75

I would like to be able to dynamically retrieve the current executing module or class name from within an imported module. Here is some code:

foo.py:

def f():
    print __name__

bar.py:

from foo import f

def b(): f()

This obviously does not work as __name__ is the name of the module that contains the function. What I would like to be access inside the foo module is the name of the current executing module that is using foo. So in the case above it would be bar but if any other module imported foo I would like foo to dynamically have access to the name of that module.

Edit: The inspect module looks quite promising but it is not exactly what I was looking for. What I was hoping for was some sort of global or environment-level variable that I could access that would contain the name of the current executing module. Not that I am unwilling to traverse the stack to find that information - I just thought that Python may have exposed that data already.

Edit: Here is how I am trying to use this. I have two different Django applications that both need to log errors to file. Lets say that they are called "AppOne" and "AppTwo". I also have a place to which I would like to log these files: "/home/hare/app_logs". In each application at any given point I would like to be able to import my logger module and call the log function which writes the log string to file. However what I would like to do is create a directory under app_logs that is the name of the current application ("AppOne" or "AppTwo") so that each application's log files will go in their respective logging directories.

In order to do this I thought that the best way would be for the logger module to have access to some sort of global variable that denotes the current application's name as it is responsible for knowing the location of the parent logging directory and creating the application's logging directory if it does not yet exist.

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9
  • Why would you want to break the allocation of responsibility like this? Commented Mar 2, 2009 at 16:06
  • 5
    I am using this within a logging module - I want to be able to tell which outer module is attempting to log without the caller having to pass a key of some sort. Commented Mar 2, 2009 at 16:10
  • Are you looking to find the module that started your program execution (ie: the module whose name == 'main'? Or literally the module that called foo.f(), which may be different each time foo.f() is called? Commented Mar 2, 2009 at 16:18
  • 1
    @Andrew Hare: are you saying that (a) you wrote your own logger, different from the built-in logging module and (b) your own logger doesn't have a simple configuration class or function? And you are tying to work around this? Commented Mar 2, 2009 at 16:46
  • 3
    @S.Lott - a) yes, b) yes, and no workaround at all - just learning the ropes a bit. Commented Mar 2, 2009 at 16:57

11 Answers 11

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76

From the comment -- not the question.

I am simply curious to see if what I am trying to do is possible.

The answer to "is it possible" is always "yes". Always. Unless your question involves time travel, anti-gravity or perpetual motion.

Since the answer is always "yes", your question is ill-formed. The real question is "what's a good way to have my logging module know the name of the client?" or something like that.

The answer is "Accept it as a parameter." Don't mess around with inspecting or looking for mysterious globals or other tricks.

Just follow the design pattern of logging.getLogger() and use explicitly-named loggers. A common idiom is the following

logger= logging.getLogger( __name__ )

That handles almost all log naming perfectly.

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7 Comments

I think you are right in that I will be better off adjusting my approach - thanks!
Though if I run the script instead of import it, __name__ will contain __main__ instead of the real path to the script (like modA.submodB.pack1). Is there a value that always return the module path regardless of how it's called?
@estani: if you run it as a script, its name will be __main__. There is no such thing as a module "real" name, or modA.submodB.pack1 when running it this way. Do not confuse module hierarchy with filenames.
"The answer to 'is it possible' is always 'yes'." Or solving NP problems in polynomial time. Then it's "maybe, but most people think not". ;)
Actually there is an antigravity module in python (though not sure for time travel and perpetual motion)
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68

This should work for referencing the current module:

import sys
sys.modules[__name__]
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1 Comment

Then you can use import sys; module_name = vars(sys.modules[__name__])['__package__'] to get the name of the current module, or None if not a module. Ex: If the code is executed with python foo/bar.py, module_name will be None; but with python -m foo.bar, module_name will be "foo".
26

To obtain a reference to the __main__ module when in another:

import sys
sys.modules['__main__']

To then obtain the module's file path, which includes its name:

sys.modules['__main__'].__file__  # type: str

If within the __main__ module, simply use: __file__

To obtain just the file name from the file path:

import os
os.path.basename(file_path)

To separate the file name from its extension:

file_name.split(".")[0]

To obtain the name of a class instance:

instance.__class__.__name__

To obtain the name of a class:

class.__name__
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1 Comment

using file does not work if you are inside a package if you have foo/__init__.py, file will return init.py instead of foo
23

The "currently executing module" clearly is foo, as that's what contains the function currently running - I think a better description as to what you want is the module of foo's immediate caller (which may itself be foo if you're calling a f() from a function in foo called by a function in bar. How far you want to go up depends on what you want this for.

In any case, assuming you want the immediate caller, you can obtain this by walking up the call stack. This can be accomplished by calling sys._getframe, with the aprropriate number of levels to walk.

def f():
    caller = sys._getframe(1)  # Obtain calling frame
    print "Called from module", caller.f_globals['__name__']

[Edit]: Actually, using the inspect module as suggested above is probably a cleaner way of obtaining the stack frame. The equivalent code is:

def f():
    caller = inspect.currentframe().f_back
    print "Called from module", caller.f_globals['__name__']

(sys._getframe is documented as being for internal use - the inspect module is a more reliable API)

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19

__file__ is the path of current module the call is made.

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1 Comment

Adding to this solution. In order to obtain the string filename on Windows. Forgive the lack of newlines. python module_name, ext = os.path.splitext(__file__) module_name = module_name.split('\\')[-1]
11

I think what you want to use is the inspect module, to inspect the python runtime stack. Check out this tutorial. I think it provides an almost exact example of what you want to do.

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7

Using __file__ alone gives you a relative path for the main module and an absolute path for imported modules. Being aware this we can get the module file constantly either way with a little help from our os.path tools.

For filename only use __file__.split(os.path.sep)[-1].

For complete path use os.path.abspath(__file__).

Demo:

/tmp $ cat f.py
from pprint import pprint
import os
import sys

pprint({
    'sys.modules[__name__]': sys.modules[__name__],
    '__file__': __file__,
    '__file__.split(os.path.sep)[-1]': __file__.split(os.path.sep)[-1],
    'os.path.abspath(__file__)': os.path.abspath(__file__),
})

/tmp $ cat i.py
import f

Results:

## on *Nix ##

/tmp $ python3 f.py
{'sys.modules[__name__]': <module '__main__' from 'f.py'>,
 '__file__': 'f.py',
 '__file__.split(os.path.sep)[-1]': 'f.py',
 'os.path.abspath(__file__)': '/tmp/f.py'}

/tmp $ python3 i.py
{'sys.modules[__name__]': <module 'f' from '/tmp/f.pyc'>,
 '__file__': '/tmp/f.pyc',
 '__file__.split(os.path.sep)[-1]': 'f.pyc',
 'os.path.abspath(__file__)': '/tmp/f.pyc'}

## on Windows ##

PS C:\tmp> python3.exe f.py
{'sys.modules[__name__]': <module '__main__' from 'f.py'>,
 '__file__': 'f.py',
 '__file__.split(os.path.sep)[-1]': 'f.py',
 'os.path.abspath(__file__)': 'C:\\tools\\cygwin\\tmp\\f.py'}

PS C:\tmp> python3.exe i.py
{'sys.modules[__name__]': <module 'f' from 'C:\\tools\\cygwin\\tmp\\f.py'>,
 '__file__': 'C:\\tools\\cygwin\\tmp\\f.py',
 '__file__.split(os.path.sep)[-1]': 'f.py',
 'os.path.abspath(__file__)': 'C:\\tools\\cygwin\\tmp\\f.py'}

If you want to strip the '.py' off the end, you can do that easily. (But don't forget that you may run a '.pyc' instead.)

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1 Comment

any reason to __file__.split(os.path.sep)[-1] instead of os.path.basename(__file__) ?
5

If you want only the name of the file:

file_name = __file__.split("/")[len(__file__.split("/"))-1]
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1 Comment

For a cross-platform solution, consider using os.path. Also, you can just use -1 as the index.
3

I don't believe that's possible since that's out of foo's scope. foo will only be aware of its internal scope since it may be being called by countless other modules and applications.

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2

It's been a while since I've done python, but I believe that you can get access to the globals and locals of a caller through its traceback.

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0

To get the current file module, containing folder, here is what worked for me:

 import os
 parts = os.path.splitext(__name__)
 module_name = parts[len(parts) - 2]
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