The answer is
import a_module
print(a_module.__file__)
Will actually give you the path to the .pyc file that was loaded, at least on Mac OS X. So I guess you can do:
import os
path = os.path.abspath(a_module.__file__)
You can also try:
path = os.path.dirname(a_module.__file__)
To get the module's directory.
If you wish to do this dynamically in a "program" try this code:
My point is, you may not know the exact name of the module to "hardcode" it.
It may be selected from a list or may not be currently running to use __file__.
(I know, it will not work in Python 3)
global modpath
modname = 'os' #This can be any module name on the fly
#Create a file called "modname.py"
f=open("modname.py","w")
f.write("import "+modname+"\n")
f.write("modpath = "+modname+"\n")
f.close()
#Call the file with execfile()
execfile('modname.py')
print modpath
<module 'os' from 'C:\Python27\lib\os.pyc'>
I tried to get rid of the "global" issue but found cases where it did not work I think "execfile()" can be emulated in Python 3 Since this is in a program, it can easily be put in a method or module for reuse.
This was trivial.
Each module has a __file__ variable that shows its relative path from where you are right now.
Therefore, getting a directory for the module to notify it is simple as:
os.path.dirname(__file__)
There is inspect module in python.
Official documentation
The inspect module provides several useful functions to help get information about live objects such as modules, classes, methods, functions, tracebacks, frame objects, and code objects. For example, it can help you examine the contents of a class, retrieve the source code of a method, extract and format the argument list for a function, or get all the information you need to display a detailed traceback.
Example:
>>> import os
>>> import inspect
>>> inspect.getfile(os)
'/usr/lib64/python2.7/os.pyc'
>>> inspect.getfile(inspect)
'/usr/lib64/python2.7/inspect.pyc'
>>> os.path.dirname(inspect.getfile(inspect))
'/usr/lib64/python2.7'
So I spent a fair amount of time trying to do this with py2exe The problem was to get the base folder of the script whether it was being run as a python script or as a py2exe executable. Also to have it work whether it was being run from the current folder, another folder or (this was the hardest) from the system's path.
Eventually I used this approach, using sys.frozen as an indicator of running in py2exe:
import os,sys
if hasattr(sys,'frozen'): # only when running in py2exe this exists
base = sys.prefix
else: # otherwise this is a regular python script
base = os.path.dirname(os.path.realpath(__file__))
I don't get why no one is talking about this, but to me the simplest solution is using imp.find_module("modulename") (documentation here):
import imp
imp.find_module("os")
It gives a tuple with the path in second position:
(<open file '/usr/lib/python2.7/os.py', mode 'U' at 0x7f44528d7540>,
'/usr/lib/python2.7/os.py',
('.py', 'U', 1))
The advantage of this method over the "inspect" one is that you don't need to import the module to make it work, and you can use a string in input. Useful when checking modules called in another script for example.
EDIT:
In python3, importlib module should do:
Doc of importlib.util.find_spec:
Return the spec for the specified module.
First, sys.modules is checked to see if the module was already imported. If so, then sys.modules[name].spec is returned. If that happens to be set to None, then ValueError is raised. If the module is not in sys.modules, then sys.meta_path is searched for a suitable spec with the value of 'path' given to the finders. None is returned if no spec could be found.
If the name is for submodule (contains a dot), the parent module is automatically imported.
The name and package arguments work the same as importlib.import_module(). In other words, relative module names (with leading dots) work.
import os
path = os.path.abspath(__file__)
dir_path = os.path.dirname(path)
I don't get why no one is talking about this, but to me the simplest solution is using imp.find_module("modulename") (documentation here):
import imp
imp.find_module("os")
It gives a tuple with the path in second position:
(<open file '/usr/lib/python2.7/os.py', mode 'U' at 0x7f44528d7540>,
'/usr/lib/python2.7/os.py',
('.py', 'U', 1))
The advantage of this method over the "inspect" one is that you don't need to import the module to make it work, and you can use a string in input. Useful when checking modules called in another script for example.
EDIT:
In python3, importlib module should do:
Doc of importlib.util.find_spec:
Return the spec for the specified module.
First, sys.modules is checked to see if the module was already imported. If so, then sys.modules[name].spec is returned. If that happens to be set to None, then ValueError is raised. If the module is not in sys.modules, then sys.meta_path is searched for a suitable spec with the value of 'path' given to the finders. None is returned if no spec could be found.
If the name is for submodule (contains a dot), the parent module is automatically imported.
The name and package arguments work the same as importlib.import_module(). In other words, relative module names (with leading dots) work.
import os
path = os.path.abspath(__file__)
dir_path = os.path.dirname(path)
I'd like to contribute with one common scenario (in Python 3) and explore a few approaches to it.
The built-in function open() accepts either relative or absolute path as its first argument. The relative path is treated as relative to the current working directory though so it is recommended to pass the absolute path to the file.
Simply said, if you run a script file with the following code, it is not guaranteed that the example.txt file will be created in the same directory where the script file is located:
with open('example.txt', 'w'):
pass
To fix this code we need to get the path to the script and make it absolute. To ensure the path to be absolute we simply use the os.path.realpath() function. To get the path to the script there are several common functions that return various path results:
os.getcwd()os.path.realpath('example.txt')sys.argv[0]__file__
Both functions os.getcwd() and os.path.realpath() return path results based on the current working directory. Generally not what we want. The first element of the sys.argv list is the path of the root script (the script you run) regardless of whether you call the list in the root script itself or in any of its modules. It might come handy in some situations. The __file__ variable contains path of the module from which it has been called.
The following code correctly creates a file example.txt in the same directory where the script is located:
filedir = os.path.dirname(os.path.realpath(__file__))
filepath = os.path.join(filedir, 'example.txt')
with open(filepath, 'w'):
pass
If you want to retrieve the package's root path from any of its modules, the following works (tested on Python 3.6):
from . import __path__ as ROOT_PATH
print(ROOT_PATH)
The main __init__.py path can also be referenced by using __file__ instead.
Hope this helps!
If you would like to know absolute path from your script you can use Path object:
from pathlib import Path
print(Path().absolute())
print(Path().resolve('.'))
print(Path().cwd())
Return a new path object representing the current directory (as returned by os.getcwd())
Make the path absolute, resolving any symlinks. A new path object is returned:
Here is a quick bash script in case it's useful to anyone. I just want to be able to set an environment variable so that I can pushd to the code.
#!/bin/bash
module=${1:?"I need a module name"}
python << EOI
import $module
import os
print os.path.dirname($module.__file__)
EOI
Shell example:
[root@sri-4625-0004 ~]# export LXML=$(get_python_path.sh lxml)
[root@sri-4625-0004 ~]# echo $LXML
/usr/lib64/python2.7/site-packages/lxml
[root@sri-4625-0004 ~]#
you can just import your module then hit its name and you'll get its full path
>>> import os
>>> os
<module 'os' from 'C:\\Users\\Hassan Ashraf\\AppData\\Local\\Programs\\Python\\Python36-32\\lib\\os.py'>
>>>
If you installed it using pip, "pip show" works great ('Location')
$ pip show detectron2
Name: detectron2
Version: 0.1
Summary: Detectron2 is FAIR next-generation research platform for object detection and segmentation.
Home-page: https://github.com/facebookresearch/detectron2
Author: FAIR
Author-email: None
License: UNKNOWN
Location: /home/ubuntu/anaconda3/envs/pytorch_p36/lib/python3.6/site-packages
Requires: yacs, tabulate, tqdm, pydot, tensorboard, Pillow, termcolor, future, cloudpickle, matplotlib, fvcore
I will try tackling a few variations on this question as well:
- finding the path of the called script
- finding the path of the currently executing script
- finding the directory of the called script
(Some of these questions have been asked on SO, but have been closed as duplicates and redirected here.)
Caveats of Using __file__
For a module that you have imported:
import something
something.__file__
will return the absolute path of the module. However, given the folowing script foo.py:
#foo.py
print '__file__', __file__
Calling it with 'python foo.py' Will return simply 'foo.py'. If you add a shebang:
#!/usr/bin/python
#foo.py
print '__file__', __file__
and call it using ./foo.py, it will return './foo.py'. Calling it from a different directory, (eg put foo.py in directory bar), then calling either
python bar/foo.py
or adding a shebang and executing the file directly:
bar/foo.py
will return 'bar/foo.py' (the relative path).
Finding the directory
Now going from there to get the directory, os.path.dirname(__file__) can also be tricky. At least on my system, it returns an empty string if you call it from the same directory as the file. ex.
# foo.py
import os
print '__file__ is:', __file__
print 'os.path.dirname(__file__) is:', os.path.dirname(__file__)
will output:
__file__ is: foo.py
os.path.dirname(__file__) is:
In other words, it returns an empty string, so this does not seem reliable if you want to use it for the current file (as opposed to the file of an imported module). To get around this, you can wrap it in a call to abspath:
# foo.py
import os
print 'os.path.abspath(__file__) is:', os.path.abspath(__file__)
print 'os.path.dirname(os.path.abspath(__file__)) is:', os.path.dirname(os.path.abspath(__file__))
which outputs something like:
os.path.abspath(__file__) is: /home/user/bar/foo.py
os.path.dirname(os.path.abspath(__file__)) is: /home/user/bar
Note that abspath() does NOT resolve symlinks. If you want to do this, use realpath() instead. For example, making a symlink file_import_testing_link pointing to file_import_testing.py, with the following content:
import os
print 'abspath(__file__)',os.path.abspath(__file__)
print 'realpath(__file__)',os.path.realpath(__file__)
executing will print absolute paths something like:
abspath(__file__) /home/user/file_test_link
realpath(__file__) /home/user/file_test.py
file_import_testing_link -> file_import_testing.py
Using inspect
@SummerBreeze mentions using the inspect module.
This seems to work well, and is quite concise, for imported modules:
import os
import inspect
print 'inspect.getfile(os) is:', inspect.getfile(os)
obediently returns the absolute path. For finding the path of the currently executing script:
inspect.getfile(inspect.currentframe())
(thanks @jbochi)
This was trivial.
Each module has a __file__ variable that shows its relative path from where you are right now.
Therefore, getting a directory for the module to notify it is simple as:
os.path.dirname(__file__)
If you wish to do this dynamically in a "program" try this code:
My point is, you may not know the exact name of the module to "hardcode" it.
It may be selected from a list or may not be currently running to use __file__.
(I know, it will not work in Python 3)
global modpath
modname = 'os' #This can be any module name on the fly
#Create a file called "modname.py"
f=open("modname.py","w")
f.write("import "+modname+"\n")
f.write("modpath = "+modname+"\n")
f.close()
#Call the file with execfile()
execfile('modname.py')
print modpath
<module 'os' from 'C:\Python27\lib\os.pyc'>
I tried to get rid of the "global" issue but found cases where it did not work I think "execfile()" can be emulated in Python 3 Since this is in a program, it can easily be put in a method or module for reuse.
From within modules of a python package I had to refer to a file that resided in the same directory as package. Ex.
some_dir/
maincli.py
top_package/
__init__.py
level_one_a/
__init__.py
my_lib_a.py
level_two/
__init__.py
hello_world.py
level_one_b/
__init__.py
my_lib_b.py
So in above I had to call maincli.py from my_lib_a.py module knowing that top_package and maincli.py are in the same directory. Here's how I get the path to maincli.py:
import sys
import os
import imp
class ConfigurationException(Exception):
pass
# inside of my_lib_a.py
def get_maincli_path():
maincli_path = os.path.abspath(imp.find_module('maincli')[1])
# top_package = __package__.split('.')[0]
# mod = sys.modules.get(top_package)
# modfile = mod.__file__
# pkg_in_dir = os.path.dirname(os.path.dirname(os.path.abspath(modfile)))
# maincli_path = os.path.join(pkg_in_dir, 'maincli.py')
if not os.path.exists(maincli_path):
err_msg = 'This script expects that "maincli.py" be installed to the '\
'same directory: "{0}"'.format(maincli_path)
raise ConfigurationException(err_msg)
return maincli_path
Based on posting by PlasmaBinturong I modified the code.
import module
print module.__path__
Packages support one more special attribute,
__path__. This is initialized to be a list containing the name of the directory holding the package’s__init__.pybefore the code in that file is executed. This variable can be modified; doing so affects future searches for modules and subpackages contained in the package.While this feature is not often needed, it can be used to extend the set of modules found in a package.
When you import a module, yo have access to plenty of information. Check out dir(a_module). As for the path, there is a dunder for that: a_module.__path__. You can also just print the module itself.
>>> import a_module
>>> print(dir(a_module))
['__builtins__', '__cached__', '__doc__', '__file__', '__loader__', '__name__', '__package__', '__path__', '__spec__']
>>> print(a_module.__path__)
['/.../.../a_module']
>>> print(a_module)
<module 'a_module' from '/.../.../a_module/__init__.py'>
If the only caveat of using __file__ is when current, relative directory is blank (ie, when running as a script from the same directory where the script is), then a trivial solution is:
import os.path
mydir = os.path.dirname(__file__) or '.'
full = os.path.abspath(mydir)
print __file__, mydir, full
And the result:
$ python teste.py
teste.py . /home/user/work/teste
The trick is in or '.' after the dirname() call. It sets the dir as ., which means current directory and is a valid directory for any path-related function.
Thus, using abspath() is not truly needed. But if you use it anyway, the trick is not needed: abspath() accepts blank paths and properly interprets it as the current directory.
As the other answers have said, the best way to do this is with __file__ (demonstrated again below). However, there is an important caveat, which is that __file__ does NOT exist if you are running the module on its own (i.e. as __main__).
For example, say you have two files (both of which are on your PYTHONPATH):
#/path1/foo.py
import bar
print(bar.__file__)
and
#/path2/bar.py
import os
print(os.getcwd())
print(__file__)
Running foo.py will give the output:
/path1 # "import bar" causes the line "print(os.getcwd())" to run
/path2/bar.py # then "print(__file__)" runs
/path2/bar.py # then the import statement finishes and "print(bar.__file__)" runs
HOWEVER if you try to run bar.py on its own, you will get:
/path2 # "print(os.getcwd())" still works fine
Traceback (most recent call last): # but __file__ doesn't exist if bar.py is running as main
File "/path2/bar.py", line 3, in <module>
print(__file__)
NameError: name '__file__' is not defined
Hope this helps. This caveat cost me a lot of time and confusion while testing the other solutions presented.
I will try tackling a few variations on this question as well:
- finding the path of the called script
- finding the path of the currently executing script
- finding the directory of the called script
(Some of these questions have been asked on SO, but have been closed as duplicates and redirected here.)
Caveats of Using __file__
For a module that you have imported:
import something
something.__file__
will return the absolute path of the module. However, given the folowing script foo.py:
#foo.py
print '__file__', __file__
Calling it with 'python foo.py' Will return simply 'foo.py'. If you add a shebang:
#!/usr/bin/python
#foo.py
print '__file__', __file__
and call it using ./foo.py, it will return './foo.py'. Calling it from a different directory, (eg put foo.py in directory bar), then calling either
python bar/foo.py
or adding a shebang and executing the file directly:
bar/foo.py
will return 'bar/foo.py' (the relative path).
Finding the directory
Now going from there to get the directory, os.path.dirname(__file__) can also be tricky. At least on my system, it returns an empty string if you call it from the same directory as the file. ex.
# foo.py
import os
print '__file__ is:', __file__
print 'os.path.dirname(__file__) is:', os.path.dirname(__file__)
will output:
__file__ is: foo.py
os.path.dirname(__file__) is:
In other words, it returns an empty string, so this does not seem reliable if you want to use it for the current file (as opposed to the file of an imported module). To get around this, you can wrap it in a call to abspath:
# foo.py
import os
print 'os.path.abspath(__file__) is:', os.path.abspath(__file__)
print 'os.path.dirname(os.path.abspath(__file__)) is:', os.path.dirname(os.path.abspath(__file__))
which outputs something like:
os.path.abspath(__file__) is: /home/user/bar/foo.py
os.path.dirname(os.path.abspath(__file__)) is: /home/user/bar
Note that abspath() does NOT resolve symlinks. If you want to do this, use realpath() instead. For example, making a symlink file_import_testing_link pointing to file_import_testing.py, with the following content:
import os
print 'abspath(__file__)',os.path.abspath(__file__)
print 'realpath(__file__)',os.path.realpath(__file__)
executing will print absolute paths something like:
abspath(__file__) /home/user/file_test_link
realpath(__file__) /home/user/file_test.py
file_import_testing_link -> file_import_testing.py
Using inspect
@SummerBreeze mentions using the inspect module.
This seems to work well, and is quite concise, for imported modules:
import os
import inspect
print 'inspect.getfile(os) is:', inspect.getfile(os)
obediently returns the absolute path. For finding the path of the currently executing script:
inspect.getfile(inspect.currentframe())
(thanks @jbochi)
If you want to retrieve the package's root path from any of its modules, the following works (tested on Python 3.6):
from . import __path__ as ROOT_PATH
print(ROOT_PATH)
The main __init__.py path can also be referenced by using __file__ instead.
Hope this helps!
When you import a module, yo have access to plenty of information. Check out dir(a_module). As for the path, there is a dunder for that: a_module.__path__. You can also just print the module itself.
>>> import a_module
>>> print(dir(a_module))
['__builtins__', '__cached__', '__doc__', '__file__', '__loader__', '__name__', '__package__', '__path__', '__spec__']
>>> print(a_module.__path__)
['/.../.../a_module']
>>> print(a_module)
<module 'a_module' from '/.../.../a_module/__init__.py'>
you can just import your module then hit its name and you'll get its full path
>>> import os
>>> os
<module 'os' from 'C:\\Users\\Hassan Ashraf\\AppData\\Local\\Programs\\Python\\Python36-32\\lib\\os.py'>
>>>
If the only caveat of using __file__ is when current, relative directory is blank (ie, when running as a script from the same directory where the script is), then a trivial solution is:
import os.path
mydir = os.path.dirname(__file__) or '.'
full = os.path.abspath(mydir)
print __file__, mydir, full
And the result:
$ python teste.py
teste.py . /home/user/work/teste
The trick is in or '.' after the dirname() call. It sets the dir as ., which means current directory and is a valid directory for any path-related function.
Thus, using abspath() is not truly needed. But if you use it anyway, the trick is not needed: abspath() accepts blank paths and properly interprets it as the current directory.
So I spent a fair amount of time trying to do this with py2exe The problem was to get the base folder of the script whether it was being run as a python script or as a py2exe executable. Also to have it work whether it was being run from the current folder, another folder or (this was the hardest) from the system's path.
Eventually I used this approach, using sys.frozen as an indicator of running in py2exe:
import os,sys
if hasattr(sys,'frozen'): # only when running in py2exe this exists
base = sys.prefix
else: # otherwise this is a regular python script
base = os.path.dirname(os.path.realpath(__file__))
Command Line Utility
You can tweak it to a command line utility,
python-which <package name>
Create /usr/local/bin/python-which
#!/usr/bin/env python
import importlib
import os
import sys
args = sys.argv[1:]
if len(args) > 0:
module = importlib.import_module(args[0])
print os.path.dirname(module.__file__)
Make it executable
sudo chmod +x /usr/local/bin/python-which
This was trivial.
Each module has a __file__ variable that shows its relative path from where you are right now.
Therefore, getting a directory for the module to notify it is simple as:
os.path.dirname(__file__)
Here is a quick bash script in case it's useful to anyone. I just want to be able to set an environment variable so that I can pushd to the code.
#!/bin/bash
module=${1:?"I need a module name"}
python << EOI
import $module
import os
print os.path.dirname($module.__file__)
EOI
Shell example:
[root@sri-4625-0004 ~]# export LXML=$(get_python_path.sh lxml)
[root@sri-4625-0004 ~]# echo $LXML
/usr/lib64/python2.7/site-packages/lxml
[root@sri-4625-0004 ~]#
There is inspect module in python.
Official documentation
The inspect module provides several useful functions to help get information about live objects such as modules, classes, methods, functions, tracebacks, frame objects, and code objects. For example, it can help you examine the contents of a class, retrieve the source code of a method, extract and format the argument list for a function, or get all the information you need to display a detailed traceback.
Example:
>>> import os
>>> import inspect
>>> inspect.getfile(os)
'/usr/lib64/python2.7/os.pyc'
>>> inspect.getfile(inspect)
'/usr/lib64/python2.7/inspect.pyc'
>>> os.path.dirname(inspect.getfile(inspect))
'/usr/lib64/python2.7'
This was trivial.
Each module has a __file__ variable that shows its relative path from where you are right now.
Therefore, getting a directory for the module to notify it is simple as:
os.path.dirname(__file__)
As the other answers have said, the best way to do this is with __file__ (demonstrated again below). However, there is an important caveat, which is that __file__ does NOT exist if you are running the module on its own (i.e. as __main__).
For example, say you have two files (both of which are on your PYTHONPATH):
#/path1/foo.py
import bar
print(bar.__file__)
and
#/path2/bar.py
import os
print(os.getcwd())
print(__file__)
Running foo.py will give the output:
/path1 # "import bar" causes the line "print(os.getcwd())" to run
/path2/bar.py # then "print(__file__)" runs
/path2/bar.py # then the import statement finishes and "print(bar.__file__)" runs
HOWEVER if you try to run bar.py on its own, you will get:
/path2 # "print(os.getcwd())" still works fine
Traceback (most recent call last): # but __file__ doesn't exist if bar.py is running as main
File "/path2/bar.py", line 3, in <module>
print(__file__)
NameError: name '__file__' is not defined
Hope this helps. This caveat cost me a lot of time and confusion while testing the other solutions presented.
import module
print module.__path__
Packages support one more special attribute,
__path__. This is initialized to be a list containing the name of the directory holding the package’s__init__.pybefore the code in that file is executed. This variable can be modified; doing so affects future searches for modules and subpackages contained in the package.While this feature is not often needed, it can be used to extend the set of modules found in a package.
If you would like to know absolute path from your script you can use Path object:
from pathlib import Path
print(Path().absolute())
print(Path().resolve('.'))
print(Path().cwd())
Return a new path object representing the current directory (as returned by os.getcwd())
Make the path absolute, resolving any symlinks. A new path object is returned:
I'd like to contribute with one common scenario (in Python 3) and explore a few approaches to it.
The built-in function open() accepts either relative or absolute path as its first argument. The relative path is treated as relative to the current working directory though so it is recommended to pass the absolute path to the file.
Simply said, if you run a script file with the following code, it is not guaranteed that the example.txt file will be created in the same directory where the script file is located:
with open('example.txt', 'w'):
pass
To fix this code we need to get the path to the script and make it absolute. To ensure the path to be absolute we simply use the os.path.realpath() function. To get the path to the script there are several common functions that return various path results:
os.getcwd()os.path.realpath('example.txt')sys.argv[0]__file__
Both functions os.getcwd() and os.path.realpath() return path results based on the current working directory. Generally not what we want. The first element of the sys.argv list is the path of the root script (the script you run) regardless of whether you call the list in the root script itself or in any of its modules. It might come handy in some situations. The __file__ variable contains path of the module from which it has been called.
The following code correctly creates a file example.txt in the same directory where the script is located:
filedir = os.path.dirname(os.path.realpath(__file__))
filepath = os.path.join(filedir, 'example.txt')
with open(filepath, 'w'):
pass
Command Line Utility
You can tweak it to a command line utility,
python-which <package name>
Create /usr/local/bin/python-which
#!/usr/bin/env python
import importlib
import os
import sys
args = sys.argv[1:]
if len(args) > 0:
module = importlib.import_module(args[0])
print os.path.dirname(module.__file__)
Make it executable
sudo chmod +x /usr/local/bin/python-which
From within modules of a python package I had to refer to a file that resided in the same directory as package. Ex.
some_dir/
maincli.py
top_package/
__init__.py
level_one_a/
__init__.py
my_lib_a.py
level_two/
__init__.py
hello_world.py
level_one_b/
__init__.py
my_lib_b.py
So in above I had to call maincli.py from my_lib_a.py module knowing that top_package and maincli.py are in the same directory. Here's how I get the path to maincli.py:
import sys
import os
import imp
class ConfigurationException(Exception):
pass
# inside of my_lib_a.py
def get_maincli_path():
maincli_path = os.path.abspath(imp.find_module('maincli')[1])
# top_package = __package__.split('.')[0]
# mod = sys.modules.get(top_package)
# modfile = mod.__file__
# pkg_in_dir = os.path.dirname(os.path.dirname(os.path.abspath(modfile)))
# maincli_path = os.path.join(pkg_in_dir, 'maincli.py')
if not os.path.exists(maincli_path):
err_msg = 'This script expects that "maincli.py" be installed to the '\
'same directory: "{0}"'.format(maincli_path)
raise ConfigurationException(err_msg)
return maincli_path
Based on posting by PlasmaBinturong I modified the code.
If you installed it using pip, "pip show" works great ('Location')
$ pip show detectron2
Name: detectron2
Version: 0.1
Summary: Detectron2 is FAIR next-generation research platform for object detection and segmentation.
Home-page: https://github.com/facebookresearch/detectron2
Author: FAIR
Author-email: None
License: UNKNOWN
Location: /home/ubuntu/anaconda3/envs/pytorch_p36/lib/python3.6/site-packages
Requires: yacs, tabulate, tqdm, pydot, tensorboard, Pillow, termcolor, future, cloudpickle, matplotlib, fvcore
Source: Stackoverflow.com