I spent some time researching this and I just cannot work this out in my head.
I run a program in its own directory home/program/core/main.py
In main.py I try and import a module called my_module.py thats located in a different directory, say home/program/modules/my_module.py
In main.py this is how I append to sys.path so the program can be run on anyone's machine (hopefully).
import os.path
import sys
# This should give the path to home/program
sys.path.append(os.path.join(os.path.abspath(os.path.dirname(__file__), '..'))
# Which it does when checking with
print os.path.join(os.path.abspath(os.path.dirname(__file__), '..')
# So now sys.path knows the location of where modules directory is, it should work right?
import modules.my_module # <----RAISES ImportError WHY?
However if I simply do:
sys.path.append('home/program/modules')
import my_module
It all works fine. But this is not ideal as it now depends on the fact that the program must exist under home/program.
that's because modules isn't a valid python package, probably because it doesn't contain any __init__.py file (You cannot traverse directories with import without them being marked with __init__.py)
So either add an empty __init__.py file or just add the path up to modules so your first snippet is equivalent to the second one:
sys.path.append(os.path.join(os.path.abspath(os.path.dirname(__file__), '..','modules'))
import my_module
note that you can also import the module by giving the full path to it, using advanced import features: How to import a module given the full path?
Although the answer can be found here, for convenience and completeness here is a quick solution:
import importlib
dirname, basename = os.path.split(pyfilepath) # pyfilepath: /my/path/mymodule.py
sys.path.append(dirname) # only directories should be added to PYTHONPATH
module_name = os.path.splitext(basename)[0] # /my/path/mymodule.py --> mymodule
module = importlib.import_module(module_name) # name space of defined module (otherwise we would literally look for "module_name")
Now you can directly use the namespace of the imported module, like this:
a = module.myvar
b = module.myfunc(a)
I would like to get the path of a library before importing the library itself.
That is, something different from:
import module, os
library_path = os.path.dirname(module.__file__)
Is that possible?
Thank you.
I think what you need is imp module:
import imp
file_handle, module_path, module_doc = imp.find_module(module_name)
The second return value is the path to actual file (assuming there is one, since requested module could be a built-in). First parameter is a file handle, already opened for you.
For as long as your use case is simple, you shouldn't have any issues. If you'll try for a generic solution you will need to read imp module documentation carefully, as there are lots of possible situations and return values for this function.
https://docs.python.org/2/library/imp.html
import commands
print commands.__file__
/usr/lib/python2.7/commands.py
import os
print os.__file__
/usr/lib/python2.7/os.pyc
Yes you can, but all modules doesn't support __ file__
Is there a universal approach in Python, to find out the path to the file that is currently executing?
Failing approaches
path = os.path.abspath(os.path.dirname(sys.argv[0]))
This does not work if you are running from another Python script in another directory, for example by using execfile in 2.x.
path = os.path.abspath(os.path.dirname(__file__))
I found that this doesn't work in the following cases:
py2exe doesn't have a __file__ attribute, although there is a workaround
When the code is run from IDLE using execute(), in which case there is no __file__ attribute
On Mac OS X v10.6 (Snow Leopard), I get NameError: global name '__file__' is not defined
Test case
Directory tree
C:.
| a.py
\---subdir
b.py
Content of a.py
#! /usr/bin/env python
import os, sys
print "a.py: sys.argv[0]=", sys.argv[0]
print "a.py: __file__=", __file__
print "a.py: os.getcwd()=", os.getcwd()
print
execfile("subdir/b.py")
Content of subdir/b.py
#! /usr/bin/env python
import os, sys
print "b.py: sys.argv[0]=", sys.argv[0]
print "b.py: __file__=", __file__
print "b.py: os.getcwd()=", os.getcwd()
print
Output of python a.py (on Windows)
a.py: __file__= a.py
a.py: os.getcwd()= C:\zzz
b.py: sys.argv[0]= a.py
b.py: __file__= a.py
b.py: os.getcwd()= C:\zzz
Related (but these answers are incomplete)
Find path to currently running file
Path to current file depends on how I execute the program
How can I know the path of the running script in Python?
Change directory to the directory of a Python script
First, you need to import from inspect and os
from inspect import getsourcefile
from os.path import abspath
Next, wherever you want to find the source file from you just use
abspath(getsourcefile(lambda:0))
You can't directly determine the location of the main script being executed. After all, sometimes the script didn't come from a file at all. For example, it could come from the interactive interpreter or dynamically generated code stored only in memory.
However, you can reliably determine the location of a module, since modules are always loaded from a file. If you create a module with the following code and put it in the same directory as your main script, then the main script can import the module and use that to locate itself.
some_path/module_locator.py:
def we_are_frozen():
# All of the modules are built-in to the interpreter, e.g., by py2exe
return hasattr(sys, "frozen")
def module_path():
encoding = sys.getfilesystemencoding()
if we_are_frozen():
return os.path.dirname(unicode(sys.executable, encoding))
return os.path.dirname(unicode(__file__, encoding))
some_path/main.py:
import module_locator
my_path = module_locator.module_path()
If you have several main scripts in different directories, you may need more than one copy of module_locator.
Of course, if your main script is loaded by some other tool that doesn't let you import modules that are co-located with your script, then you're out of luck. In cases like that, the information you're after simply doesn't exist anywhere in your program. Your best bet would be to file a bug with the authors of the tool.
This solution is robust even in executables:
import inspect, os.path
filename = inspect.getframeinfo(inspect.currentframe()).filename
path = os.path.dirname(os.path.abspath(filename))
I was running into a similar problem, and I think this might solve the problem:
def module_path(local_function):
''' returns the module path without the use of __file__. Requires a function defined
locally in the module.
from http://stackoverflow.com/questions/729583/getting-file-path-of-imported-module'''
return os.path.abspath(inspect.getsourcefile(local_function))
It works for regular scripts and in IDLE. All I can say is try it out for others!
My typical usage:
from toolbox import module_path
def main():
pass # Do stuff
global __modpath__
__modpath__ = module_path(main)
Now I use _modpath_ instead of _file_.
You have simply called:
path = os.path.abspath(os.path.dirname(sys.argv[0]))
instead of:
path = os.path.dirname(os.path.abspath(sys.argv[0]))
abspath() gives you the absolute path of sys.argv[0] (the filename your code is in) and dirname() returns the directory path without the filename.
The short answer is that there is no guaranteed way to get the information you want, however there are heuristics that work almost always in practice. You might look at How do I find the location of the executable in C?. It discusses the problem from a C point of view, but the proposed solutions are easily transcribed into Python.
See my answer to the question Importing modules from parent folder for related information, including why my answer doesn't use the unreliable __file__ variable. This simple solution should be cross-compatible with different operating systems as the modules os and inspect come as part of Python.
First, you need to import parts of the inspect and os modules.
from inspect import getsourcefile
from os.path import abspath
Next, use the following line anywhere else it's needed in your Python code:
abspath(getsourcefile(lambda:0))
How it works:
From the built-in module os (description below), the abspath tool is imported.
OS routines for Mac, NT, or Posix depending on what system we're on.
Then getsourcefile (description below) is imported from the built-in module inspect.
Get useful information from live Python objects.
abspath(path) returns the absolute/full version of a file path
getsourcefile(lambda:0) somehow gets the internal source file of the lambda function object, so returns '<pyshell#nn>' in the Python shell or returns the file path of the Python code currently being executed.
Using abspath on the result of getsourcefile(lambda:0) should make sure that the file path generated is the full file path of the Python file.
This explained solution was originally based on code from the answer at How do I get the path of the current executed file in Python?.
This should do the trick in a cross-platform way (so long as you're not using the interpreter or something):
import os, sys
non_symbolic=os.path.realpath(sys.argv[0])
program_filepath=os.path.join(sys.path[0], os.path.basename(non_symbolic))
sys.path[0] is the directory that your calling script is in (the first place it looks for modules to be used by that script). We can take the name of the file itself off the end of sys.argv[0] (which is what I did with os.path.basename). os.path.join just sticks them together in a cross-platform way. os.path.realpath just makes sure if we get any symbolic links with different names than the script itself that we still get the real name of the script.
I don't have a Mac; so, I haven't tested this on one. Please let me know if it works, as it seems it should. I tested this in Linux (Xubuntu) with Python 3.4. Note that many solutions for this problem don't work on Macs (since I've heard that __file__ is not present on Macs).
Note that if your script is a symbolic link, it will give you the path of the file it links to (and not the path of the symbolic link).
You can use Path from the pathlib module:
from pathlib import Path
# ...
Path(__file__)
You can use call to parent to go further in the path:
Path(__file__).parent
Simply add the following:
from sys import *
path_to_current_file = sys.argv[0]
print(path_to_current_file)
Or:
from sys import *
print(sys.argv[0])
If the code is coming from a file, you can get its full name
sys._getframe().f_code.co_filename
You can also retrieve the function name as f_code.co_name
The main idea is, somebody will run your python code, but you need to get the folder nearest the python file.
My solution is:
import os
print(os.path.dirname(os.path.abspath(__file__)))
With
os.path.dirname(os.path.abspath(__file__))
You can use it with to save photos, output files, ...etc
import os
current_file_path=os.path.dirname(os.path.realpath('__file__'))
How can I get the absolute path of an imported module?
As the other answers have said, you can use __file__. However, note that this won't give the full path if the other module is in the same directory as the program. So to be safe, do something like this:
>>> import os
>>> import math
>>> os.path.abspath(math.__file__)
'/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/lib-dynload/math.so'
Here's an example with a module I made called checkIP to illustrate why you need to get the abspath (checkIP.py is in the current directory):
>>> import os
>>> import checkIP
>>> os.path.abspath(checkIP.__file__)
'/Users/Matthew/Programs/checkIP.py'
>>> checkIP.__file__
'checkIP.py'
If it is a module within your PYTHONPATH directory tree (and reachable by placing a __init__.py in its directory and parent directories), then call its path attribute.
>>>import sample_module
>>>sample_module.__path__
['/absolute/path/to/sample/module']
You can try:
import os
print os.__file__
to see where the module is located.
I have a Python module and I'd like to get that modules directory from inside itself. I want to do this because I have some files that I'd like to reference relative to the module.
First you need to get a reference to the module inside itself.
mod = sys.__modules__[__name__]
Then you can use __file__ to get to the module file.
mod.__file__
Its directory is a dirname of that.
As you are inside the module all you need is this:
import os
path_to_this_module = os.path.dirname(__file__)
However, if the module in question is actually your programs entry point, then __file__ will only be the name of the file and you'll need to expand the path:
import os
path_to_this_module = os.path.dirname(os.path.abspath(__file__))
I think this is what you are looking for:
import <module>
import os
print os.path.dirname(<module>.__file__)
You should be using pkg_resources for this, the resource* family of functions do just about everything you need without having to muck about with the filesystem.
import pkg_resources
data = pkg_resources.resource_string(__name__, "some_file")