Related
I have a python file, converted from a Jupiter Notebook, and there is a subfolder called 'datasets' inside this file folder. When I'm trying to open a file that is inside that 'datasets' folder, with this code:
import pandas as pd
# Load the CSV data into DataFrames
super_bowls = pd.read_csv('/datasets/super_bowls.csv')
It says that there is no such file or folder. Then I add this line
os.getcwd()
And the output is the top-level folder of the project, and not the subfolder when is this python file. And I think maybe that's the reason why it's not working.
So, how can I open that csv file with relative paths? I don't want to use absolute path because this code is going to be used in another computers.
Why os.getcwd() is not getting the actual folder path?
My observation, the dot (.) notation to move to the parent directory sometimes does not work depending on the operating system. What I generally do to make it os agnostic is this:
import pandas as pd
import os
__location__ = os.path.realpath(os.path.join(os.getcwd(), os.path.dirname(__file__)))
super_bowls = pd.read_csv(__location__ + '/datasets/super_bowls.csv')
This works on my windows and ubantu machine equally well.
I am not sure if there are other and better ways to achieve this. Would like to hear back if there are.
(edited)
Per your comment below, the current working directory is
/Users/ivanparra/AprendizajePython/
while the file is in
/Users/ivanparra/AprendizajePython/Jupyter/datasets/super_bowls.csv
For that reason, going to the datasets subfolder of the current working directory (CWD) takes you to /Users/ivanparra/AprendizajePython/datasets which either doesn't exist or doesn't contain the file you're looking for.
You can do one of two things:
(1) Use an absolute path to the file, as in
super_bowls = pd.read_csv("/Users/ivanparra/AprendizajePython/Jupyter/datasets/super_bowls.csv")
(2) use the right relative path, as in
super_bowls = pd.read_csv("./Jupyter/datasets/super_bowls.csv")
There's also (3) - use os.path.join to contact the CWD to the relative path - it's basically the same as (2).
(you can also use
The answer really lies in the response by user2357112:
os.getcwd() is working fine. The problem is in your expectations. The current working directory is the directory where Python is running, not the directory of any particular source file. – user2357112 supports Monica May 22 at 6:03
The solution is:
data_dir = os.path.dirname(__file__)
Try this code
super_bowls = pd.read_csv( os.getcwd() + '/datasets/super_bowls.csv')
I noticed this problem a few years ago. I think it's a matter of design style. The problem is that: your workspace folder is just a folder, not a project folder. Most of the time, your relative reference is based on the current file.
VSCode actually supports the dynamic setting of cwd, but that's not the default. If your work folder is not a rigorous and professional project, I recommend you adding the following settings to launch.json. This is the simplest answer you need.
"cwd": "${fileDirname}"
Thanks to everyone that tried to help me. Thanks to the Roy2012 response, I got a code that works for me.
import pandas as pd
import os
currentPath = os.path.dirname(__file__)
# Load the CSV data into DataFrames
super_bowls = pd.read_csv(currentPath + '/datasets/super_bowls.csv')
The os.path.dirname gives me the path of the current file, and let me work with relative paths.
'/Users/ivanparra/AprendizajePython/Jupyter'
and with that it works like a charm!!
P.S.: As a side note, the behavior of os.getcwd() is quite different in a Jupyter Notebook than a python file. Inside the notebook, that function gives the current file path, but in a python file, gives the top folder path.
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 am trying to use shutils.py , make_archive function.
here: https://docs.python.org/2/library/shutil.html#archiving-operations
but I can't understand the difference between root_dir and base_dir.
Here's a simple code using make_archive:
#!user/bin/python
from os import path
from os import curdir
from shutil import make_archive
# Setting current Directory
current = path.realpath(curdir)
# Now Compressing
make_archive("Backup", "gztar", current)
This will create an archive named Backup.tar.gz which contains a . Directory inside it.
I don't want the . directory but the whole thing inside in the archive.
root_dir refers to base directory of output file, or working directory for your working script.
base_dir refers to content you want pack.
For example, if you have a directory tree like:
/home/apast/git/someproject
And you want to build a package for someproject folder, you can set:
root_dir="/home/apast/git"
base_dir="someproject"
If the contents of your tree is like following, for example:
/home/apast/git/someproject/test.py
/home/apast/git/someproject/model.py
The content of your package will acquire following structure:
someproject/test.py
someproject/model.py
And your package file will be stored at:
/home/apast/git/<packfile-name>
Like doc shows, by default, root_dir and base_dir are initialized for your current working directory (cwd, or curdir). But, you can use it in a more flexible way.
Let's consider following dir structure:
/home/apast/git/web/tornado.py
/home/apast/git/web/setup.py
/home/apast/git/core/service.py
/home/apast/git/mobile/gui.py
/home/apast/git/mobile/restfulapi.py
We will try two snippets to clarify examples:
1. Defining base_dir
2. Without defined base_dir
Defining base_dir, we specify which directory we will include on our file:
from shutil import make_archive
root_dir = "/home/apast/git/"
make_archive(base_name="/tmp/outputfile",
format="gztar",
root_dir=root_dir,
base_dir="web")
This code will generate a file called /tmp/outputfile.tar.gz with following struct:
web/tornado.py
web/setup.py
Running without base_dir, like following:
from shutil import make_archive
root_dir = "/home/apast/git/"
make_archive(base_name="/tmp/outputfile",
format="gztar",
root_dir=root_dir)
It will product a file containing:
web/tornado.py
web/setup.py
core/service.py
mobile/gui.py
mobile/restfulapi.py
To define specific folders, maybe it will be necessary use some other technique, like gzip lib directly.
cd in the root_dir... then tar the base_dir...
the docs makes me confuse too, read the code, that will make u clear.
It's a bit confusing if you read the documentation but if you see it visually, it can help quite a bit.
The root_dir is the directory that you will be storing the file in.
If I were storing a file in C:\Users\Elipzer\Desktop\MyFolder\, That would be my root_dir.
The base_dir is the part added onto the root_dir so if I were storing it under my MyFolder in ...\MyFolder\MySubFolder\, I would put that as the base_dir.
In many cases, there is no need to use these since you can just change the default directory to the directory that you want to store the file in and the make_archive function will just use the default directory as the root_dir and base_dir.
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)