listdir( training_data_dir2 + '/' + f):
oldfile= Path(training_data_dir2 + '/' + f + '/' + file)
newfile = Path(training_data_dir2 + '/' + f + '/' + file[49:])
p=oldfile
p.rename(newfile)
You can use os.system to invoke terminal to accomplish the task:
os.system('mv oldfile newfile')