Separate namespace from file hierarchy?

Steven D'Aprano steven at REMOVE.THIS.cybersource.com.au
Tue Oct 20 23:58:28 CEST 2009


On Tue, 20 Oct 2009 08:51:44 -0500, Peng Yu wrote:

> Suppose I have the dirname/both.py, which has the definitions of classes
> A and B. I can use this module in the following code.
> 
> #################
> import dirname.both
> 
> a=dirname.both.A()
> b=dirname.both.B()


Have you tried this, or are you just assuming it will work?

You can't specify an arbitrary directory location for a module. For the 
above to work, "dirname" has to be a package, which means it needs a 
__init__.py file.

I'll assume from this point on that "dirname" is the name of a package.


> ####################
> 
> When the definitions of A and B become too long, it is better that I put
> them in two different files (to improve the maintainability), for
> example, dirname/A.py and dirname/B.py. Now, the code becomes
> 
> #################
> import dirname.A
> import dirname.B
> 
> a=dirname.A.A() #two A seems redundant 
> b=dirname.B.B() #two B seems redundant 
> ####################
>
> However, the last two lines are annoying to me, as 'A' and 'B' appears
> twice, which seems redundant.

Saying it is redundant three times is redundant.

If that worries you, then don't do it. You don't *have* to put the class 
A in a module A.py, you can call the module anything appropriate:

import dirname.internal
a = dirname.internal.A()

You can arrange your package in whatever way seems sensible to you. What 
I tend to do is something like this:

dirname/
+-- __init__.py
+-- A.py
+-- B.py


and then in __init__.py I have this:


from A import A
from B import B


so the caller can do this:

import dirname
a = dirname.A()
b = dirname.B()


without worrying about the internal structure of the package (the 
submodules A and B).



-- 
Steven



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