This question already has answers here:
How do you properly determine the current script directory?
(16 answers)
How to know/change current directory in Python shell?
(7 answers)
Closed 5 years ago.
How do I determine:
the current directory (where I was in the shell when I ran the Python script), and
where the Python file I am executing is?
To get the full path to the directory a Python file is contained in, write this in that file:
import os
dir_path = os.path.dirname(os.path.realpath(__file__))
(Note that the incantation above won't work if you've already used os.chdir() to change your current working directory, since the value of the __file__ constant is relative to the current working directory and is not changed by an os.chdir() call.)
To get the current working directory use
import os
cwd = os.getcwd()
Documentation references for the modules, constants and functions used above:
The os and os.path modules.
The __file__ constant
os.path.realpath(path) (returns "the canonical path of the specified filename, eliminating any symbolic links encountered in the path")
os.path.dirname(path) (returns "the directory name of pathname path")
os.getcwd() (returns "a string representing the current working directory")
os.chdir(path) ("change the current working directory to path")
Current working directory: os.getcwd()
And the __file__ attribute can help you find out where the file you are executing is located. This Stack Overflow post explains everything: How do I get the path of the current executed file in Python?
You may find this useful as a reference:
import os
print("Path at terminal when executing this file")
print(os.getcwd() + "\n")
print("This file path, relative to os.getcwd()")
print(__file__ + "\n")
print("This file full path (following symlinks)")
full_path = os.path.realpath(__file__)
print(full_path + "\n")
print("This file directory and name")
path, filename = os.path.split(full_path)
print(path + ' --> ' + filename + "\n")
print("This file directory only")
print(os.path.dirname(full_path))
The pathlib module, introduced in Python 3.4 (PEP 428 — The pathlib module — object-oriented filesystem paths), makes the path-related experience much much better.
pwd
/home/skovorodkin/stack
tree
.
└── scripts
├── 1.py
└── 2.py
In order to get the current working directory, use Path.cwd():
from pathlib import Path
print(Path.cwd()) # /home/skovorodkin/stack
To get an absolute path to your script file, use the Path.resolve() method:
print(Path(__file__).resolve()) # /home/skovorodkin/stack/scripts/1.py
And to get the path of a directory where your script is located, access .parent (it is recommended to call .resolve() before .parent):
print(Path(__file__).resolve().parent) # /home/skovorodkin/stack/scripts
Remember that __file__ is not reliable in some situations: How do I get the path of the current executed file in Python?.
Please note, that Path.cwd(), Path.resolve() and other Path methods return path objects (PosixPath in my case), not strings. In Python 3.4 and 3.5 that caused some pain, because open built-in function could only work with string or bytes objects, and did not support Path objects, so you had to convert Path objects to strings or use the Path.open() method, but the latter option required you to change old code:
File scripts/2.py
from pathlib import Path
p = Path(__file__).resolve()
with p.open() as f: pass
with open(str(p)) as f: pass
with open(p) as f: pass
print('OK')
Output
python3.5 scripts/2.py
Traceback (most recent call last):
File "scripts/2.py", line 11, in <module>
with open(p) as f:
TypeError: invalid file: PosixPath('/home/skovorodkin/stack/scripts/2.py')
As you can see, open(p) does not work with Python 3.5.
PEP 519 — Adding a file system path protocol, implemented in Python 3.6, adds support of PathLike objects to the open function, so now you can pass Path objects to the open function directly:
python3.6 scripts/2.py
OK
To get the current directory full path
>>import os
>>print os.getcwd()
Output: "C :\Users\admin\myfolder"
To get the current directory folder name alone
>>import os
>>str1=os.getcwd()
>>str2=str1.split('\\')
>>n=len(str2)
>>print str2[n-1]
Output: "myfolder"
Pathlib can be used this way to get the directory containing the current script:
import pathlib
filepath = pathlib.Path(__file__).resolve().parent
If you are trying to find the current directory of the file you are currently in:
OS agnostic way:
dirname, filename = os.path.split(os.path.abspath(__file__))
If you're using Python 3.4, there is the brand new higher-level pathlib module which allows you to conveniently call pathlib.Path.cwd() to get a Path object representing your current working directory, along with many other new features.
More info on this new API can be found here.
To get the current directory full path:
os.path.realpath('.')
Answer to #1:
If you want the current directory, do this:
import os
os.getcwd()
If you want just any folder name and you have the path to that folder, do this:
def get_folder_name(folder):
'''
Returns the folder name, given a full folder path
'''
return folder.split(os.sep)[-1]
Answer to #2:
import os
print os.path.abspath(__file__)
I think the most succinct way to find just the name of your current execution context would be:
current_folder_path, current_folder_name = os.path.split(os.getcwd())
If you're searching for the location of the currently executed script, you can use sys.argv[0] to get the full path.
For question 1, use os.getcwd() # Get working directory and os.chdir(r'D:\Steam\steamapps\common') # Set working directory
I recommend using sys.argv[0] for question 2 because sys.argv is immutable and therefore always returns the current file (module object path) and not affected by os.chdir(). Also you can do like this:
import os
this_py_file = os.path.realpath(__file__)
# vvv Below comes your code vvv #
But that snippet and sys.argv[0] will not work or will work weird when compiled by PyInstaller, because magic properties are not set in __main__ level and sys.argv[0] is the way your executable was called (it means that it becomes affected by the working directory).
Related
I mistakenly, typed the following code:
f = open('\TestFiles\'sample.txt', 'w')
f.write('I just wrote this line')
f.close()
I ran this code and even though I have mistakenly typed the above code, it is however a valid code because the second backslash ignores the single-quote and what I should get, according to my knowledge is a .txt file named "\TestFiles'sample" in my project folder. However when I navigated to the project folder, I could not find a file there.
However, if I do the same thing with a different filename for example. Like,
f = open('sample1.txt', 'w')
f.write('test')
f.close()
I find the 'sample.txt' file created in my folder. Is there a reason for the file to not being created even though the first code was valid according to my knowledge?
Also is there a way to mention a file relative to my project folder rather than mentioning the absolute path to a file? (For example I want to create a file called 'sample.txt' in a folder called 'TestFiles' inside my project folder. So without mentioning the absolute path to TestFiles folder, is there a way to mention the path to TestFiles folder relative to the project folder in Python when opening files?)
I am a beginner in Python and I hope someone could help me.
Thank you.
What you're looking for are relative paths, long story short, if you want to create a file called 'sample.txt' in a folder 'TestFiles' inside your project folder, you can do:
import os
f = open(os.path.join('TestFiles', 'sample1.txt'), 'w')
f.write('test')
f.close()
Or using the more recent pathlib module:
from pathlib import Path
f = open(Path('TestFiles', 'sample1.txt'), 'w')
f.write('test')
f.close()
But you need to keep in mind that it depends on where you started your Python interpreter (which is probably why you're not able to find "\TestFiles'sample" in your project folder, it's created elsewhere), to make sure everything works fine, you can do something like this instead:
from pathlib import Path
sample_path = Path(Path(__file__).parent, 'TestFiles', 'sample1.txt')
with open(sample_path, "w") as f:
f.write('test')
By using a [context manager]{https://book.pythontips.com/en/latest/context_managers.html} you can avoid using f.close()
When you create a file you can specify either an absolute filename or a relative filename.
If you start the file path with '\' (on Win) or '/' it will be an absolute path. So in your first case you specified an absolute path, which is in fact:
from pathlib import Path
Path('\Testfile\'sample.txt').absolute()
WindowsPath("C:/Testfile'sample.txt")
Whenever you run some code in python, the relative paths that will be generate will be composed by your current folder, which is the folder from which you started the python interpreter, which you can check with:
import os
os.getcwd()
and the relative path that you added afterwards, so if you specify:
Path('Testfiles\sample.txt').absolute()
WindowsPath('C:/Users/user/Testfiles/sample.txt')
In general I suggest you use pathlib to handle paths. That makes it safer and cross platform. For example let's say that your scrip is under:
project
src
script.py
testfiles
and you want to store/read a file in project/testfiles. What you can do is get the path for script.py with __file__ and build the path to project/testfiles
from pathlib import Path
src_path = Path(__file__)
testfiles_path = src_path.parent / 'testfiles'
sample_fname = testfiles_path / 'sample.txt'
with sample_fname.open('w') as f:
f.write('yo')
As I am running the first code example in vscode, I'm getting a warning
Anomalous backslash in string: '\T'. String constant might be missing an r prefix.
And when I am running the file, it is also creating a file with the name \TestFiles'sample.txt. And it is being created in the same directory where the .py file is.
now, if your working tree is like this:
project_folder
-testfiles
-sample.txt
-something.py
then you can just say: open("testfiles//hello.txt")
I hope you find it helpful.
This question already has answers here:
How do you properly determine the current script directory?
(16 answers)
Closed 6 months ago.
I'm building a simple helper script for work that will copy a couple of template files in our code base to the current directory. I don't, however, have the absolute path to the directory where the templates are stored. I do have a relative path from the script but when I call the script it treats that as a path relative to the current working directory. Is there a way to specify that this relative url is from the location of the script instead?
In the file that has the script, you want to do something like this:
import os
dirname = os.path.dirname(__file__)
filename = os.path.join(dirname, 'relative/path/to/file/you/want')
This will give you the absolute path to the file you're looking for. Note that if you're using setuptools, you should probably use its package resources API instead.
UPDATE: I'm responding to a comment here so I can paste a code sample. :-)
Am I correct in thinking that __file__ is not always available (e.g. when you run the file directly rather than importing it)?
I'm assuming you mean the __main__ script when you mention running the file directly. If so, that doesn't appear to be the case on my system (python 2.5.1 on OS X 10.5.7):
#foo.py
import os
print os.getcwd()
print __file__
#in the interactive interpreter
>>> import foo
/Users/jason
foo.py
#and finally, at the shell:
~ % python foo.py
/Users/jason
foo.py
However, I do know that there are some quirks with __file__ on C extensions. For example, I can do this on my Mac:
>>> import collections #note that collections is a C extension in Python 2.5
>>> collections.__file__
'/System/Library/Frameworks/Python.framework/Versions/2.5/lib/python2.5/lib-
dynload/collections.so'
However, this raises an exception on my Windows machine.
It's 2018 now, and Python has already evolved to the __future__ long time ago. So how about using the amazing pathlib coming with Python 3.4 to accomplish the task instead of struggling with os, os.path, glob , shutil, etc.
So we have 3 paths here (possibly duplicated):
mod_path: which is the path of the simple helper script
src_path: which contains a couple of template files waiting to be copied.
cwd: current directory, the destination of those template files.
and the problem is: we don't have the full path of src_path, only know its relative path to the mod_path.
Now let's solve this with the amazing pathlib:
# Hope you don't be imprisoned by legacy Python code :)
from pathlib import Path
# `cwd`: current directory is straightforward
cwd = Path.cwd()
# `mod_path`: According to the accepted answer and combine with future power
# if we are in the `helper_script.py`
mod_path = Path(__file__).parent
# OR if we are `import helper_script`
mod_path = Path(helper_script.__file__).parent
# `src_path`: with the future power, it's just so straightforward
relative_path_1 = 'same/parent/with/helper/script/'
relative_path_2 = '../../or/any/level/up/'
src_path_1 = (mod_path / relative_path_1).resolve()
src_path_2 = (mod_path / relative_path_2).resolve()
In the future, it's just that simple.
Moreover, we can select and check and copy/move those template files with pathlib:
if src_path != cwd:
# When we have different types of files in the `src_path`
for template_path in src_path.glob('*.ini'):
fname = template_path.name
target = cwd / fname
if not target.exists():
# This is the COPY action
with target.open(mode='wb') as fd:
fd.write(template_path.read_bytes())
# If we want MOVE action, we could use:
# template_path.replace(target)
you need os.path.realpath (sample below adds the parent directory to your path)
import sys,os
sys.path.append(os.path.realpath('..'))
As mentioned in the accepted answer
import os
dir = os.path.dirname(__file__)
filename = os.path.join(dir, '/relative/path/to/file/you/want')
I just want to add that
the latter string can't begin with the backslash , infact no string
should include a backslash
It should be something like
import os
dir = os.path.dirname(__file__)
filename = os.path.join(dir, 'relative','path','to','file','you','want')
The accepted answer can be misleading in some cases , please refer to this link for details
Consider my code:
import os
def readFile(filename):
filehandle = open(filename)
print filehandle.read()
filehandle.close()
fileDir = os.path.dirname(os.path.realpath('__file__'))
print fileDir
#For accessing the file in the same folder
filename = "same.txt"
readFile(filename)
#For accessing the file in a folder contained in the current folder
filename = os.path.join(fileDir, 'Folder1.1/same.txt')
readFile(filename)
#For accessing the file in the parent folder of the current folder
filename = os.path.join(fileDir, '../same.txt')
readFile(filename)
#For accessing the file inside a sibling folder.
filename = os.path.join(fileDir, '../Folder2/same.txt')
filename = os.path.abspath(os.path.realpath(filename))
print filename
readFile(filename)
See sys.path
As initialized upon program startup, the first item of this list, path[0], is the directory containing the script that was used to invoke the Python interpreter.
Use this path as the root folder from which you apply your relative path
>>> import sys
>>> import os.path
>>> sys.path[0]
'C:\\Python25\\Lib\\idlelib'
>>> os.path.relpath(sys.path[0], "path_to_libs") # if you have python 2.6
>>> os.path.join(sys.path[0], "path_to_libs")
'C:\\Python25\\Lib\\idlelib\\path_to_libs'
Instead of using
import os
dirname = os.path.dirname(__file__)
filename = os.path.join(dirname, 'relative/path/to/file/you/want')
as in the accepted answer, it would be more robust to use:
import inspect
import os
dirname = os.path.dirname(os.path.abspath(inspect.stack()[0][1]))
filename = os.path.join(dirname, 'relative/path/to/file/you/want')
because using __file__ will return the file from which the module was loaded, if it was loaded from a file, so if the file with the script is called from elsewhere, the directory returned will not be correct.
These answers give more detail: https://stackoverflow.com/a/31867043/5542253 and https://stackoverflow.com/a/50502/5542253
From what suggest others and from pathlib documentation, a simple (but not ideal) solution is the following (suppose the file we need to refer to is Test/data/users.csv):
# Current file location: Tests/src/long/module/subdir/some_script.py
from pathlib import Path
# back to Tests/
PROJECT_ROOT = Path(__file__).parents[4]
# then down to Test/data/users.csv
CSV_USERS_PATH = PROJECT_ROOT / 'data' / 'users.csv'
with CSV_USERS_PATH.open() as users:
print(users.read())
This works but looks a bit odd because if you move some_script.py around, the path to the root of our project may change (and we would therefore need to change the parents[4] part).
I think I found a better solution that, based on the same idea.
We will use a file paths.py to store where the root of the project is, this file will remain at the same location compared to the root directory.
Tests
├── data
│ └── users.csv
└── src
├── long
│ └── module
│ └── subdir
│ └── some_script.py
├── main.py
└── paths.py
Where paths.py's only responsability is to provide PROJECT_ROOT:
from pathlib import Path
PROJECT_ROOT = Path(__file__).parents[1]
All scripts can now use paths.PROJECT_ROOT to express absolute paths from the root of the project. For example in src/long/module/subdir/some_script.py we could have:
from paths import PROJECT_ROOT
CSV_USERS_PATH = PROJECT_ROOT / 'data' / 'users.csv'
def hello():
with CSV_USERS_PATH.open() as f:
print(f.read())
And everything goes as expected:
~/Tests/src/$ python main.py
/Users/cglacet/Tests/data/users.csv
hello, user
~/Tests/$ python src/main.py
/Users/cglacet/Tests/data/users.csv
hello, user
The main.py script simply is:
from long.module.subdir import some_script
some_script.hello()
summary of the most important commands
>>> import os
>>> os.path.join('/home/user/tmp', 'subfolder')
'/home/user/tmp/subfolder'
>>> os.path.normpath('/home/user/tmp/../test/..')
'/home/user'
>>> os.path.relpath('/home/user/tmp', '/home/user')
'tmp'
>>> os.path.isabs('/home/user/tmp')
True
>>> os.path.isabs('/tmp')
True
>>> os.path.isabs('tmp')
False
>>> os.path.isabs('./../tmp')
False
>>> os.path.realpath('/home/user/tmp/../test/..') # follows symbolic links
'/home/user'
A detailed description is found in the docs.
These are linux paths. Windows should work analogous.
Hi first of all you should understand functions os.path.abspath(path) and os.path.relpath(path)
In short os.path.abspath(path) makes a relative path to absolute path. And if the path provided is itself a absolute path then the function returns the same path.
similarly os.path.relpath(path) makes a absolute path to relative path. And if the path provided is itself a relative path then the function returns the same path.
Below example can let you understand the above concept properly:
suppose i have a file input_file_list.txt which contains list of input files to be processed by my python script.
D:\conc\input1.dic
D:\conc\input2.dic
D:\Copyioconc\input_file_list.txt
If you see above folder structure, input_file_list.txt is present in Copyofconc folder and the files to be processed by the python script are present in conc folder
But the content of the file input_file_list.txt is as shown below:
..\conc\input1.dic
..\conc\input2.dic
And my python script is present in D: drive.
And the relative path provided in the input_file_list.txt file are relative to the path of input_file_list.txt file.
So when python script shall executed the current working directory (use os.getcwd() to get the path)
As my relative path is relative to input_file_list.txt, that is "D:\Copyofconc", i have to change the current working directory to "D:\Copyofconc".
So i have to use os.chdir('D:\Copyofconc'), so the current working directory shall be "D:\Copyofconc".
Now to get the files input1.dic and input2.dic, i will read the lines "..\conc\input1.dic" then shall use the command
input1_path= os.path.abspath('..\conc\input1.dic') (to change relative path to absolute path. Here as current working directory is "D:\Copyofconc", the file ".\conc\input1.dic" shall be accessed relative to "D:\Copyofconc")
so input1_path shall be "D:\conc\input1.dic"
This code will return the absolute path to the main script.
import os
def whereAmI():
return os.path.dirname(os.path.realpath(__import__("__main__").__file__))
This will work even in a module.
An alternative which works for me:
this_dir = os.path.dirname(__file__)
filename = os.path.realpath("{0}/relative/file.path".format(this_dir))
Example
Here's an example, tested in Python '3.9.5`:
your current directory: 'c:\project1\code\'
and you want to access the following folder: 'c:\project1\dataset\train\'.
Then you can access the folder using the following address: '../dataset/train/'
References
If you want some more information about path in Python, read this:
PEP - 355
PEP - 519
What worked for me is using sys.path.insert. Then I specified the directory I needed to go. For example I just needed to go up one directory.
import sys
sys.path.insert(0, '../')
I think to work with all systems use "ntpath" instead of "os.path". Today, it works well with Windows, Linux and Mac OSX.
import ntpath
import os
dirname = ntpath.dirname(__file__)
filename = os.path.join(dirname, 'relative/path/to/file/you/want')
A simple solution would be
import os
os.chdir(os.path.dirname(__file__))
From C:\Users\xyz\myFolder to C:\Users\xyz\testdata :
import os
working_dir = os.path.abspath(os.path.dirname(os.path.dirname(__file__)))
# C:\Users\xyz\myFolder
print(working_dir)
updated_working_dir = os.path.join(os.path.realpath(working_dir + '/../'), 'testdata')
# C:\Users\xyz\testdata
print(updated_working_dir)
Output
C:\Users\xyz\myFolder
C:\Users\xyz\testdata
Here is my sumup:
First, define the tool function named relpath, which convert a relative path to current file into a relative path to cwd
import os
relpath = lambda p: os.path.normpath(os.path.join(os.path.dirname(__file__), p))
Then we use it to wrap paths which is relative to current file
path1 = relpath('../src/main.py')
And you can also call sys.path.append() to import file relative to current file position
sys.path.append(relpath('..')) # so that you can import from upper dir
The full example code : https://gist.github.com/luochen1990/9b1ffa30f5c4a721dab5991e040e3eb1
Say the current archive named "Helper" and the upper directory named "Workshop", and the template files are in \Workshop\Templates, then the relative path in Python is "..\Templates".
This a simple way to add a relative path to the system path set . For example, for frequent case when the target directory is one level above (thus, '/../') the working directory:
import os
import sys
workingDir = os.getcwd()
targetDir = os.path.join(os.path.relpath(workingDir + '/../'),'target_directory')
sys.path.insert(0,targetDir)
This solution was tested for:
Python 3.9.6 | packaged by conda-forge | (default, Jul 11 2021,
03:37:25) [MSC v.1916 64 bit (AMD64)]
I'm not sure if this applies to some of the older versions, but I believe Python 3.3 has native relative path support.
For example the following code should create a text file in the same folder as the python script:
open("text_file_name.txt", "w+t")
(note that there shouldn't be a forward or backslash at the beginning if it's a relative path)
I have Directory structure like this
projectfolder/fold1/fold2/fold3/script.py
now I'm giving script.py a path as commandline argument of a file which is there in
fold1/fold_temp/myfile.txt
So basically I want to be able to give path in this way
../../fold_temp/myfile.txt
>>python somepath/pythonfile.py -input ../../fold_temp/myfile.txt
Here problem is that I might be given full path or relative path so I should be able to decide and based on that I should be able to create absolute path.
I already have knowledge of functions related to path.
Question 1
Question 2
Reference questions are giving partial answer but I don't know how to build full path using the functions provided in them.
try os.path.abspath, it should do what you want ;)
Basically it converts any given path to an absolute path you can work with, so you do not need to distinguish between relative and absolute paths, just normalize any of them with this function.
Example:
from os.path import abspath
filename = abspath('../../fold_temp/myfile.txt')
print(filename)
It will output the absolute path to your file.
EDIT:
If you are using Python 3.4 or newer you may also use the resolve() method of pathlib.Path. Be aware that this will return a Path object and not a string. If you need a string you can still use str() to convert it to a string.
Example:
from pathlib import Path
filename = Path('../../fold_temp/myfile.txt').resolve()
print(filename)
A practical example:
sys.argv[0] gives you the name of the current script
os.path.dirname() gives you the relative directory name
thus, the next line, gives you the absolute working directory of the current executing file.
cwd = os.path.abspath(os.path.dirname(sys.argv[0]))
Personally, I always use this instead of os.getcwd() since it gives me the script absolute path, independently of the directory from where the script was called.
For Python3, you can use pathlib's resolve functionality to resolve symlinks and .. components.
You need to have a Path object however it is very simple to do convert between str and Path.
I recommend for anyone using Python3 to drop os.path and its messy long function names and stick to pathlib Path objects.
import os
dir = os.path.dirname(__file__)
path = raw_input()
if os.path.isabs(path):
print "input path is absolute"
else:
path = os.path.join(dir, path)
print "absolute path is %s" % path
Use os.path.isabs to judge if input path is absolute or relative, if it is relative, then use os.path.join to convert it to absolute
How can I set the current path of my python file "myproject.py" to the file itself?
I do not want something like this:
path = "the path of myproject.py"
In mathematica I can set:
SetDirectory[NotebookDirectory[]]
The advantage with the code in Mathematica is that if I change the path of my Mathematica file, for example if I give it to someone else or I put it in another folder, I do not need to do anything extra. Each time Mathematica automatically set the directory to the current folder.
I want something similar to this in Python.
The right solution is not to change the current working directory, but to get the full path to the directory containing your script or module then use os.path.join to build your files path:
import os
ROOT_PATH = os.path.dirname(os.path.abspath(__file__))
# then:
myfile_path = os.path.join(ROOT_PATH, "myfile.txt")
This is safer than messing with current working directory (hint : what would happen if another module changes the current working directory after you did but before you access your files ?)
I want to set the directory in which the python file is, as working directory
There are two step:
Find out path to the python file
Set its parent directory as the working directory
The 2nd is simple:
import os
os.chdir(module_dir) # set working directory
The 1st might be complex if you want to support a general case (python file that is run as a script directly, python file that is imported in another module, python file that is symlinked, etc). Here's one possible solution:
import inspect
import os
module_path = inspect.getfile(inspect.currentframe())
module_dir = os.path.realpath(os.path.dirname(module_path))
Use the os.getcwd() function from the built in os module also there's os.getcwdu() which returns a unicode object of the current working directory
Example usage:
import os
path = os.getcwd()
print path
#C:\Users\KDawG\Desktop\Python
This question already has answers here:
How do you properly determine the current script directory?
(16 answers)
Closed 6 months ago.
I'm building a simple helper script for work that will copy a couple of template files in our code base to the current directory. I don't, however, have the absolute path to the directory where the templates are stored. I do have a relative path from the script but when I call the script it treats that as a path relative to the current working directory. Is there a way to specify that this relative url is from the location of the script instead?
In the file that has the script, you want to do something like this:
import os
dirname = os.path.dirname(__file__)
filename = os.path.join(dirname, 'relative/path/to/file/you/want')
This will give you the absolute path to the file you're looking for. Note that if you're using setuptools, you should probably use its package resources API instead.
UPDATE: I'm responding to a comment here so I can paste a code sample. :-)
Am I correct in thinking that __file__ is not always available (e.g. when you run the file directly rather than importing it)?
I'm assuming you mean the __main__ script when you mention running the file directly. If so, that doesn't appear to be the case on my system (python 2.5.1 on OS X 10.5.7):
#foo.py
import os
print os.getcwd()
print __file__
#in the interactive interpreter
>>> import foo
/Users/jason
foo.py
#and finally, at the shell:
~ % python foo.py
/Users/jason
foo.py
However, I do know that there are some quirks with __file__ on C extensions. For example, I can do this on my Mac:
>>> import collections #note that collections is a C extension in Python 2.5
>>> collections.__file__
'/System/Library/Frameworks/Python.framework/Versions/2.5/lib/python2.5/lib-
dynload/collections.so'
However, this raises an exception on my Windows machine.
It's 2018 now, and Python has already evolved to the __future__ long time ago. So how about using the amazing pathlib coming with Python 3.4 to accomplish the task instead of struggling with os, os.path, glob , shutil, etc.
So we have 3 paths here (possibly duplicated):
mod_path: which is the path of the simple helper script
src_path: which contains a couple of template files waiting to be copied.
cwd: current directory, the destination of those template files.
and the problem is: we don't have the full path of src_path, only know its relative path to the mod_path.
Now let's solve this with the amazing pathlib:
# Hope you don't be imprisoned by legacy Python code :)
from pathlib import Path
# `cwd`: current directory is straightforward
cwd = Path.cwd()
# `mod_path`: According to the accepted answer and combine with future power
# if we are in the `helper_script.py`
mod_path = Path(__file__).parent
# OR if we are `import helper_script`
mod_path = Path(helper_script.__file__).parent
# `src_path`: with the future power, it's just so straightforward
relative_path_1 = 'same/parent/with/helper/script/'
relative_path_2 = '../../or/any/level/up/'
src_path_1 = (mod_path / relative_path_1).resolve()
src_path_2 = (mod_path / relative_path_2).resolve()
In the future, it's just that simple.
Moreover, we can select and check and copy/move those template files with pathlib:
if src_path != cwd:
# When we have different types of files in the `src_path`
for template_path in src_path.glob('*.ini'):
fname = template_path.name
target = cwd / fname
if not target.exists():
# This is the COPY action
with target.open(mode='wb') as fd:
fd.write(template_path.read_bytes())
# If we want MOVE action, we could use:
# template_path.replace(target)
you need os.path.realpath (sample below adds the parent directory to your path)
import sys,os
sys.path.append(os.path.realpath('..'))
As mentioned in the accepted answer
import os
dir = os.path.dirname(__file__)
filename = os.path.join(dir, '/relative/path/to/file/you/want')
I just want to add that
the latter string can't begin with the backslash , infact no string
should include a backslash
It should be something like
import os
dir = os.path.dirname(__file__)
filename = os.path.join(dir, 'relative','path','to','file','you','want')
The accepted answer can be misleading in some cases , please refer to this link for details
Consider my code:
import os
def readFile(filename):
filehandle = open(filename)
print filehandle.read()
filehandle.close()
fileDir = os.path.dirname(os.path.realpath('__file__'))
print fileDir
#For accessing the file in the same folder
filename = "same.txt"
readFile(filename)
#For accessing the file in a folder contained in the current folder
filename = os.path.join(fileDir, 'Folder1.1/same.txt')
readFile(filename)
#For accessing the file in the parent folder of the current folder
filename = os.path.join(fileDir, '../same.txt')
readFile(filename)
#For accessing the file inside a sibling folder.
filename = os.path.join(fileDir, '../Folder2/same.txt')
filename = os.path.abspath(os.path.realpath(filename))
print filename
readFile(filename)
See sys.path
As initialized upon program startup, the first item of this list, path[0], is the directory containing the script that was used to invoke the Python interpreter.
Use this path as the root folder from which you apply your relative path
>>> import sys
>>> import os.path
>>> sys.path[0]
'C:\\Python25\\Lib\\idlelib'
>>> os.path.relpath(sys.path[0], "path_to_libs") # if you have python 2.6
>>> os.path.join(sys.path[0], "path_to_libs")
'C:\\Python25\\Lib\\idlelib\\path_to_libs'
Instead of using
import os
dirname = os.path.dirname(__file__)
filename = os.path.join(dirname, 'relative/path/to/file/you/want')
as in the accepted answer, it would be more robust to use:
import inspect
import os
dirname = os.path.dirname(os.path.abspath(inspect.stack()[0][1]))
filename = os.path.join(dirname, 'relative/path/to/file/you/want')
because using __file__ will return the file from which the module was loaded, if it was loaded from a file, so if the file with the script is called from elsewhere, the directory returned will not be correct.
These answers give more detail: https://stackoverflow.com/a/31867043/5542253 and https://stackoverflow.com/a/50502/5542253
From what suggest others and from pathlib documentation, a simple (but not ideal) solution is the following (suppose the file we need to refer to is Test/data/users.csv):
# Current file location: Tests/src/long/module/subdir/some_script.py
from pathlib import Path
# back to Tests/
PROJECT_ROOT = Path(__file__).parents[4]
# then down to Test/data/users.csv
CSV_USERS_PATH = PROJECT_ROOT / 'data' / 'users.csv'
with CSV_USERS_PATH.open() as users:
print(users.read())
This works but looks a bit odd because if you move some_script.py around, the path to the root of our project may change (and we would therefore need to change the parents[4] part).
I think I found a better solution that, based on the same idea.
We will use a file paths.py to store where the root of the project is, this file will remain at the same location compared to the root directory.
Tests
├── data
│ └── users.csv
└── src
├── long
│ └── module
│ └── subdir
│ └── some_script.py
├── main.py
└── paths.py
Where paths.py's only responsability is to provide PROJECT_ROOT:
from pathlib import Path
PROJECT_ROOT = Path(__file__).parents[1]
All scripts can now use paths.PROJECT_ROOT to express absolute paths from the root of the project. For example in src/long/module/subdir/some_script.py we could have:
from paths import PROJECT_ROOT
CSV_USERS_PATH = PROJECT_ROOT / 'data' / 'users.csv'
def hello():
with CSV_USERS_PATH.open() as f:
print(f.read())
And everything goes as expected:
~/Tests/src/$ python main.py
/Users/cglacet/Tests/data/users.csv
hello, user
~/Tests/$ python src/main.py
/Users/cglacet/Tests/data/users.csv
hello, user
The main.py script simply is:
from long.module.subdir import some_script
some_script.hello()
summary of the most important commands
>>> import os
>>> os.path.join('/home/user/tmp', 'subfolder')
'/home/user/tmp/subfolder'
>>> os.path.normpath('/home/user/tmp/../test/..')
'/home/user'
>>> os.path.relpath('/home/user/tmp', '/home/user')
'tmp'
>>> os.path.isabs('/home/user/tmp')
True
>>> os.path.isabs('/tmp')
True
>>> os.path.isabs('tmp')
False
>>> os.path.isabs('./../tmp')
False
>>> os.path.realpath('/home/user/tmp/../test/..') # follows symbolic links
'/home/user'
A detailed description is found in the docs.
These are linux paths. Windows should work analogous.
Hi first of all you should understand functions os.path.abspath(path) and os.path.relpath(path)
In short os.path.abspath(path) makes a relative path to absolute path. And if the path provided is itself a absolute path then the function returns the same path.
similarly os.path.relpath(path) makes a absolute path to relative path. And if the path provided is itself a relative path then the function returns the same path.
Below example can let you understand the above concept properly:
suppose i have a file input_file_list.txt which contains list of input files to be processed by my python script.
D:\conc\input1.dic
D:\conc\input2.dic
D:\Copyioconc\input_file_list.txt
If you see above folder structure, input_file_list.txt is present in Copyofconc folder and the files to be processed by the python script are present in conc folder
But the content of the file input_file_list.txt is as shown below:
..\conc\input1.dic
..\conc\input2.dic
And my python script is present in D: drive.
And the relative path provided in the input_file_list.txt file are relative to the path of input_file_list.txt file.
So when python script shall executed the current working directory (use os.getcwd() to get the path)
As my relative path is relative to input_file_list.txt, that is "D:\Copyofconc", i have to change the current working directory to "D:\Copyofconc".
So i have to use os.chdir('D:\Copyofconc'), so the current working directory shall be "D:\Copyofconc".
Now to get the files input1.dic and input2.dic, i will read the lines "..\conc\input1.dic" then shall use the command
input1_path= os.path.abspath('..\conc\input1.dic') (to change relative path to absolute path. Here as current working directory is "D:\Copyofconc", the file ".\conc\input1.dic" shall be accessed relative to "D:\Copyofconc")
so input1_path shall be "D:\conc\input1.dic"
This code will return the absolute path to the main script.
import os
def whereAmI():
return os.path.dirname(os.path.realpath(__import__("__main__").__file__))
This will work even in a module.
An alternative which works for me:
this_dir = os.path.dirname(__file__)
filename = os.path.realpath("{0}/relative/file.path".format(this_dir))
Example
Here's an example, tested in Python '3.9.5`:
your current directory: 'c:\project1\code\'
and you want to access the following folder: 'c:\project1\dataset\train\'.
Then you can access the folder using the following address: '../dataset/train/'
References
If you want some more information about path in Python, read this:
PEP - 355
PEP - 519
What worked for me is using sys.path.insert. Then I specified the directory I needed to go. For example I just needed to go up one directory.
import sys
sys.path.insert(0, '../')
I think to work with all systems use "ntpath" instead of "os.path". Today, it works well with Windows, Linux and Mac OSX.
import ntpath
import os
dirname = ntpath.dirname(__file__)
filename = os.path.join(dirname, 'relative/path/to/file/you/want')
A simple solution would be
import os
os.chdir(os.path.dirname(__file__))
From C:\Users\xyz\myFolder to C:\Users\xyz\testdata :
import os
working_dir = os.path.abspath(os.path.dirname(os.path.dirname(__file__)))
# C:\Users\xyz\myFolder
print(working_dir)
updated_working_dir = os.path.join(os.path.realpath(working_dir + '/../'), 'testdata')
# C:\Users\xyz\testdata
print(updated_working_dir)
Output
C:\Users\xyz\myFolder
C:\Users\xyz\testdata
Here is my sumup:
First, define the tool function named relpath, which convert a relative path to current file into a relative path to cwd
import os
relpath = lambda p: os.path.normpath(os.path.join(os.path.dirname(__file__), p))
Then we use it to wrap paths which is relative to current file
path1 = relpath('../src/main.py')
And you can also call sys.path.append() to import file relative to current file position
sys.path.append(relpath('..')) # so that you can import from upper dir
The full example code : https://gist.github.com/luochen1990/9b1ffa30f5c4a721dab5991e040e3eb1
Say the current archive named "Helper" and the upper directory named "Workshop", and the template files are in \Workshop\Templates, then the relative path in Python is "..\Templates".
This a simple way to add a relative path to the system path set . For example, for frequent case when the target directory is one level above (thus, '/../') the working directory:
import os
import sys
workingDir = os.getcwd()
targetDir = os.path.join(os.path.relpath(workingDir + '/../'),'target_directory')
sys.path.insert(0,targetDir)
This solution was tested for:
Python 3.9.6 | packaged by conda-forge | (default, Jul 11 2021,
03:37:25) [MSC v.1916 64 bit (AMD64)]
I'm not sure if this applies to some of the older versions, but I believe Python 3.3 has native relative path support.
For example the following code should create a text file in the same folder as the python script:
open("text_file_name.txt", "w+t")
(note that there shouldn't be a forward or backslash at the beginning if it's a relative path)