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I try to get to the directory of where the current python script is located [duplicate]

How do I get the current file's directory path?
I tried:
>>> os.path.abspath(__file__)
'C:\\python27\\test.py'
But I want:
'C:\\python27\\'

The special variable __file__ contains the path to the current file. From that we can get the directory using either pathlib or the os.path module.
Python 3
For the directory of the script being run:
import pathlib
pathlib.Path(__file__).parent.resolve()
For the current working directory:
import pathlib
pathlib.Path().resolve()
Python 2 and 3
For the directory of the script being run:
import os
os.path.dirname(os.path.abspath(__file__))
If you mean the current working directory:
import os
os.path.abspath(os.getcwd())
Note that before and after file is two underscores, not just one.
Also note that if you are running interactively or have loaded code from something other than a file (eg: a database or online resource), __file__ may not be set since there is no notion of "current file". The above answer assumes the most common scenario of running a python script that is in a file.
References
pathlib in the python documentation.
os.path - Python 2.7, os.path - Python 3
os.getcwd - Python 2.7, os.getcwd - Python 3
what does the __file__ variable mean/do?

Using Path from pathlib is the recommended way since Python 3:
from pathlib import Path
print("File Path:", Path(__file__).absolute())
print("Directory Path:", Path().absolute()) # Directory of current working directory, not __file__
Note: If using Jupyter Notebook, __file__ doesn't return expected value, so Path().absolute() has to be used.

In Python 3.x I do:
from pathlib import Path
path = Path(__file__).parent.absolute()
Explanation:
Path(__file__) is the path to the current file.
.parent gives you the directory the file is in.
.absolute() gives you the full absolute path to it.
Using pathlib is the modern way to work with paths. If you need it as a string later for some reason, just do str(path).

Try this:
import os
dir_path = os.path.dirname(os.path.realpath(__file__))

import os
print(os.path.dirname(__file__))

I found the following commands return the full path of the parent directory of a Python 3 script.
Python 3 Script:
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
from pathlib import Path
#Get the absolute path of a Python3.6 and above script.
dir1 = Path().resolve() #Make the path absolute, resolving any symlinks.
dir2 = Path().absolute() #See #RonKalian answer
dir3 = Path(__file__).parent.absolute() #See #Arminius answer
dir4 = Path(__file__).parent
print(f'dir1={dir1}\ndir2={dir2}\ndir3={dir3}\ndir4={dir4}')
REMARKS !!!!
dir1 and dir2 works only when running a script located in the current working directory, but will break in any other case.
Given that Path(__file__).is_absolute() is True, the use of the .absolute() method in dir3 appears redundant.
The shortest command that works is dir4.
Explanation links: .resolve(), .absolute(), Path(file).parent().absolute()

USEFUL PATH PROPERTIES IN PYTHON:
from pathlib import Path
#Returns the path of the current directory
mypath = Path().absolute()
print('Absolute path : {}'.format(mypath))
#if you want to go to any other file inside the subdirectories of the directory path got from above method
filePath = mypath/'data'/'fuel_econ.csv'
print('File path : {}'.format(filePath))
#To check if file present in that directory or Not
isfileExist = filePath.exists()
print('isfileExist : {}'.format(isfileExist))
#To check if the path is a directory or a File
isadirectory = filePath.is_dir()
print('isadirectory : {}'.format(isadirectory))
#To get the extension of the file
fileExtension = mypath/'data'/'fuel_econ.csv'
print('File extension : {}'.format(filePath.suffix))
OUTPUT:
ABSOLUTE PATH IS THE PATH WHERE YOUR PYTHON FILE IS PLACED
Absolute path : D:\Study\Machine Learning\Jupitor Notebook\JupytorNotebookTest2\Udacity_Scripts\Matplotlib and seaborn Part2
File path : D:\Study\Machine Learning\Jupitor Notebook\JupytorNotebookTest2\Udacity_Scripts\Matplotlib and seaborn Part2\data\fuel_econ.csv
isfileExist : True
isadirectory : False
File extension : .csv

works also if __file__ is not available (jupyter notebooks)
import sys
from pathlib import Path
path_file = Path(sys.path[0])
print(path_file)
Also uses pathlib, which is the object oriented way of handling paths in python 3.

IPython has a magic command %pwd to get the present working directory. It can be used in following way:
from IPython.terminal.embed import InteractiveShellEmbed
ip_shell = InteractiveShellEmbed()
present_working_directory = ip_shell.magic("%pwd")
On IPython Jupyter Notebook %pwd can be used directly as following:
present_working_directory = %pwd

I have made a function to use when running python under IIS in CGI in order to get the current folder:
import os
def getLocalFolder():
path=str(os.path.dirname(os.path.abspath(__file__))).split(os.sep)
return path[len(path)-1]

Python 2 and 3
You can simply also do:
from os import sep
print(__file__.rsplit(sep, 1)[0] + sep)
Which outputs something like:
C:\my_folder\sub_folder\

This can be done without a module.
def get_path():
return (__file__.replace(f"<your script name>.py", ""))
print(get_path())

Relative path not pointing to desired directory

I have the following code:
import datetime as date
import os
import pdfkit
import getpass #Gets me current username
username = getpass.getuser()
path = f"/home/{username}/Data"
relative_path = os.path.relpath(path, os.getcwd())
destination = os.path.join(relative_path, 'data.pdf')
pdfkit.from_url('www.google.com', f'{destination}/data.pdf')
I want the pdf to be saved in windows equivalent of /home/[username]/datafolder. I don't really need to use use linux or mac but for academic reasons i have decided to use the relative path method.
This code makes sense to me but for some reason it is not the directory i want it to be because when i specify the path this way the pdf generator, generates an error.
Error: Unable to write to destination
Exit with code 1, due to unknown error.
I know the error is in the last line of code where i have specified '/relative_path/data.pdf'. Could you please advise how i can resolve this issue?
Update 1:
As suggested by #Matthias I have updated the code but I am still getting the same error
Update 2:
I tried:
from pathlib import Path
destination = Path.home()
try:
os.mkdir(destination\Data)
except OSError as error:
print(error)
But it is still not pointing to the directory Data
Update 3
I know i am getting closer:
import pdfkit
import datetime as date
import calendar
import os.path
import getpass
username = getpass.getuser()
path = f"/home/{username}/Data"
os.makedirs(relative_path, exist_ok=True)
#start = os.getcwd()
relative_path = os.path.relpath(path, os.getcwd())
destination = os.path.join(relative_path, 'data.pdf')
pdfkit.from_url('www.google.com', f'{relative_path}/data.pdf')
At this point the code is executes but the folder Data was not created not am i able to locate data.pdf. I did get sucessful run though:
Loading pages (1/6)
Counting pages (2/6)
Resolving links (4/6)
Loading headers and footers (5/6)
Printing pages (6/6)
Done
Any ideas on how i can get this working correctly? The code does not produce the folder or the file?

Just check by putting
relative_path line before os.makedirs
As below
import pdfkit
import datetime as date
import calendar
import os.path
import getpass
username = getpass.getuser()
#path = os.path.join("home","{username}","Data")
# in case of window you will need to add drive "c:" or "d:" before os.path.sep
path = os.path.join(,"home",username,"Data")
relative_path = os.path.relpath(path, os.getcwd())
os.makedirs(relative_path, exist_ok=True)
#start = os.getcwd()
destination = os.path.join(relative_path, 'data.pdf')
pdfkit.from_url('www.google.com', f'{relative_path}/data.pdf')

Maybe you could change your last line to:
pdfkit.from_url('www.google.com', f'{relative_path}/data.pdf')
in order to get it to save to the home directory.

Perhaps the issue is that the directory doesn't exist. You could use os.makedirs to create the directory, using the exist_ok=True flag in case the directory already exists. Like so:
import datetime as date
import os
import pdfkit
import getpass #Gets me current username
username = getpass.getuser()
path = f"/home/{username}/Data"
os.makedirs(path, exist_ok=True)
pdfkit.from_url('www.google.com', f'{path}/data.pdf')

You can use os.environ. Run this little script on your machine:
import os
for key, value in os.environ.items():
print(key, '-->', value)
and see for yourself what you need exactly. It's portable as well.
Let's say you want to get the path of the user's home directory. You could get it from os.environ['HOME'] and then create the path to the target directory using os.path.join(os.environ['HOME'], 'target_directory_name').
You won't be able to create files in a directory if you don't have the required permissions, though.

User folders in windows are stored in "/Users/{username}/*". I don't know if you are trying to make this compatible for multiple OSs but if you just want to make this run on windows try:
path = f"/Users/{username}/Data"
start = f"/Users/{username}"
Hope it works.:)
Edit:
To get the home directory of a user regardless of OS you could use
from pathlib import Path
home = str(Path.home())
sorry for the late edit.

Python 2.7.10 Prompt for user input

I'm working on an online tutorial for Python, & I'm trying to go a little farther for an example problem than it calls for.
The objective is to rename all the files in a folder. My addition is to prompt the user for the folder, rather than hardcoding it.
I've tried the suggestions in Python: user input and commandline arguments, but when I run the script no prompt text is displayed.
As it stands my script looks like this:
import os
import sys
import optparse
def RName_Files():
#Get the folder to user
Fol = raw_input("Please enter the folder whose files should have numbers stripped from their name: ") #I've never run past this point
#Iterate through the files in the folder
for f in ListDir(f):
print("Current file is '" + f)
I imagine I'm misunderstanding the answers in the question I linked to, and was hoping someone could clarify the responses for me. Especially since that thread mixes 2.7 and 3.x.
Thanks!

f is undefined when you loop through it. Did you mean ListDir(Fol)? And also ListDir is undefined too.
But above all you are not calling the RName_Files function in your program, try addding RName_Files() at the end of the script.
What could work
import os
ListDir = os.listdir
def RName_Files():
#Get the folder to user
Fol = raw_input("Please enter the folder whose files should have numbers stripped from their name: ")
#Iterate through the files in the folder
for f in ListDir(Fol):
print("Current file is '" + f)
if __name__ == "__main__":
RName_Files()
You should also follow the PEP8 naming conventions for variables and function names. In python variables and functions are snake_case, while class names are CamelCase. And you can also be more clear with your names, rename_files instead of RName_Files, folder or path instead of Fol, file_name instead of f.
Which will look like this:
from os import listdir
def rename_files():
#Get the folder to user
path = raw_input("Please enter the folder whose files should have numbers stripped from their name: ")
#Iterate through the files in the folder
for file_name in listdir(path):
print("Current file is " + file_name )
# do something with file_name
if __name__ == "__main__":
rename_files()

You need to call your method
import os
import sys
import optparse
def RName_Files():
#Get the folder to user
fol = raw_input("Please enter the folder whose files should have numbers stripped from their name: ") #I've never run past this point
#Iterate through the files in the folder
for f in os.listdir(fol):
print("Current file is '" + f)
RName_Files()

The "pythonic" way to do this would be like this:
import os
import sys
import optparse
def RName_Files():
#Get the folder to user
Fol = raw_input("Please enter the folder whose files should have numbers stripped from their name: ") #I've never run past this point
#Iterate through the files in the folder
for f in ListDir(Fol):
print("Current file is '" + f)
def main():
RName_Files()
if __name__ == "__main__":
main()
Basically you defined your function using def but never actually called it. Its good practice to have a main function to call the ones you've created and to call them in this fashion.

Change File extensions for entire folder

I am trying to create a script where all the file extensions are changed from .bmp to .jpg. There have been some helpful answers I have found but I still can't get my script to work. If anyone had any suggestions on what I might need to change, it would be greatly appreciated. The script doesn't come up with any errors but none of the file extensions in the folder change.
import os, glob, sys
folder = '\\stsosage\...\LegalCardsTest'
for filename in glob.iglob(os.path.join(folder, '*.bmp')):
os.rename(filename, filename[:-4] + '.jpg')

try this:
import os
def walk_dir(targetdir,topdown=True):
for root, dirs, files in os.walk(targetdir, topdown):
for name in files:
os.rename(targetdir+name, targetdir+name.replace(".bmp",".jpg"))
walk_dir(FOLDER_PATH_END_WITH_\)

My Code:
import os
os.chdir(input("Enter dir: "))
endv = input("File extension before: ")
endn = input("File extension after: ")
for s in os.listdir(pfad):
if "."+endv in s:
try:os.rename(s,s.replace("."+endv,"."+endn)
except BaseException e:print(e)
Hope it works fine.

filename is only the file name and not the entire path that you need to provide to you mv function.
Try to concat folder with filename in your mv command.

Rename multiple files in a directory in Python

I'm trying to rename some files in a directory using Python.
Say I have a file called CHEESE_CHEESE_TYPE.*** and want to remove CHEESE_ so my resulting filename would be CHEESE_TYPE
I'm trying to use the os.path.split but it's not working properly. I have also considered using string manipulations, but have not been successful with that either.

Use os.rename(src, dst) to rename or move a file or a directory.
$ ls
cheese_cheese_type.bar cheese_cheese_type.foo
$ python
>>> import os
>>> for filename in os.listdir("."):
... if filename.startswith("cheese_"):
... os.rename(filename, filename[7:])
...
>>>
$ ls
cheese_type.bar cheese_type.foo

Here's a script based on your newest comment.
#!/usr/bin/env python
from os import rename, listdir
badprefix = "cheese_"
fnames = listdir('.')
for fname in fnames:
if fname.startswith(badprefix*2):
rename(fname, fname.replace(badprefix, '', 1))

The following code should work. It takes every filename in the current directory, if the filename contains the pattern CHEESE_CHEESE_ then it is renamed. If not nothing is done to the filename.
import os
for fileName in os.listdir("."):
os.rename(fileName, fileName.replace("CHEESE_CHEESE_", "CHEESE_"))

Assuming you are already in the directory, and that the "first 8 characters" from your comment hold true always. (Although "CHEESE_" is 7 characters... ? If so, change the 8 below to 7)
from glob import glob
from os import rename
for fname in glob('*.prj'):
rename(fname, fname[8:])

I have the same issue, where I want to replace the white space in any pdf file to a dash -.
But the files were in multiple sub-directories. So, I had to use os.walk().
In your case for multiple sub-directories, it could be something like this:
import os
for dpath, dnames, fnames in os.walk('/path/to/directory'):
for f in fnames:
os.chdir(dpath)
if f.startswith('cheese_'):
os.rename(f, f.replace('cheese_', ''))

Try this:
import os
import shutil
for file in os.listdir(dirpath):
newfile = os.path.join(dirpath, file.split("_",1)[1])
shutil.move(os.path.join(dirpath,file),newfile)
I'm assuming you don't want to remove the file extension, but you can just do the same split with periods.

This sort of stuff is perfectly fitted for IPython, which has shell integration.
In [1] files = !ls
In [2] for f in files:
newname = process_filename(f)
mv $f $newname
Note: to store this in a script, use the .ipy extension, and prefix all shell commands with !.
See also: http://ipython.org/ipython-doc/stable/interactive/shell.html

Here is a more general solution:
This code can be used to remove any particular character or set of characters recursively from all filenames within a directory and replace them with any other character, set of characters or no character.
import os
paths = (os.path.join(root, filename)
for root, _, filenames in os.walk('C:\FolderName')
for filename in filenames)
for path in paths:
# the '#' in the example below will be replaced by the '-' in the filenames in the directory
newname = path.replace('#', '-')
if newname != path:
os.rename(path, newname)

It seems that your problem is more in determining the new file name rather than the rename itself (for which you could use the os.rename method).
It is not clear from your question what the pattern is that you want to be renaming. There is nothing wrong with string manipulation. A regular expression may be what you need here.

import os
import string
def rename_files():
#List all files in the directory
file_list = os.listdir("/Users/tedfuller/Desktop/prank/")
print(file_list)
#Change current working directory and print out it's location
working_location = os.chdir("/Users/tedfuller/Desktop/prank/")
working_location = os.getcwd()
print(working_location)
#Rename all the files in that directory
for file_name in file_list:
os.rename(file_name, file_name.translate(str.maketrans("","",string.digits)))
rename_files()

This command will remove the initial "CHEESE_" string from all the files in the current directory, using renamer:
$ renamer --find "/^CHEESE_/" *

I was originally looking for some GUI which would allow renaming using regular expressions and which had a preview of the result before applying changes.
On Linux I have successfully used krename, on Windows Total Commander does renaming with regexes, but I found no decent free equivalent for OSX, so I ended up writing a python script which works recursively and by default only prints the new file names without making any changes. Add the '-w' switch to actually modify the file names.
#!/usr/bin/python
# -*- coding: utf-8 -*-
import os
import fnmatch
import sys
import shutil
import re
def usage():
print """
Usage:
%s <work_dir> <search_regex> <replace_regex> [-w|--write]
By default no changes are made, add '-w' or '--write' as last arg to actually rename files
after you have previewed the result.
""" % (os.path.basename(sys.argv[0]))
def rename_files(directory, search_pattern, replace_pattern, write_changes=False):
pattern_old = re.compile(search_pattern)
for path, dirs, files in os.walk(os.path.abspath(directory)):
for filename in fnmatch.filter(files, "*.*"):
if pattern_old.findall(filename):
new_name = pattern_old.sub(replace_pattern, filename)
filepath_old = os.path.join(path, filename)
filepath_new = os.path.join(path, new_name)
if not filepath_new:
print 'Replacement regex {} returns empty value! Skipping'.format(replace_pattern)
continue
print new_name
if write_changes:
shutil.move(filepath_old, filepath_new)
else:
print 'Name [{}] does not match search regex [{}]'.format(filename, search_pattern)
if __name__ == '__main__':
if len(sys.argv) < 4:
usage()
sys.exit(-1)
work_dir = sys.argv[1]
search_regex = sys.argv[2]
replace_regex = sys.argv[3]
write_changes = (len(sys.argv) > 4) and sys.argv[4].lower() in ['--write', '-w']
rename_files(work_dir, search_regex, replace_regex, write_changes)
Example use case
I want to flip parts of a file name in the following manner, i.e. move the bit m7-08 to the beginning of the file name:
# Before:
Summary-building-mobile-apps-ionic-framework-angularjs-m7-08.mp4
# After:
m7-08_Summary-building-mobile-apps-ionic-framework-angularjs.mp4
This will perform a dry run, and print the new file names without actually renaming any files:
rename_files_regex.py . "([^\.]+?)-(m\\d+-\\d+)" "\\2_\\1"
This will do the actual renaming (you can use either -w or --write):
rename_files_regex.py . "([^\.]+?)-(m\\d+-\\d+)" "\\2_\\1" --write

You can use os.system function for simplicity and to invoke bash to accomplish the task:
import os
os.system('mv old_filename new_filename')

This works for me.
import os
for afile in os.listdir('.'):
filename, file_extension = os.path.splitext(afile)
if not file_extension == '.xyz':
os.rename(afile, filename + '.abc')

What about this :
import re
p = re.compile(r'_')
p.split(filename, 1) #where filename is CHEESE_CHEESE_TYPE.***