Related
I'm writing a Python application that takes a command as an argument, for example:
$ python myapp.py command1
I want the application to be extensible, that is, to be able to add new modules that implement new commands without having to change the main application source. The tree looks something like:
myapp/
__init__.py
commands/
__init__.py
command1.py
command2.py
foo.py
bar.py
So I want the application to find the available command modules at runtime and execute the appropriate one.
Python defines an __import__() function, which takes a string for a module name:
__import__(name, globals=None, locals=None, fromlist=(), level=0)
The function imports the module name, potentially using the given globals and locals to determine how to interpret the name in a package context. The fromlist gives the names of objects or submodules that should be imported from the module given by name.
Source: https://docs.python.org/3/library/functions.html#__import__
So currently I have something like:
command = sys.argv[1]
try:
command_module = __import__("myapp.commands.%s" % command, fromlist=["myapp.commands"])
except ImportError:
# Display error message
command_module.run()
This works just fine, I'm just wondering if there is possibly a more idiomatic way to accomplish what we are doing with this code.
Note that I specifically don't want to get in to using eggs or extension points. This is not an open-source project and I don't expect there to be "plugins". The point is to simplify the main application code and remove the need to modify it each time a new command module is added.
See also: How do I import a module given the full path?
With Python older than 2.7/3.1, that's pretty much how you do it.
For newer versions, see importlib.import_module for Python 2 and Python 3.
Or using __import__ you can import a list of modules by doing this:
>>> moduleNames = ['sys', 'os', 're', 'unittest']
>>> moduleNames
['sys', 'os', 're', 'unittest']
>>> modules = map(__import__, moduleNames)
Ripped straight from Dive Into Python.
The recommended way for Python 2.7 and 3.1 and later is to use importlib module:
importlib.import_module(name, package=None)
Import a module. The name argument specifies what module to import in absolute or relative terms (e.g. either pkg.mod or ..mod). If the name is specified in relative terms, then the package argument must be set to the name of the package which is to act as the anchor for resolving the package name (e.g. import_module('..mod', 'pkg.subpkg') will import pkg.mod).
e.g.
my_module = importlib.import_module('os.path')
Note: imp is deprecated since Python 3.4 in favor of importlib
As mentioned the imp module provides you loading functions:
imp.load_source(name, path)
imp.load_compiled(name, path)
I've used these before to perform something similar.
In my case I defined a specific class with defined methods that were required.
Once I loaded the module I would check if the class was in the module, and then create an instance of that class, something like this:
import imp
import os
def load_from_file(filepath):
class_inst = None
expected_class = 'MyClass'
mod_name,file_ext = os.path.splitext(os.path.split(filepath)[-1])
if file_ext.lower() == '.py':
py_mod = imp.load_source(mod_name, filepath)
elif file_ext.lower() == '.pyc':
py_mod = imp.load_compiled(mod_name, filepath)
if hasattr(py_mod, expected_class):
class_inst = getattr(py_mod, expected_class)()
return class_inst
Using importlib
Importing a source file
Here is a slightly adapted example from the documentation:
import sys
import importlib.util
file_path = 'pluginX.py'
module_name = 'pluginX'
spec = importlib.util.spec_from_file_location(module_name, file_path)
module = importlib.util.module_from_spec(spec)
spec.loader.exec_module(module)
# Verify contents of the module:
print(dir(module))
From here, module will be a module object representing the pluginX module (the same thing that would be assigned to pluginX by doing import pluginX). Thus, to call e.g. a hello function (with no parameters) defined in pluginX, use module.hello().
To get the effect "importing" functionality from the module instead, store it in the in-memory cache of loaded modules, and then do the corresponding from import:
sys.modules[module_name] = module
from pluginX import hello
hello()
Importing a package
To import a package instead, calling import_module is sufficient. Suppose there is a package folder pluginX in the current working directory; then just do
import importlib
pkg = importlib.import_module('pluginX')
# check if it's all there..
print(dir(pkg))
Use the imp module, or the more direct __import__() function.
You can use exec:
exec("import myapp.commands.%s" % command)
If you want it in your locals:
>>> mod = 'sys'
>>> locals()['my_module'] = __import__(mod)
>>> my_module.version
'2.6.6 (r266:84297, Aug 24 2010, 18:46:32) [MSC v.1500 32 bit (Intel)]'
same would work with globals()
Similar as #monkut 's solution but reusable and error tolerant described here http://stamat.wordpress.com/dynamic-module-import-in-python/:
import os
import imp
def importFromURI(uri, absl):
mod = None
if not absl:
uri = os.path.normpath(os.path.join(os.path.dirname(__file__), uri))
path, fname = os.path.split(uri)
mname, ext = os.path.splitext(fname)
if os.path.exists(os.path.join(path,mname)+'.pyc'):
try:
return imp.load_compiled(mname, uri)
except:
pass
if os.path.exists(os.path.join(path,mname)+'.py'):
try:
return imp.load_source(mname, uri)
except:
pass
return mod
The below piece worked for me:
>>>import imp;
>>>fp, pathname, description = imp.find_module("/home/test_module");
>>>test_module = imp.load_module("test_module", fp, pathname, description);
>>>print test_module.print_hello();
if you want to import in shell-script:
python -c '<above entire code in one line>'
The following worked for me:
import sys, glob
sys.path.append('/home/marc/python/importtest/modus')
fl = glob.glob('modus/*.py')
modulist = []
adapters=[]
for i in range(len(fl)):
fl[i] = fl[i].split('/')[1]
fl[i] = fl[i][0:(len(fl[i])-3)]
modulist.append(getattr(__import__(fl[i]),fl[i]))
adapters.append(modulist[i]())
It loads modules from the folder 'modus'. The modules have a single class with the same name as the module name. E.g. the file modus/modu1.py contains:
class modu1():
def __init__(self):
self.x=1
print self.x
The result is a list of dynamically loaded classes "adapters".
I'm writing a Python application that takes a command as an argument, for example:
$ python myapp.py command1
I want the application to be extensible, that is, to be able to add new modules that implement new commands without having to change the main application source. The tree looks something like:
myapp/
__init__.py
commands/
__init__.py
command1.py
command2.py
foo.py
bar.py
So I want the application to find the available command modules at runtime and execute the appropriate one.
Python defines an __import__() function, which takes a string for a module name:
__import__(name, globals=None, locals=None, fromlist=(), level=0)
The function imports the module name, potentially using the given globals and locals to determine how to interpret the name in a package context. The fromlist gives the names of objects or submodules that should be imported from the module given by name.
Source: https://docs.python.org/3/library/functions.html#__import__
So currently I have something like:
command = sys.argv[1]
try:
command_module = __import__("myapp.commands.%s" % command, fromlist=["myapp.commands"])
except ImportError:
# Display error message
command_module.run()
This works just fine, I'm just wondering if there is possibly a more idiomatic way to accomplish what we are doing with this code.
Note that I specifically don't want to get in to using eggs or extension points. This is not an open-source project and I don't expect there to be "plugins". The point is to simplify the main application code and remove the need to modify it each time a new command module is added.
See also: How do I import a module given the full path?
With Python older than 2.7/3.1, that's pretty much how you do it.
For newer versions, see importlib.import_module for Python 2 and Python 3.
Or using __import__ you can import a list of modules by doing this:
>>> moduleNames = ['sys', 'os', 're', 'unittest']
>>> moduleNames
['sys', 'os', 're', 'unittest']
>>> modules = map(__import__, moduleNames)
Ripped straight from Dive Into Python.
The recommended way for Python 2.7 and 3.1 and later is to use importlib module:
importlib.import_module(name, package=None)
Import a module. The name argument specifies what module to import in absolute or relative terms (e.g. either pkg.mod or ..mod). If the name is specified in relative terms, then the package argument must be set to the name of the package which is to act as the anchor for resolving the package name (e.g. import_module('..mod', 'pkg.subpkg') will import pkg.mod).
e.g.
my_module = importlib.import_module('os.path')
Note: imp is deprecated since Python 3.4 in favor of importlib
As mentioned the imp module provides you loading functions:
imp.load_source(name, path)
imp.load_compiled(name, path)
I've used these before to perform something similar.
In my case I defined a specific class with defined methods that were required.
Once I loaded the module I would check if the class was in the module, and then create an instance of that class, something like this:
import imp
import os
def load_from_file(filepath):
class_inst = None
expected_class = 'MyClass'
mod_name,file_ext = os.path.splitext(os.path.split(filepath)[-1])
if file_ext.lower() == '.py':
py_mod = imp.load_source(mod_name, filepath)
elif file_ext.lower() == '.pyc':
py_mod = imp.load_compiled(mod_name, filepath)
if hasattr(py_mod, expected_class):
class_inst = getattr(py_mod, expected_class)()
return class_inst
Using importlib
Importing a source file
Here is a slightly adapted example from the documentation:
import sys
import importlib.util
file_path = 'pluginX.py'
module_name = 'pluginX'
spec = importlib.util.spec_from_file_location(module_name, file_path)
module = importlib.util.module_from_spec(spec)
spec.loader.exec_module(module)
# Verify contents of the module:
print(dir(module))
From here, module will be a module object representing the pluginX module (the same thing that would be assigned to pluginX by doing import pluginX). Thus, to call e.g. a hello function (with no parameters) defined in pluginX, use module.hello().
To get the effect "importing" functionality from the module instead, store it in the in-memory cache of loaded modules, and then do the corresponding from import:
sys.modules[module_name] = module
from pluginX import hello
hello()
Importing a package
To import a package instead, calling import_module is sufficient. Suppose there is a package folder pluginX in the current working directory; then just do
import importlib
pkg = importlib.import_module('pluginX')
# check if it's all there..
print(dir(pkg))
Use the imp module, or the more direct __import__() function.
You can use exec:
exec("import myapp.commands.%s" % command)
If you want it in your locals:
>>> mod = 'sys'
>>> locals()['my_module'] = __import__(mod)
>>> my_module.version
'2.6.6 (r266:84297, Aug 24 2010, 18:46:32) [MSC v.1500 32 bit (Intel)]'
same would work with globals()
Similar as #monkut 's solution but reusable and error tolerant described here http://stamat.wordpress.com/dynamic-module-import-in-python/:
import os
import imp
def importFromURI(uri, absl):
mod = None
if not absl:
uri = os.path.normpath(os.path.join(os.path.dirname(__file__), uri))
path, fname = os.path.split(uri)
mname, ext = os.path.splitext(fname)
if os.path.exists(os.path.join(path,mname)+'.pyc'):
try:
return imp.load_compiled(mname, uri)
except:
pass
if os.path.exists(os.path.join(path,mname)+'.py'):
try:
return imp.load_source(mname, uri)
except:
pass
return mod
The below piece worked for me:
>>>import imp;
>>>fp, pathname, description = imp.find_module("/home/test_module");
>>>test_module = imp.load_module("test_module", fp, pathname, description);
>>>print test_module.print_hello();
if you want to import in shell-script:
python -c '<above entire code in one line>'
The following worked for me:
import sys, glob
sys.path.append('/home/marc/python/importtest/modus')
fl = glob.glob('modus/*.py')
modulist = []
adapters=[]
for i in range(len(fl)):
fl[i] = fl[i].split('/')[1]
fl[i] = fl[i][0:(len(fl[i])-3)]
modulist.append(getattr(__import__(fl[i]),fl[i]))
adapters.append(modulist[i]())
It loads modules from the folder 'modus'. The modules have a single class with the same name as the module name. E.g. the file modus/modu1.py contains:
class modu1():
def __init__(self):
self.x=1
print self.x
The result is a list of dynamically loaded classes "adapters".
I am trying to dynamically import modules in python.
I need something like a plugin import system.
I use the following code for import of a module, and it works fine, as long as the entire code of the module is in the same file.
caller.py code:
PluginFolder = "./plugins"
MainModule = "__init__"
possibleplugins = os.listdir(PluginFolder)
for possible_plugin in possibleplugins:
location = os.path.abspath(os.path.join(PluginFolder, possible_plugin))
if not os.path.isdir(location) or not MainModule + ".py" in os.listdir(location):
continue
info = imp.find_module(MainModule, [location])
plugin = {"name": possible_plugin, "info": info}
After that, I use the load_module method to load the module:
module = imp.load_module(MainModule, *plugin["info"])
The structure of the directories is as follows:
Project/
--plugins/
----plugin_1/
-------__init__.py
-------X.py
caller.py
Everything works great when all the methods are in the same file (__init__.py).
When the method I use in __init__.py calls another method from another file (in the same package) than an error saying "No module named X"
the __init__.py code fails at the line:
import X
I also tried
import plugin_1.X
and other variations.
Just to make sure you guys understand- importing and using the module in a normal way (not dynamically) works fine.
What am I missing?
Is there another way to do this? Maybe use the __import__ method, or something else?
I usually use importlib. It's load_module method does the job. You can put the importing code into plugins/__init__.py to do the following:
import importlib
import os
import logging
skip = set(("__init__.py",))
plugins = []
cwd = os.getcwd()
os.chdir(os.path.dirname(__file__))
for mod in glob.glob("*.py"):
if mod in skip:
continue
try:
mod = mod.replace(".py", "")
plugin = importlib.import_module("." + mod, __name__)
plugin.append(plugin)
except Exception as e:
logging.warn("Failed to load %s: %s. Skipping", mod, e)
os.chdir(cwd)
File setup:
...\Project_Folder
...\Project_Folder\Project.py
...\Project_folder\Script\TestScript.py
I'm attempting to have Project.py import modules from the folder Script based on user input.
Python Version: 3.4.2
Ideally, the script would look something like
q = str(input("Input: "))
from Script import q
However, python does not recognize q as a variable when using import.
I've tried using importlib, however I cannot figure out how to import from the Script folder mentioned above.
import importlib
q = str(input("Input: "))
module = importlib.import_module(q, package=None)
I'm not certain where I would implement the file path.
Repeat of my answer originally posted at How to import a module given the full path?
as this is a Python 3.4 specific question:
This area of Python 3.4 seems to be extremely tortuous to understand, mainly because the documentation doesn't give good examples! This was my attempt using non-deprecated modules. It will import a module given the path to the .py file. I'm using it to load "plugins" at runtime.
def import_module_from_file(full_path_to_module):
"""
Import a module given the full path/filename of the .py file
Python 3.4
"""
module = None
try:
# Get module name and path from full path
module_dir, module_file = os.path.split(full_path_to_module)
module_name, module_ext = os.path.splitext(module_file)
# Get module "spec" from filename
spec = importlib.util.spec_from_file_location(module_name,full_path_to_module)
module = spec.loader.load_module()
except Exception as ec:
# Simple error printing
# Insert "sophisticated" stuff here
print(ec)
finally:
return module
# load module dynamically
path = "<enter your path here>"
module = import_module_from_file(path)
# Now use the module
# e.g. module.myFunction()
I did this by defining the entire import line as a string, formatting the string with q and then using the exec command:
imp = 'from Script import %s' %q
exec imp
How can I know if a Python module exists, without importing it?
Importing something that might not exist (not what I want) results in:
try:
import eggs
except ImportError:
pass
TL;DR) Use importlib.util.find_spec(module_name) (Python 3.4+).
Python2: imp.find_module
To check if import can find something in Python 2, using imp:
import imp
try:
imp.find_module('eggs')
found = True
except ImportError:
found = False
To find dotted imports, you need to do more:
import imp
try:
spam_info = imp.find_module('spam')
spam = imp.load_module('spam', *spam_info)
imp.find_module('eggs', spam.__path__) # __path__ is already a list
found = True
except ImportError:
found = False
You can also use pkgutil.find_loader (more or less the same as the Python 3 part:
import pkgutil
eggs_loader = pkgutil.find_loader('eggs')
found = eggs_loader is not None
Python 3
Python 3 ≤ 3.3: importlib.find_loader
You should use importlib. I went about doing this like:
import importlib
spam_loader = importlib.find_loader('spam')
found = spam_loader is not None
My expectation being, if you can find a loader for it, then it exists. You can also be a bit more smart about it, like filtering out what loaders you will accept. For example:
import importlib
spam_loader = importlib.find_loader('spam')
# only accept it as valid if there is a source file for the module - no bytecode only.
found = issubclass(type(spam_loader), importlib.machinery.SourceFileLoader)
Python 3 ≥ 3.4: importlib.util.find_spec
In Python 3.4 importlib.find_loader Python documentation was deprecated in favour of importlib.util.find_spec. The recommended method is the importlib.util.find_spec. There are others like importlib.machinery.FileFinder, which is useful if you're after a specific file to load. Figuring out how to use them is beyond the scope of this.
import importlib
spam_spec = importlib.util.find_spec("spam")
found = spam_spec is not None
This also works with relative imports, but you must supply the starting package, so you could also do:
import importlib
spam_spec = importlib.util.find_spec("..spam", package="eggs.bar")
found = spam_spec is not None
spam_spec.name == "eggs.spam"
While I'm sure there exists a reason for doing this - I'm not sure what it would be.
Warning
When trying to find a submodule, it will import the parent module (for ALL of the above methods)!
food/
|- __init__.py
|- eggs.py
## __init__.py
print("module food loaded")
## eggs.py
print("module eggs")
were you then to run
>>> import importlib
>>> spam_spec = importlib.util.find_spec("food.eggs")
module food loaded
ModuleSpec(name='food.eggs', loader=<_frozen_importlib.SourceFileLoader object at 0x10221df28>, origin='/home/user/food/eggs.py')
Comments are welcome on getting around this
Acknowledgements
#rvighne for importlib
#lucas-guido for Python 3.3+ deprecating find_loader
#enpenax for pkgutils.find_loader behaviour in Python 2.7
Python 3 >= 3.6: ModuleNotFoundError
The ModuleNotFoundError has been introduced in Python 3.6 and can be used for this purpose:
try:
import eggs
except ModuleNotFoundError:
# Error handling
pass
The error is raised when a module or one of its parents cannot be found. So
try:
import eggs.sub
except ModuleNotFoundError as err:
# Error handling
print(err)
would print a message that looks like No module named 'eggs' if the eggs module cannot be found; but it would print something like No module named 'eggs.sub' if only the sub module couldn't be found, but the eggs package could be found.
See the documentation of the import system for more information on the ModuleNotFoundError.
After using yarbelk's response, I've made this so I don't have to import ìmp.
try:
__import__('imp').find_module('eggs')
# Make things with a supposed existing module
except ImportError:
pass
It is useful in Django's settings.py file, for example.
Python 2, without relying on ImportError
Until the current answer is updated, here is the way for Python 2
import pkgutil
import importlib
if pkgutil.find_loader(mod) is not None:
return importlib.import_module(mod)
return None
Why another answer?
A lot of answers make use of catching an ImportError. The problem with that is, that we cannot know what throws the ImportError.
If you import your existent module and there happens to be an ImportError in your module (e.g., typo on line 1), the result will be that your module does not exist.
It will take you quite the amount of backtracking to figure out that your module exists and the ImportError is caught and makes things fail silently.
go_as's answer as a one-liner:
python -c "help('modules');" | grep module
Use one of the functions from pkgutil, for example:
from pkgutil import iter_modules
def module_exists(module_name):
return module_name in (name for loader, name, ispkg in iter_modules())
I wrote this helper function:
def is_module_available(module_name):
if sys.version_info < (3, 0):
# python 2
import importlib
torch_loader = importlib.find_loader(module_name)
elif sys.version_info <= (3, 3):
# python 3.0 to 3.3
import pkgutil
torch_loader = pkgutil.find_loader(module_name)
elif sys.version_info >= (3, 4):
# python 3.4 and above
import importlib
torch_loader = importlib.util.find_spec(module_name)
return torch_loader is not None
Here is a way to check if a module is loaded from the command line:
Linux/UNIX script file method: make a file module_help.py:
#!/usr/bin/env python
help('modules')
Then make sure it's executable: chmod u+x module_help.py
And call it with a pipe to grep:
./module_help.py | grep module_name
Invoke the built-in help system. (This function is intended for interactive use.) If no argument is given, the interactive help system starts on the interpreter console. If the argument is a string, then the string is looked up as the name of a module, function, class, method, keyword, or documentation topic, and a help page is printed on the console. If the argument is any other kind of object, a help page on the object is generated.
Interactive method: in the console, load python
>>> help('module_name')
If found, quit reading by typing q.
To exit the Python interpreter interactive session, press Ctrl + D
Windows script file method, also Linux/UNIX compatible, and better overall:
#!/usr/bin/env python
import sys
help(sys.argv[1])
Calling it from the command like:
python module_help.py site
Would output:
Help on module site:
NAME
site - Append module search paths for third-party packages to sys.path.
FILE
/usr/lib/python2.7/site.py
MODULE DOCS
http://docs.python.org/library/site
DESCRIPTION
...
:
And you'd have to press q to exit interactive mode.
Using it for an unknown module, e.g.,
python module_help.py lkajshdflkahsodf
Would output:
no Python documentation found for 'lkajshdflkahsodf'
and exit.
You could just write a little script that would try to import all the modules and tell you which ones are failing and which ones are working:
import pip
if __name__ == '__main__':
for package in pip.get_installed_distributions():
pack_string = str(package).split(" ")[0]
try:
if __import__(pack_string.lower()):
print(pack_string + " loaded successfully")
except Exception as e:
print(pack_string + " failed with error code: {}".format(e))
Output:
zope.interface loaded successfully
zope.deprecation loaded successfully
yarg loaded successfully
xlrd loaded successfully
WMI loaded successfully
Werkzeug loaded successfully
WebOb loaded successfully
virtualenv loaded successfully
...
A word of warning: this will try to import everything, so you'll see things like PyYAML failed with error code: No module named pyyaml, because the actual import name is just yaml. So as long as you know your imports, this should do the trick for you.
There isn't any way to reliably check if "dotted module" is importable without importing its parent package. Saying this, there are many solutions to problem "how to check if a Python module exists".
The below solution addresses the problem that an imported module can raise an ImportError even if it exists. We want to distinguish that situation from such in which the module does not exist.
Python 2:
import importlib
import pkgutil
import sys
def find_module(full_module_name):
"""
Returns module object if module `full_module_name` can be imported.
Returns None if module does not exist.
Exception is raised if (existing) module raises exception during its import.
"""
module = sys.modules.get(full_module_name)
if module is None:
module_path_tail = full_module_name.split('.')
module_path_head = []
loader = True
while module_path_tail and loader:
module_path_head.append(module_path_tail.pop(0))
module_name = ".".join(module_path_head)
loader = bool(pkgutil.find_loader(module_name))
if not loader:
# Double check if module realy does not exist
# (case: full_module_name == 'paste.deploy')
try:
importlib.import_module(module_name)
except ImportError:
pass
else:
loader = True
if loader:
module = importlib.import_module(full_module_name)
return module
Python 3:
import importlib
def find_module(full_module_name):
"""
Returns module object if module `full_module_name` can be imported.
Returns None if module does not exist.
Exception is raised if (existing) module raises exception during its import.
"""
try:
return importlib.import_module(full_module_name)
except ImportError as exc:
if not (full_module_name + '.').startswith(exc.name + '.'):
raise
In django.utils.module_loading.module_has_submodule:
import sys
import os
import imp
def module_has_submodule(package, module_name):
"""
check module in package
django.utils.module_loading.module_has_submodule
"""
name = ".".join([package.__name__, module_name])
try:
# None indicates a cached miss; see mark_miss() in Python/import.c.
return sys.modules[name] is not None
except KeyError:
pass
try:
package_path = package.__path__ # No __path__, then not a package.
except AttributeError:
# Since the remainder of this function assumes that we're dealing with
# a package (module with a __path__), so if it's not, then bail here.
return False
for finder in sys.meta_path:
if finder.find_module(name, package_path):
return True
for entry in package_path:
try:
# Try the cached finder.
finder = sys.path_importer_cache[entry]
if finder is None:
# Implicit import machinery should be used.
try:
file_, _, _ = imp.find_module(module_name, [entry])
if file_:
file_.close()
return True
except ImportError:
continue
# Else see if the finder knows of a loader.
elif finder.find_module(name):
return True
else:
continue
except KeyError:
# No cached finder, so try and make one.
for hook in sys.path_hooks:
try:
finder = hook(entry)
# XXX Could cache in sys.path_importer_cache
if finder.find_module(name):
return True
else:
# Once a finder is found, stop the search.
break
except ImportError:
# Continue the search for a finder.
continue
else:
# No finder found.
# Try the implicit import machinery if searching a directory.
if os.path.isdir(entry):
try:
file_, _, _ = imp.find_module(module_name, [entry])
if file_:
file_.close()
return True
except ImportError:
pass
# XXX Could insert None or NullImporter
else:
# Exhausted the search, so the module cannot be found.
return False
In case you know the location of file and want to check that the respective Python code file has that module or not, you can simply check via the astor package in Python. Here is a quick example:
"""
Check if a module function exists or not without importing a Python package file
"""
import ast
import astor
tree = astor.parse_file('handler.py')
method_to_check = 'handle'
for item in tree.body:
if isinstance(item, ast.FunctionDef):
if item.name == method_to_check:
print('method exists')
break
A simpler if statement from Ask Ubuntu, How do I check whether a module is installed in Python?:
import sys
print('eggs' in sys.modules)
You can also use importlib.util directly
import importlib.util
def module_exists_without_import(module_name):
spec = importlib.util.find_spec(module_name)
return spec is not None