Module name different than directory name? - python

Let's assume I have a python package called bestpackage.
Convention dictates that bestpacakge would also be a directory on sys.path that contains an __init__.py to make the interpreter assume it can be imported from.
Is there any way I can set a variable for the package name so the directory could be named something different than the directive I import it with? Is there any way to make the namespacing not care about the directory name and honor some other config instead?
My super trendy client-side devs are just so much in love with these sexy something.otherthing.js project names for one of our smaller side projects!
EDIT:
To clarify, the main purpose of my question was to allow my client side guys continue to call the directories in their "projects" (the one we all have added to our paths) folder using their existing convention (some.app.js), even though in some cases they are in fact python packages that will be on the path and sourced to import statements internally. I realize this is in practice a pretty horrible thing and so I ask more out of curiosity. So part of the big problem here is circumventing the fact that the . in the directory name (and thereby the assumed package name) implies attribute access. It doesn't really surprise me that this cannot be worked around, I was just curious if it was possible deeper in the "magic" behind import.
There's some great responses here, but all rely on doing a classical import of some kind where the attribute accessor . will clash with the directory names.

A directory with a __init__.py file is called a package.
And no, the package name is always the same as the directory. That's how Python can discover packages, it matches it against directory names found on the search path, and if there is a __init__.py file in that directory it has found a match and imports the __init__.py file contained.
You can always import something into your local module namespace under a shorter, easier to use name using the from module import something or the import module as alias syntax:
from something.otherthing.js import foo
from something.otherthing import js as bar
import something.otherthing.js as hamspam

There is one solution wich needs one initial import somewhere
>>> import sys
>>> sys.modules['blinot_existing_blubb'] = sys
>>> import blinot_existing_blubb
>>> blinot_existing_blubb
<module 'sys' (built-in)>
Without a change to the import mechanism you can not import from an other name. This is intended, I think, to make Python easier to understand.
However if you want to change the import mechanism I recommend this: Getting the Most Out of Python Imports

Well, first I would say that Python is not Java/Javascript/C/C++/Cobol/YourFavoriteLanguageThatIsntPython. Of course, in the real world, some of us have to answer to bosses who disagree. So if all you want is some indirection, use smoke and mirrors, as long as they don't pay too much attention to what's under the hood. Write your module the Python way, then provide an API on the side in the dot-heavy style that your coworkers want. Ex:
pythonic_module.py
def func_1():
pass
def func_2():
pass
def func_3():
pass
def func_4():
pass
indirection
/dotty_api_1/__init__.py
from pythonic_module import func_1 as foo, func_2 as bar
/dotty_api_2/__init__.py
from pythonic_module import func_3 as foo, func_4 as bar
Now they can dot to their hearts' content, but you can write things the Pythonic way under the hood.

Actually yes!
you could do a canonical import Whatever or newmodulename = __import__("Whatever")
python keeps track of your modules and you can inspect that by doing:
import sys
print sys.modules
See this article more details
But that's maybe not your problem? let's guess: you have a module in a different path, which your current project can't access because it's not in the sys-path?
well the just add:
import sys
sys.path.append('path_to_the_other_package_or_module_directory')
prior to your import statement or see this SO-post for a more permanent solution.

I was looking for this to happen with setup.py at sdist and install time, rather than runtime, and found the directive package_dir:
https://docs.python.org/3.5/distutils/setupscript.html#listing-whole-packages

Related

How to get source file which python class/function/variable is imported from?

Let's say I have imported two modules like this:
from module0 import hello_func
from directory.module1 import hello_var
Where in module0:
def hello_func(): return "hello from module0"
And module1:
hello_var = "hello from module1"
How can I know from which file is each object being imported?
I tried to check locals() function but nothing in there giving reference to the file...
Actually, you kind of answered your question yourself:
Let's say I have imported two modules
(insert "from xxx import *" here)
How can I know from which file is each object being imported?
One of the reasons of NOT using wildcard imports is precisely to make it clear where names are imported from (the other one being to avoind one imported name to shadow a previously imported one - something that tends to break you code in the most unexpected - and sometimes quite hard to spot - ways).
Note that in your edited question:
from module0 import hello_func
from directory.module1 import hello_var
you already have a much better idea where a name comes from. Not the exact files path yet, but at least the name of the package/module.
And that's one of the main reasons why one should NOT use wildcard imports.
Now if you want to know the exact files path, you have two distinct cases.
Some objects keep trace of where they were created (mostly modules, classes, functions, etc - cf the list of types supported by inspect.getfile()), and then, well, you already know the answer (use inspect.getfile() xD).
But most types wont (because there's no reason for it). In this case, you have to know which module they were imported from and call inspect.getfile() on the module itself. In this case, if you used wildcard imports, you will have to manually inspect all the modules you imported from to find out which one defined this name. Enjoy. Specially if one of those modules did also use wildcard imports...
one question please: where they does keep traces? and how these traces look like?
Modules keep it in their __file__ attribute. Functions and classes keep a reference to their module's name in their __module__ attribute, and from this you can use it to retrieve the module from the sys.modules dict (a cache of all modules already imported in the current process), which will gives you the file.
I never had a need to search this info for tracebacks, frames, code objects etc so you'll have to check it yourself I'm afraid ;-)
You can define in each module a constant with its path, something like this should work:
import os
FILE_PATH = os.path.abspath(__file__)
When you import that module you can access its location like this:
import module
print(module.FILE_PATH)
Another solution using the inspect and os modules.
import module0
import os
import inspect
print(os.path.abspath(inspect.getfile(module0.hello_func)))
If you are looking for the absolute path of where the script is being run, this should work for sure:
import os
abs_path = os.path.dirname(os.path.abspath(__file__))
print(abs_path)

Should I define __all__ even if I prefix hidden functions and variables with underscores in modules?

From the perspective of an external user of the module, are both necessary?
From my understanding, by correctly prefix hidden functions with an underscore it essentially does the same thing as explicitly define __all__, but I keep seeing developers doing both in their code. Why is that?
When importing from a module with from modulename import * names starting with underscores are indeed skipped.
However, a module rarely contains only public API objects. Usually you've made imports to support the code as well, and those names are global in the module as well. Without __all__, those names would be part of the import too.
In other words, unless you want to 'export' os in the following example you should use __all__:
import os
from .implementation import some_other_api_call
_module_path = os.path.dirname(os.path.abspath(__file__))
_template = open(os.path.join(_module_path, 'templates/foo_template.txt')).read()
VERSION = '1.0.0'
def make_bar(baz, ham, spam):
return _template.format(baz, ham, spam)
__all__ = ['some_other_api_call', 'make_bar']
because without the __all__ list, Python cannot distinguish between some_other_api_call and os here and divine which one should not be imported when using from ... import *.
You could work around this by renaming all your imports, so import os as _os, but that'd just make your code less readable.
And an explicit export list is always nice. Explicit is better than implicit, as the Zen of Python tells you.
I also use __all__: that explictly tells module users what you intend to export. Searching the module for names is tedious, even if you are careful to do, e.g., import os as _os, etc. A wise man once wrote "explicit is better than implicit" ;-)
Defining all will overide the default behaviour. There is actually might be one reason to define __all__
When importing a module, you might want that doing from mod import * will import only a minimal amount of things. Even if you prefix everything correctly, there could be reasons not to import everything.
The other problem that I had once was defining a gettext shortcut. The translation function was _ which would not get imported. Even if it is prefixed "in theory" I still wanted it to get exported.
One other reason as stated above is importing a module that imports a lot of thing. As python cannot make the difference between symbols created by imports and the one defined in the actual module. It will automatically reexport everything that can be reexported. For that reason, it could be wise to explicitely limit the thing exported to the things you want to export.
With that in mind, you might want to have some prefixed symbols exported by default. Usually, you don't redefine __all__. Whenever you need it to do something unusual then it may make sense to do it.

Python module: how to prevent importing modules called by the new module

I am new in Python and I am creating a module to re-use some code.
My module (impy.py) looks like this (it has one function so far)...
import numpy as np
def read_image(fname):
....
and it is stored in the following directory:
custom_modules/
__init.py__
impy.py
As you can see it uses the module numpy. The problem is that when I import it from another script, like this...
import custom_modules.impy as im
and I type im. I get the option of calling not only the function read_image() but also the module np.
How can I do to make it only available the functions I am writing in my module and not the modules that my module is calling (numpy in this case)?
Thank you very much for your help.
I've got a proposition, that could maybe answer the following concern: "I do not want to mess class/module attributes with class/module imports". Because, Idle also proposes access to imported modules within a class or module.
This simply consists in taking the conventional name that coders normally don't want to access and IDE not to propose: name starting with underscore. This is also known as "weak « internal use » indicator", as described in PEP 8 / Naming styles.
class C(object):
import numpy as _np # <-- here
def __init__(self):
# whatever we need
def do(self, arg):
# something useful
Now, in Idle, auto-completion will only propose do function; imported module is not proposed.
By the way, you should change the title of your question: you do not want to avoid imports of your imported modules (that would make them unusable), so it should rather be "how to prevent IDE to show imported modules of an imported module" or something similar.
You could import numpy inside your function
def read_image(fname):
import numpy as np
....
making it locally available to the read_image code, but not globally available.
Warning though, this might cause a performance hit (as numpy would be imported each time the code is run rather than just once on the initial import) - especially if you run read_image multiple times.
If you really want to hide it, then I suggest creating a new directory such that your structure looks like this:
custom_modules/
__init__.py
impy/
__init__.py
impy.py
and let the new impy/__init__.py contain
from impy import read_image
This way, you can control what ends up in the custom_modules.impy namespace.

Accessing classes from arbitrary modules inside of package directly

Not sure if there's a neat way of dealing with it, it just makes sense to me visually to lay out each object/class into it's own module under a common package.
For instance:
/Settings/
/Settings/__init__.py
/Settings/AbstractSetting.py
/Settings/Float.py
/Settings/String.py
Each class inside of every module has the same name as the module and at the moment I keep doing this:
import Settings
mysetting = Settings.Float.Float()
..which is giving me these double "Float" names.
I could do, in the __init__.py of the package:
from Float import Float
..so that I could then do:
import Settings
mysetting = Settings.Float()
But I'd like this package to be dynamically updating to whatever modules I put inside of it. So that the next day, when I've added "Knob.py" to this package, I could do:
import Settings
myknob = Settings.Knob()
Makes sense?
But again, I haven't worked with packages before and are still trying to wrap my head around it and try and make it as easy as possible. At this point, I found it easier having all classes inside one big master module which is getting increasingly cumbersome.
Maybe packages isn't the way to go? What alternatives do I have?
Thanks a bunch.
EDIT: Main reason I want to do this is to let users write their own modules that will integrate with the rest of the application. A native "plugin" architeture, if you will.
Each module will contain a class inherited by a superclass with default values. The app then has a browser with available modules that, when clicked, displays relevant information found under the modules attributes. Each class contained then has a similar interface with which the application can use.
I did some further reading and apparently this is not the way to go. I'd love to hear your ideas though on what the benefits/disadvantages of this approach could be.
You should be aware that this is not the Python way. "One class per file" is a Java philosphy that does not apply in the Python world. We usually name modules in lowercase and stick related classes into the same file (in your example, all of the classes would go into settings.py or would be explicitely imported from there). But I guess the fact that you want users to provide plugins is a legitimate reason for your approach (immdbg does it the same way, I think).
So, if you really want to do this, you could put something like this into your Settings/__init__.py:
import os
import glob
import imp
for f in glob.glob(os.path.join(os.path.dirname(__file__), '*.py')):
modname = os.path.basename(f)[:-3]
if modname.startswith('__'): continue
mod = imp.load_source(modname, f)
globals()[modname] = getattr(mod, modname)
# or if you just want to import everything (even worse):
#for name in dir(mod):
# if name.startswith('__'): continue
# globals()[name] = getattr(mod, name)
Can you feel how the Python developers don't want you to do this? :)
There are many plugin systems. It is exemplified by the name of one such system yapsy (yet another plugin system).
You could create an object that provides necessary interface:
class Settings(object):
def __getattr__(self, attr):
return load_plugin(attr)
settings = Settings()
In your code:
from settings import settings
knob = settings.Knob()
You can use whatever implementation you like for load_plugin() e.g., for the code from the question:
from importlib import import_module
def load_plugin(name):
m = import_module('Settings.'+name)
return getattr(m, name)

Why import when you need to use the full name?

In python, if you need a module from a different package you have to import it. Coming from a Java background, that makes sense.
import foo.bar
What doesn't make sense though, is why do I need to use the full name whenever I want to use bar? If I wanted to use the full name, why do I need to import? Doesn't using the full name immediately describe which module I'm addressing?
It just seems a little redundant to have from foo import bar when that's what import foo.bar should be doing. Also a little vague why I had to import when I was going to use the full name.
The thing is, even though Python's import statement is designed to look similar to Java's, they do completely different things under the hood. As you know, in Java an import statement is really little more than a hint to the compiler. It basically sets up an alias for a fully qualified class name. For example, when you write
import java.util.Set;
it tells the compiler that throughout that file, when you write Set, you mean java.util.Set. And if you write s.add(o) where s is an object of type Set, the compiler (or rather, linker) goes out and finds the add method in Set.class and puts in a reference to it.
But in Python,
import util.set
(that is a made-up module, by the way) does something completely different. See, in Python, packages and modules are not just names, they're actual objects, and when you write util.set in your code, that instructs Python to access an object named util and look for an attribute on it named set. The job of Python's import statement is to create that object and attribute. The way it works is that the interpreter looks for a file named util/__init__.py, uses the code in it to define properties of an object, and binds that object to the name util. Similarly, the code in util/set.py is used to initialize an object which is bound to util.set. There's a function called __import__ which takes care of all of this, and in fact the statement import util.set is basically equivalent to
util = __import__('util.set')
The point is, when you import a Python module, what you get is an object corresponding to the top-level package, util. In order to get access to util.set you need to go through that, and that's why it seems like you need to use fully qualified names in Python.
There are ways to get around this, of course. Since all these things are objects, one simple approach is to just bind util.set to a simpler name, i.e. after the import statement, you can have
set = util.set
and from that point on you can just use set where you otherwise would have written util.set. (Of course this obscures the built-in set class, so I don't recommend actually using the name set.) Or, as mentioned in at least one other answer, you could write
from util import set
or
import util.set as set
This still imports the package util with the module set in it, but instead of creating a variable util in the current scope, it creates a variable set that refers to util.set. Behind the scenes, this works kind of like
_util = __import__('util', fromlist='set')
set = _util.set
del _util
in the former case, or
_util = __import__('util.set')
set = _util.set
del _util
in the latter (although both ways do essentially the same thing). This form is semantically more like what Java's import statement does: it defines an alias (set) to something that would ordinarily only be accessible by a fully qualified name (util.set).
You can shorten it, if you would like:
import foo.bar as whateveriwant
Using the full name prevents two packages with the same-named submodules from clobbering each other.
There is a module in the standard library called io:
In [84]: import io
In [85]: io
Out[85]: <module 'io' from '/usr/lib/python2.6/io.pyc'>
There is also a module in scipy called io:
In [95]: import scipy.io
In [96]: scipy.io
Out[96]: <module 'scipy.io' from '/usr/lib/python2.6/dist-packages/scipy/io/__init__.pyc'>
If you wanted to use both modules in the same script, then namespaces are a convenient way to distinguish the two.
In [97]: import this
The Zen of Python, by Tim Peters
...
Namespaces are one honking great idea -- let's do more of those!
in Python, importing doesn't just indicate you might use something. The import actually executes code at the module level. You can think of the import as being the moment where the functions are 'interpreted' and created. Any code that is in the _____init_____.py level or not inside a function or class definition happens then.
The import also makes an inexpensive copy of the whole module's namespace and puts it inside the namespace of the file / module / whatever where it is imported. An IDE then has a list of the functions you might be starting to type for command completion.
Part of the Python philosophy is explicit is better than implicit. Python could automatically import the first time you try to access something from a package, but that's not explicit.
I'm also guessing that package initialization would be much more difficult if the imports were automatic, as it wouldn't be done consistently in the code.
You're a bit confused about how Python imports work. (I was too when I first started.) In Python, you can't simply refer to something within a module by the full name, unlike in Java; you HAVE to import the module first, regardless of how you plan on referring to the imported item. Try typing math.sqrt(5) in the interpreter without importing math or math.sqrt first and see what happens.
Anyway... the reason import foo.bar has you required to use foo.bar instead of just bar is to prevent accidental namespace conflicts. For example, what if you do import foo.bar, and then import baz.bar?
You could, of course, choose to do import foo.bar as bar (i.e. aliasing), but if you're doing that you may as well just use from foo import bar. (EDIT: except when you want to import methods and variables. Then you have to use the from ... import ... syntax. This includes instances where you want to import a method or variable without aliasing, i.e. you can't simply do import foo.bar if bar is a method or variable.)
Other than in Java, in Python import foo.bar declares, that you are going to use the thing referred to by foo.bar.
This matches with Python's philosophy that explicit is better than implicit. There are more programming languages that make inter-module dependencies more explicit than Java, for example Ada.
Using the full name makes it possible to disambiguate definitions with the same name coming from different modules.
You don't have to use the full name. Try one of these
from foo import bar
import foo.bar as bar
import foo.bar
bar = foo.bar
from foo import *
A few reasons why explicit imports are good:
They help signal to humans and tools what packages your module depends on.
They avoid the overhead of dynamically determining which packages have to be loaded (and possibly compiled) at run time.
They (along with sys.path) unambiguously distinguish symbols with conflicting names from different namespaces.
They give the programmer some control of what enters the namespace within which he is working.

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