Python imports by folder module

问题:

I have a directory structure:

example.py
templates/
    __init__.py
    a.py
    b.py

a.py and b.py have only one class, named the same as the file (because they are cheetah templates). For purely style reasons, I want to be able to import and use these classes in example.py like so:

import templates

t = templates.a()

Right now I do that by having this in the template folder's __init__.py:

__all__ = ["a", "b"]
from . import *

However, this seems pretty poor (and maybe superfluous), and doesn't even do what I want, as I have to use the classes like this:

t = templates.a.a()

Thoughts?

回答1:

To avoid repeating from <whatever> import * 25 times, you need a loop, such as:

import sys

def _allimports(modnames)
  thismod = sys.modules[__name__]

  for modname in modnames:
    submodname = '%s.%s' % (thismod, modname)
    __import__(submodname)
    submod = sys.modules[submodname]
    thismod.__dict__.update(submod.__dict__)

_allimports('a b c d e'.split())  # or whatever

I'm putting the meaningful code in a function because (a) it's always best [[for performance and to avoid polluting the module's namespace]], (b) in this particular case it also avoids accidents (e.g., some submodule might define a name thismod or modnames... so it's important to keep those names that we're using in the loop local to the function, not module globals, so they can't be accidentally trampled;-).

If you want to enforce the fact that a module named modname only has one class (or other global) with the same name, change the last statement of the loop to:

    setattr(thismod, modname, getattr(submod, modname))

回答2:

In your __init__.py:

from a import *
from b import *

Now all of a's contents will be in templates, as will all of b's contents.

回答3:

I didn't even know you could have from . import *. My python interpreter complains about such statements. Still, to your problem, you could do:

# __init__.py
from . import a, b
a = a.a
b = a.b

you can now use

# example.py
import templates
t = templates.a()

other solution:

# __init__.py
from a import *
from b import *