Getting script filepath from an imported python module - python-3.x

Let say I have a file a/b/c.py with following content.
import mymodule
a = mymodule.MyClass()
print(a.scriptpath)
I want a.scriptpath to print a/b/c.py. Essentially, what I am asking: Is there a way for MyClass.__init__ or mymodule.__init__.py to know from which file they are being called?

Why not just instantiate the class with that information:
a = mymodule.MyClass(__file__)

Related

Importing a daily changing variable name in python [duplicate]

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".

Loading a class of unknown name in a dynamic location

Currently I am extracting files to the temp directory of the operating system. One of the files is a Python file containing a class which I need to get a handle of. The Python's file is known, but the name of the class inside the file is unknown. But it is safe to assume, that the there is only one single class, and that the class is a subclass of another.
I tried to work with importlib, but I am not able to get a handle of the class.
So far I tried:
# Assume
# module_name contains the name of the class and -> "MyClass"
# path_module contains the path to the python file -> "../Module.py"
spec = spec_from_file_location(module_name, path_module)
module = module_from_spec(spec)
for pair in inspect.getmembers(module):
print(f"{pair[1]} is class: {inspect.isclass(pair[1])}")
When I iterate over the members of the module, none of them get printed as a class.
My class in this case is called BasicModel and the Output on the console looks like this:
BasicModel is class: False
What is the correct approach to this?
Edit:
As the content of the file was requested, here you go:
class BasicModel(Sequential):
def __init__(self, class_count: int, input_shape: tuple):
Sequential.__init__(self)
self.add(Input(shape=input_shape))
self.add(Flatten())
self.add(Dense(128, activation=nn.relu))
self.add(Dense(128, activation=nn.relu))
self.add(Dense(class_count, activation=nn.softmax))
Use dir() to get the attributes of the file and inspect to check if the attribute is a class. If so, you can create an object.
Assuming that your file's path is /tmp/mysterious you can do this:
import importlib
import inspect
from pathlib import Path
import sys
path_pyfile = Path('/tmp/mysterious.py')
sys.path.append(str(path_pyfile.parent))
mysterious = importlib.import_module(path_pyfile.stem)
for name_local in dir(mysterious):
if inspect.isclass(getattr(mysterious, name_local)):
print(f'{name_local} is a class')
MysteriousClass = getattr(mysterious, name_local)
mysterious_object = MysteriousClass()

accessing variable from one file to another file

i am try to import a varaiable from sectionc.py to sectionB.py file.I have used import SectionC but it is giving an error of unresolved reference. How do i import the variable from File C to File B?
from SectionC import action #----------FileB
def action(max_prov_id):
print(max_prov_id) #-------------Filec
I want value of max_prov_id to be accessed in sectionB file
max_prov_id is an argument. You no need to import that. You should add return to the function and get the value by calling it.

Call a function dynamically from a variable - Python

Ok so I have two files, filename1.py and filename2.py and they both have a function with same name funB. The third file process.py has function that calls function from either files. I seem to be struggling in calling the correct function.
In process.py:
from directoryA.filename1 import funB
from directoryA.filename2 import funB
def funA:
#do stuff to determine which filename and save it in variable named 'd'
d = 'filename2'
# here i want to call funB with *args based on what 'd' is
So i have tried eval() like so:
call_right_funB = eval(d.funB(*args))
but it seems not to work.
Any help is appreciated.
The problem is, you can't use eval() with a combination of a string and a method like that. What you have written is:
call_right_funB = eval('filename'.funB(*args))
What you can do is:
call_right_funB = eval(d + '.funB(*args)')
But this is not very pythonic approach.
I would recommend creating a dictionary switch. Even though you have to import entire module:
import directoryA.filename1
import directoryA.filename2
dic_switch = {1: directoryA.filename1, 2: directoryA.filename2}
switch_variable = 1
call_right_funB = dic_switch[switch_variable].funB(*args)
Hope it helps.

Load R package from character string

I want to create a function which includes loading a package that I make within the function. A short example (which doesn't run!):
loadMe <- function(name){
genLib(xxx, libName = name) #make a new library with name "name"
library(name) #load the new library...
}
This does not work! A bit of reproducible code which illustrates my main problem:
library(ggplot) #this works fine
load.this <- "ggplot"
library(load.this) #I want this to load ggplot!
I know the problem is that library() and require() take as an argument an object name which does not exist yet. I have tried wrapping my character string with parse(), deparse(), substitute(), expression(), quote(), etc etc. These all return the same problem:
library(load.this)
# Error in library(loadss) : there is no package called 'loadss'
library(deparse(load.this))
# Error in library(deparse(loadss)) : 'package' must be of length 1
Is there a way to do this?
Use the character.only argument
foo <- "ggplot2"
library(foo,character.only=TRUE)
You say that you have tried using parse(). The following seems to work for me:
eval(parse(text = 'library(MASS)')[1])

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