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Module/package hierarchy and its separation from file structure

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Peter Schuller

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Jan 23, 2008, 4:49:56 AM1/23/08
to
Hello,

In writing some non-trivial amount of Python code I keep running into
an organizational issue. I will try to state the problem fairly
generally, and follow up with a (contrived) example.

The root cause of my difficulties is that by default, the relationship
between a module hierarchy and the structure of files on disk is too
strong for my taste. I want to separate the two as much as possible,
but I do not want to resort to non-conventional "hacks" to do it. I am
posting this in an attempt to present what I perceive to be a
practical problem, and to get suggestions for solutions, or opinions
on the most practical policy for how to deal with it.

Like I said, I would like a weaker relationship between file system
structure and module hierarchy. In particular there are two things I
would like:

* Least importantly, I don't like jamming code into __init__.py,
as a personal preference.
* Most importantly, I do not like to jam large amounts of code
into a single source file, just for the purpose of keeping
the public interface in the same package.

An contrived but hopefully illustrative example:

We have an organization "Org", which has a library, and as part of
that library is code that relates to doing something with animals. As
a result, the interesting top-level package for this example is:

org.lib.animal

Suppose now that I want an initial implementation of the most
important animal. I want to create the class (but see [1]):

org.lib.animal.Monkey

The public interface consists of that class only (and possibly a small
handful of functions). The implementation is quite significant however
- it is 500 lines of code long.

At this point, we had to jam those 500 lines of code into
__init__.py. Let's ignore my personal preference of not liking to put
code in __init__.py; the fact remains that we have 500 lines of code
in a single source file.

Now, we want to continue working on this library, adding ten
additional animals.

At this point, we have these choices (it seems to me):

(1) Simply add these to __init__.py, resulting in
__init__.py being 5000 lines long[2].

(2) Put each animal into its own file, resulting in
org.lib.animal.Monkey now becoming
org.lib.animal.monkey.Monkey, and animal X becoming
org.lib.animal.x.X.

The problem I have is that both of these solutions are, in my opinion,
very ugly:

* (1) is ugly from a source code management perspective, because jamming
5000 lines of code for ten different animals into a single file
is bad for obvious reasons.

* (2) is ugly because we introduce org.lib.animal.x.X for
animal X, which:
(a) is redundant in terms of naming
(b) redundant in function since we have a single package for
each animal containing nothing but a single class of
the same name

Clearly, (1) is bad due to file/source structure reasons, and (2) is
bad for module organizational reasons. So we are back to my original
wish - I want to separate the two, so that I can solve (1)
indepeendently of (2).

Now, I realize that __init__.py can contain arbitrary code, and that
one can override __import__. However, I do not want to resort to
"hacks" just to solve this problem; I would prefer some established
convention in the community, or at least something that is elegant.

Does are people's thoughts on this problem?

Let me just shoot down one possible suggestion right away, to show you
what I am trying to accomplish:

I do *not* want to simply break out X into org.lib.animal.x, and have
org.lib.animal import org.lib.animal.x.X as X. While this naively
solves the problem of being able to refer to X as org.lib.animal.X,
the solution is anything but consistent because the *identity* of X is
still org.lib.animal.x.X. Examples of way this breaks things:

* X().__class__.__name__ gives unexpected results.
* Automatically generated documentation will document using the "real"
package name.
* Moving the *actual* classes around by way of this aliasing would
break things like pickled data structure as a result of the change
of actual identity, unless one *always* pre-emptively maintains
this shadow hierarchy (which is a problem in and of itself).

Thus, it's not clean. It breaks the module abstraction and as a result
has unintended consequences. I am looking for some kind of clean
solution. What do people do about this in practice?

[1] Optionally, we might introduce an "animals" package such that it
would become org.lib.animal.animals.Monkey, if we thought we were
going to have a lot of public API outside of the animals themselves.
This does not affect this dicussion however, as the exact same thing
would apply to org.lib.animal.animals as applies to org.lib.animal in
the above example.

[2] Ignoring for now that it may not be realistic that every animal
implementation would be that long; in many cases a lot of code would
be in common. But feel free to substitude for something else (a Zoo
say).

--
/ Peter Schuller

PGP userID: 0xE9758B7D or 'Peter Schuller <peter.s...@infidyne.com>'
Key retrieval: Send an E-Mail to getp...@scode.org
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Marc 'BlackJack' Rintsch

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Jan 23, 2008, 5:55:03 AM1/23/08
to
On Wed, 23 Jan 2008 03:49:56 -0600, Peter Schuller wrote:

> Let me just shoot down one possible suggestion right away, to show you
> what I am trying to accomplish:
>
> I do *not* want to simply break out X into org.lib.animal.x, and have
> org.lib.animal import org.lib.animal.x.X as X.

Then you shoot down the idiomatic answer I guess. That's what most people
do.

Ciao,
Marc 'BlackJack' Rintsch

Ben Finney

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Jan 23, 2008, 4:32:14 PM1/23/08
to
Peter Schuller <peter.s...@infidyne.com> writes:

> Let me just shoot down one possible suggestion right away, to show
> you what I am trying to accomplish:
>
> I do *not* want to simply break out X into org.lib.animal.x, and
> have org.lib.animal import org.lib.animal.x.X as X.

Nevertheless, that seems the best (indeed, the Pythonic) solution to
your problem as stated. Rather than just shooting it down, we'll have
to know more about ehat actual problem you're trying to solve to
understand why this solution doesn't fit.

> While this naively solves the problem of being able to refer to X as
> org.lib.animal.X, the solution is anything but consistent because
> the *identity* of X is still org.lib.animal.x.X.

The term "identity" in Python means something separate from this
concept; you seem to mean "the name of X".

> Examples of way this breaks things:
>
> * X().__class__.__name__ gives unexpected results.

Who is expecting them otherwise, and why is that a problem?

> * Automatically generated documentation will document using the
> "real" package name.

Here I lose all track of what problem you're trying to solve. You want
the documentation to say exactly where the class "is" (by name), but
you don't want the class to actually be defined at that location? I
can't make sense of that, so probably I don't understand the
requirement.

--
\ "If it ain't bust don't fix it is a very sound principle and |
`\ remains so despite the fact that I have slavishly ignored it |
_o__) all my life." —Douglas Adams |
Ben Finney

Peter Schuller

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Jan 24, 2008, 2:16:51 AM1/24/08
to
>> I do *not* want to simply break out X into org.lib.animal.x, and
>> have org.lib.animal import org.lib.animal.x.X as X.
>
> Nevertheless, that seems the best (indeed, the Pythonic) solution to
> your problem as stated. Rather than just shooting it down, we'll have
> to know more about ehat actual problem you're trying to solve to
> understand why this solution doesn't fit.

That is exactly what my original post was trying very hard to
explain. The problem is the discrepancy that I described between the
organization desired in terms of file system structure, and the
organization required in terms of module hierarchy. The reason it is a
problem is that, by default, there is an (in my opinion) too strong
connection between file system structure and module hierarchy in
Python.

>> While this naively solves the problem of being able to refer to X as
>> org.lib.animal.X, the solution is anything but consistent because
>> the *identity* of X is still org.lib.animal.x.X.
>
> The term "identity" in Python means something separate from this
> concept; you seem to mean "the name of X".

Not necessarily. In part it is the name, in that __name__ will be
different. But to the extent that calling code can potentially import
them under differents names, it's identity. Because importing the same
module under two names results in two distinct modules (two distinct
module objects) that have no realation with each other. So for
example, if a module has a single global protected by a mutex, there
are suddenly two copies of that. In short: identity matters.

>> Examples of way this breaks things:
>>
>> * X().__class__.__name__ gives unexpected results.
>
> Who is expecting them otherwise, and why is that a problem?

Depends on situation. One example is that if your policy is that
instances log using a logger named by the fully qualified name of the
class, than someone importing and using x.y.z.Class will expect to be
able to grep for x.y.z.Class in the output of the log file.

>> * Automatically generated documentation will document using the
>> "real" package name.
>
> Here I lose all track of what problem you're trying to solve. You want
> the documentation to say exactly where the class "is" (by name), but
> you don't want the class to actually be defined at that location? I
> can't make sense of that, so probably I don't understand the
> requirement.

You are baffled that what I seem to want is that the definition of the
class (file on disk) be different from the location inferred by the
module name. Well, this is *exactly* what I want because, like I said,
I do not want the strong connection beteween file system structure and
module hierarchy. The fact that this connection exists, is what is
causing my problems.

Please note that this is not any kind of crazy-brained idea; lots of
languages have absolutely zero relationship between file location and
modules/namespaces.

I realize that technically Python does not have this either. Like I
said in the original post, I do realize that I can override __import__
with any arbitrary function, and/or do magic in __init__. But I also
did not want to resort to hacks, and would prefer that there be some
kind of well-established solution to the problem.

Although I was originally hesitant to use an actual example for fear
of giving the sense that I was trying to start a language war, your
answer above prompts me to do so anyway, to show in concrete terms
what I mean, for those that wonder why/how it would work.

So for example, in Ruby, there is no problem having:

File monkey.rb:

module Org
module Lib
module Animal
class Monkey ...
..
end
end
end
end

File tiger.rb:

module Org
module Lib
module Animal
class Tiger ...
..
end
end
end
end

This is possible because the act of addressing code to be loaded into
the interpreter is not connected to the namespace/module system, but
rather to the file system.

Some languages avoid (but does not eliminate) the problem I am having
without having this disconnect. For example, Java does have a strong
connection between file system structure and class names. However the
critical difference is that in Java, everything is modeled around
classes, and class names map directly to the file system structure. So
in Java, you would have the class

org.lib.animal.Monkey

in

<wherever>/org/lib/animal/Monkey.java

and

org.lib.animal.Tiger

in

<wherever>/org/lib/animal/Tiger.java

In other words, introducing a separate file does not introduce a new
package. This works well as long as you are fine with having
everything related to a class in the same file.

The problem is that with Python, everything is not a classes, and a
file translates to a module, not a class. So you cannot have your
source in different files without introducing as many packages as you
introduce files.

Carl Banks

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Jan 24, 2008, 3:01:02 AM1/24/08
to
On Jan 23, 4:49 am, Peter Schuller <peter.schul...@infidyne.com>
wrote:

> I do *not* want to simply break out X into org.lib.animal.x, and have
> org.lib.animal import org.lib.animal.x.X as X. While this naively
> solves the problem of being able to refer to X as org.lib.animal.X,
> the solution is anything but consistent because the *identity* of X is
> still org.lib.animal.x.X. Examples of way this breaks things:
>
> * X().__class__.__name__ gives unexpected results.
> * Automatically generated documentation will document using the "real"
> package name.
> * Moving the *actual* classes around by way of this aliasing would
> break things like pickled data structure as a result of the change
> of actual identity, unless one *always* pre-emptively maintains
> this shadow hierarchy (which is a problem in and of itself).


You can reassign the class's module:

from org.lib.animal.monkey import Monkey
Monkey.__module__ = 'org.lib.animal'


(Which, I must admit, is not a bad idea in some cases.)


Carl Banks

Gabriel Genellina

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Jan 24, 2008, 3:17:12 AM1/24/08
to pytho...@python.org
En Thu, 24 Jan 2008 05:16:51 -0200, Peter Schuller
<peter.s...@infidyne.com> escribió:

>>> I do *not* want to simply break out X into org.lib.animal.x, and
>>> have org.lib.animal import org.lib.animal.x.X as X.
>>

>>> While this naively solves the problem of being able to refer to X as
>>> org.lib.animal.X, the solution is anything but consistent because
>>> the *identity* of X is still org.lib.animal.x.X.
>>
>> The term "identity" in Python means something separate from this
>> concept; you seem to mean "the name of X".
>
> Not necessarily. In part it is the name, in that __name__ will be
> different. But to the extent that calling code can potentially import
> them under differents names, it's identity. Because importing the same
> module under two names results in two distinct modules (two distinct
> module objects) that have no realation with each other. So for
> example, if a module has a single global protected by a mutex, there
> are suddenly two copies of that. In short: identity matters.

That's not true. It doesn't matter if you Import a module several times
at different places and with different names, it's always the same module
object.

py> from xml.etree import ElementTree
py> import xml.etree.ElementTree as ET2
py> import xml.etree
py> ET3 = getattr(xml.etree, 'ElementTree')
py> ElementTree is ET2
True
py> ET2 is ET3
True

Ok, there is one exception: the main script is loaded as __main__, but if
you import it using its own file name, you get a duplicate module.
You could confuse Python adding a package root to sys.path and doing
imports from inside that package and from the outside with different
names, but... just don't do that!

> I realize that technically Python does not have this either. Like I
> said in the original post, I do realize that I can override __import__
> with any arbitrary function, and/or do magic in __init__. But I also
> did not want to resort to hacks, and would prefer that there be some
> kind of well-established solution to the problem.

I don't really understand what your problem is exactly, but I think you
don't require any __import__ magic or arcane hacks. Perhaps the __path__
package attribute may be useful to you. You can add arbitrary directories
to this list, which are searched for submodules of the package. This way
you can (partially) decouple the file structure from the logical package
structure. But I don't think it's a good thing...

> in Java, you would have the class
>
> org.lib.animal.Monkey
>
> in
>
> <wherever>/org/lib/animal/Monkey.java
>
> and
>
> org.lib.animal.Tiger
>
> in
>
> <wherever>/org/lib/animal/Tiger.java
>
> In other words, introducing a separate file does not introduce a new
> package. This works well as long as you are fine with having
> everything related to a class in the same file.
>
> The problem is that with Python, everything is not a classes, and a
> file translates to a module, not a class. So you cannot have your
> source in different files without introducing as many packages as you
> introduce files.

Isn't org.lib.animal a package, reflected as a directory on disk? That's
the same both for Java and Python. Monkey.py and Tiger.py would be modules
inside that directory, just like Monkey.java and Tiger.java. Aren't the
same thing?

--
Gabriel Genellina

Peter Schuller

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Jan 24, 2008, 8:57:49 AM1/24/08
to
>> Not necessarily. In part it is the name, in that __name__ will be
>> different. But to the extent that calling code can potentially import
>> them under differents names, it's identity. Because importing the same
>> module under two names results in two distinct modules (two distinct
>> module objects) that have no realation with each other. So for
>> example, if a module has a single global protected by a mutex, there
>> are suddenly two copies of that. In short: identity matters.
>
> That's not true. It doesn't matter if you Import a module several times
> at different places and with different names, it's always the same module
> object.

Sorry, this is all my stupidity. I was being daft. When I said
importing under different names, I meant exactly that. As in, applying
hacks to import a module under a different name by doing it relative
to a different root directory. This is however not what anyone is
suggesting in this discussion. I got my wires crossed. I fully
understand that "import x.y.z" or "import x.y.z as B", and so one do
not affect the identity of the module.

> Ok, there is one exception: the main script is loaded as __main__, but if
> you import it using its own file name, you get a duplicate module.
> You could confuse Python adding a package root to sys.path and doing
> imports from inside that package and from the outside with different
> names, but... just don't do that!

Right :)

> I don't really understand what your problem is exactly, but I think you
> don't require any __import__ magic or arcane hacks. Perhaps the __path__
> package attribute may be useful to you. You can add arbitrary directories
> to this list, which are searched for submodules of the package. This way
> you can (partially) decouple the file structure from the logical package
> structure. But I don't think it's a good thing...

That sounds useful if I want to essentially put the contents of a
directory somewhere else, without using a symlink. In this case my
problem is more related to the "file == module" and "directory ==
module" semantics, since I want to break contents in a single module
out into several files.

> Isn't org.lib.animal a package, reflected as a directory on disk? That's
> the same both for Java and Python. Monkey.py and Tiger.py would be modules
> inside that directory, just like Monkey.java and Tiger.java. Aren't the
> same thing?

No, because in Java Monkey.java is a class. So we have class Monkey in
package org.lib.animal. In Python we would have class Monkey in module
org.lib.animal.monkey, which is redundant and does not reflect the
intended hierarchy. I have to either live with this, or put Monkey in
.../animal/__init__.py. Neither option is what I would want, ideally.

Java does still suffer from the same problem since it forces "class ==
file" (well, "public class == file"). However it is less of a problem
since you tend to want to keep a single class in a single file, while
I have a lot more incentive to split up a module into different files
(because you may have a lot of code hiding behind the public interface
of a module).

So essentially, Java and Python have the same problem, but certain
aspects of Java happens to mitigate the effects of it. Languages like
Ruby do not have the problem at all, because the relationship between
files and modules is non-existent.

Gabriel Genellina

unread,
Jan 25, 2008, 5:03:55 PM1/25/08
to pytho...@python.org
En Thu, 24 Jan 2008 11:57:49 -0200, Peter Schuller
<peter.s...@infidyne.com> escribió:

> In this case my


> problem is more related to the "file == module" and "directory ==
> module" semantics, since I want to break contents in a single module
> out into several files.

You already can do that, just import the public interfase of those several
files onto the desired container module. See below for an example.

>> Isn't org.lib.animal a package, reflected as a directory on disk? That's
>> the same both for Java and Python. Monkey.py and Tiger.py would be
>> modules
>> inside that directory, just like Monkey.java and Tiger.java. Aren't the
>> same thing?
>
> No, because in Java Monkey.java is a class. So we have class Monkey in
> package org.lib.animal. In Python we would have class Monkey in module
> org.lib.animal.monkey, which is redundant and does not reflect the
> intended hierarchy. I have to either live with this, or put Monkey in
> .../animal/__init__.py. Neither option is what I would want, ideally.

You can also put, in animal/__init__.py:
from monkey import Monkey
and now you can refer to it as org.lib.animal.Monkey, but keep the
implementation of Monkey class and all related stuff into
.../animal/monkey.py

--
Gabriel Genellina

Ben Finney

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Jan 25, 2008, 6:45:38 PM1/25/08
to
"Gabriel Genellina" <gags...@yahoo.com.ar> writes:

> You can also put, in animal/__init__.py:
> from monkey import Monkey
> and now you can refer to it as org.lib.animal.Monkey, but keep the
> implementation of Monkey class and all related stuff into
> .../animal/monkey.py

This (as far as I can understand) is exactly the solution the original
poster desired to "shoot down", for reasons I still don't understand.

--
\ "Reichel's Law: A body on vacation tends to remain on vacation |
`\ unless acted upon by an outside force." -- Carol Reichel |
_o__) |
Ben Finney

Carl Banks

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Jan 25, 2008, 7:48:38 PM1/25/08
to
On Jan 25, 6:45 pm, Ben Finney <bignose+hates-s...@benfinney.id.au>
wrote:

> "Gabriel Genellina" <gagsl-...@yahoo.com.ar> writes:
> > You can also put, in animal/__init__.py:
> > from monkey import Monkey
> > and now you can refer to it as org.lib.animal.Monkey, but keep the
> > implementation of Monkey class and all related stuff into
> > .../animal/monkey.py
>
> This (as far as I can understand) is exactly the solution the original
> poster desired to "shoot down", for reasons I still don't understand.

Come on, the OP explained it quite clearly in his original post. Did
you guys even read it?

The module where org.lib.animal.Monkey is actually defined should be
an implementation detail of the library, but simply importing Monkey
into org.lib.animal doesn't quite make it one.

If a user pickles a Monkey class, and then the OP decides to refactor
the Monkey class into a new module (say
org.lib.animal.primate.monkey), then the user would not be able to
unpickle it. Because, you see, pickles record the module a class is
defined in. So, now the user has to worry about where Monkey is
actually defined. It is not an implementation detail.

The solution is to modify the class's __module__ attribute as well as
importing it, as I've already pointed out:

from org.lib.animal.monkey import Monkey
Monkey.__module__ = 'org.lib.animal'

This should be enough to satisfy the OP's requirements, at least for
classes, without softening the one-to-one module-to-file relationship,
or using "hacks".

In fact, I'd say this is good practice.


Carl Banks

Ben Finney

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Jan 26, 2008, 6:47:52 PM1/26/08
to
Carl Banks <pavlove...@gmail.com> writes:

> On Jan 25, 6:45 pm, Ben Finney <bignose+hates-s...@benfinney.id.au>
> wrote:
> > "Gabriel Genellina" <gagsl-...@yahoo.com.ar> writes:
> > > You can also put, in animal/__init__.py:
> > > from monkey import Monkey
> > > and now you can refer to it as org.lib.animal.Monkey, but keep the
> > > implementation of Monkey class and all related stuff into
> > > .../animal/monkey.py
> >
> > This (as far as I can understand) is exactly the solution the
> > original poster desired to "shoot down", for reasons I still don't
> > understand.
>

> The solution is to modify the class's __module__ attribute as well as
> importing it, as I've already pointed out:
>
> from org.lib.animal.monkey import Monkey
> Monkey.__module__ = 'org.lib.animal'

Thanks, that makes it clear.

> This should be enough to satisfy the OP's requirements, at least for
> classes, without softening the one-to-one module-to-file
> relationship, or using "hacks".
>
> In fact, I'd say this is good practice.

I've not seen that before, but it seems an elegant way to address what
the OP is asking for.

--
\ "Madness is rare in individuals, but in groups, parties, |
`\ nations and ages it is the rule." -- Friedrich Nietzsche |
_o__) |
Ben Finney

Peter Schuller

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Jan 29, 2008, 7:48:59 AM1/29/08
to
> You can also put, in animal/__init__.py:
> from monkey import Monkey
> and now you can refer to it as org.lib.animal.Monkey, but keep the
> implementation of Monkey class and all related stuff into
> .../animal/monkey.py

The problem is that we are now back to the identity problem. The class
won't actually *BE* org.lib.animal.Monkey. Perhaps manipulating
__module__ is enough; perhaps not (for example, what about
sys.modules?). Looks like I'll just live with putting more than I
would like in the same file.

Peter Schuller

unread,
Jan 29, 2008, 7:51:58 AM1/29/08
to
> You can reassign the class's module:
>
> from org.lib.animal.monkey import Monkey
> Monkey.__module__ = 'org.lib.animal'
>
>
> (Which, I must admit, is not a bad idea in some cases.)

Is there a sense whether this is truly a supported way of doing this,
in terms of not running into various unintended side-effects? One
example would be sys.modules that I mentioned in the previous
post. Another, possibly related, might be interaction with the import
keyword and its implementation.

I will probably have to read up more on the semantics of __import__
and related machinery.

Carl Banks

unread,
Jan 29, 2008, 12:02:45 PM1/29/08
to
On Jan 29, 7:48 am, Peter Schuller <peter.schul...@infidyne.com>
wrote:

> > You can also put, in animal/__init__.py:
> > from monkey import Monkey
> > and now you can refer to it as org.lib.animal.Monkey, but keep the
> > implementation of Monkey class and all related stuff into
> > .../animal/monkey.py
>
> The problem is that we are now back to the identity problem. The class
> won't actually *BE* org.lib.animal.Monkey.

The usage is the same; it works in all cases once you redefine
__module__. Who cares what it really is?


> Perhaps manipulating
> __module__ is enough; perhaps not (for example, what about
> sys.modules?).

It's enough. It satisfies the criteria you listed. sys.modules has
nothing to do with it. Monkey is a class, not a module.

If you set __module__, the only remaining discernable difference is
that the global variables accessed from the Monkey class will be in
org.lib.animal.monkey instead of org.lib.animal. This has no ill
effects when unpickling or instantiating the class from
org.lib.animal.

> Looks like I'll just live with putting more than I
> would like in the same file.

Whatever. ISTM you came here looking for a particular means and not a
particular end. Python already has the power to meet your stated
needs, but you won't use that solution because it's "hacky".
Apparently all you really wanted was the loosened file structure in
the first place.


Carl Banks

Robert Kern

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Jan 29, 2008, 2:44:33 PM1/29/08
to pytho...@python.org
Carl Banks wrote:
> On Jan 29, 7:48 am, Peter Schuller <peter.schul...@infidyne.com>
> wrote:
>>> You can also put, in animal/__init__.py:
>>> from monkey import Monkey
>>> and now you can refer to it as org.lib.animal.Monkey, but keep the
>>> implementation of Monkey class and all related stuff into
>>> .../animal/monkey.py
>> The problem is that we are now back to the identity problem. The class
>> won't actually *BE* org.lib.animal.Monkey.
>
> The usage is the same; it works in all cases once you redefine
> __module__. Who cares what it really is?

The inspect module.

[animals]$ ls
animals
[animals]$ rm animals/*.pyc
[animals]$ ls
animals
[animals]$ ls animals
__init__.py monkey.py
[animals]$ cat animals/monkey.py
class Monkey(object):
pass
[animals]$ cat animals/__init__.py
from animals.monkey import Monkey
Monkey.__module__ = 'animals'
[animals]$ python
Python 2.5.1 (r251:54869, Apr 18 2007, 22:08:04)
[GCC 4.0.1 (Apple Computer, Inc. build 5367)] on darwin
Type "help", "copyright", "credits" or "license" for more information.
>>> from animals import Monkey
>>> import inspect
>>> inspect.getsource(Monkey)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File
"/Library/Frameworks/Python.framework/Versions/2.5/lib/python2.5/inspect.py",
line 629, in getsource
lines, lnum = getsourcelines(object)
File
"/Library/Frameworks/Python.framework/Versions/2.5/lib/python2.5/inspect.py",
line 618, in getsourcelines
lines, lnum = findsource(object)
File
"/Library/Frameworks/Python.framework/Versions/2.5/lib/python2.5/inspect.py",
line 494, in findsource
raise IOError('could not find class definition')
IOError: could not find class definition
>>>

--
Robert Kern

"I have come to believe that the whole world is an enigma, a harmless enigma
that is made terrible by our own mad attempt to interpret it as though it had
an underlying truth."
-- Umberto Eco

Peter Schuller

unread,
Jan 30, 2008, 4:31:16 AM1/30/08
to
>> The problem is that we are now back to the identity problem. The class
>> won't actually *BE* org.lib.animal.Monkey.
>
> The usage is the same; it works in all cases once you redefine
> __module__. Who cares what it really is?

The cases I listed were just examples. My point was that I wanted it
to *be* the right class, to avoid unintended consequences. If I knew
what all those possible consequences were, there would not be a
problem to begin with.

The other follow-up to your E-Mail points out a possible problem for
example. I would not have come up with that, but that does not mean
the effect does not exist. And committing to a solution that "seems to
work", only to break massively for some particular use case in the
future, is exactly why I don't want a "hack" for a solution.

I don't know Python internals enough to state of believe with any
authority wither, let's say, stomping __module__ and hacking
sys.modules would be enough to *truly* do it correctly in a proper way
such that it is entirely transparent. This is why I care about whether
it truly changes the real identity of the class; it's not about
satisfying my particular list of examples (because they *were* just
examples).

> Whatever. ISTM you came here looking for a particular means and not a
> particular end.

My particular preferred end is to be able to separate file hierarchy
from module hierarchy without causing unforseen consequences. This was
the stated goal all along.

> Python already has the power to meet your stated
> needs, but you won't use that solution because it's "hacky".
> Apparently all you really wanted was the loosened file structure in
> the first place.

Yes, or failing that an alternative that mitigates the problem. And it
*is* hacky, in my opinion, if things break as a result of it (such as
the other poster's inspect example).

Steven D'Aprano

unread,
Jan 30, 2008, 9:00:19 AM1/30/08
to
On Tue, 29 Jan 2008 13:44:33 -0600, Robert Kern wrote:

> Carl Banks wrote:
>> On Jan 29, 7:48 am, Peter Schuller <peter.schul...@infidyne.com> wrote:
>>>> You can also put, in animal/__init__.py:
>>>> from monkey import Monkey
>>>> and now you can refer to it as org.lib.animal.Monkey, but keep the
>>>> implementation of Monkey class and all related stuff into
>>>> .../animal/monkey.py
>>> The problem is that we are now back to the identity problem. The class
>>> won't actually *BE* org.lib.animal.Monkey.
>>
>> The usage is the same; it works in all cases once you redefine
>> __module__. Who cares what it really is?
>
> The inspect module.

[snip example]

I call that a bug in the inspect module. In fact, looking at the source
for the findsource() function, I can see no fewer than two bugs, just in
the way it handles classes:

(1) it assumes that the only way to create a class is with a class
statement, which is wrong; and

(2) it assumes that the first occurrence of "class <name>" must be the
correct definition, which is also wrong.


It isn't hard to break the inspect module. Here's an example:


>>> import broken
>>> import inspect
>>> lines, lineno = inspect.findsource(broken.Parrot)
>>> lines[lineno]
'class Parrot which will be defined later.\n'
>>>
>>> lines, lineno = inspect.findsource(broken.Wensleydale)
>>> lines[lineno]
'class Wensleydale: # THIS IS GONE\n'

Here's the source of broken.py:


$ cat broken.py
"""Here is a doc string, where I happen to discuss the
class Parrot which will be defined later.
"""
class Parrot:
pass

class Wensleydale: # THIS IS GONE
pass

del Wensleydale
class Wensleydale(object): # but this exists
pass

It isn't often that I would come right out and say that part of the
Python standard library is buggy, but this is one of those cases.


--
Steven

Steven D'Aprano

unread,
Jan 30, 2008, 9:06:26 AM1/30/08
to
On Tue, 29 Jan 2008 06:48:59 -0600, Peter Schuller wrote:

>> You can also put, in animal/__init__.py:
>> from monkey import Monkey
>> and now you can refer to it as org.lib.animal.Monkey, but keep the
>> implementation of Monkey class and all related stuff into
>> .../animal/monkey.py
>
> The problem is that we are now back to the identity problem. The class
> won't actually *BE* org.lib.animal.Monkey.

It what sense will it not be? Why do you care so much about where the
source code for Monkey is defined? If you actually want to read the
source, you might need to follow the chain from "animal", see that Monkey
is imported from "monkey", and go look at that. But the rest of the time,
why would you care?

There is a very good reason to care *in practice*: if there is code out
there that assumes that the source code from Monkey is in the file it was
found in. In practice, you might be stuck needing to work around that.
But that's not a good reason to care *in principle*. In principle, the
actual location of the source code should be an implementation detail of
which we care nothing. It's possible that the source for Monkey doesn't
exist *anywhere*.

It is important to deal with buggy tools. But the correct way to do so is
to fix the bugs, not to throw away perfectly good abstractions.

--
Steven

Gabriel Genellina

unread,
Jan 30, 2008, 9:27:29 AM1/30/08
to
On 30 ene, 12:00, Steven D'Aprano <st...@REMOVE-THIS-
cybersource.com.au> wrote:

> I call that a bug in the inspect module. In fact, looking at the source
> for the findsource() function, I can see no fewer than two bugs, just in
> the way it handles classes:
>
> (1) it assumes that the only way to create a class is with a class
> statement, which is wrong; and
>
> (2) it assumes that the first occurrence of "class <name>" must be the
> correct definition, which is also wrong.

Yes, it's broken. But I'm afraid that's the only available thing to
do.
Python stores filename and line number information in code objects
(only). If you have a reference to any code object (a method, a
function, a traceback...) inspect can use it to retrieve that
information.
Once a class is defined, there is no code object attached to it. (The
class statement is executed when the module is loaded and initialized,
but that code object is discarded afterwards because it's not required
anymore).
If you *know* that a certain method is defined in a class, you can use
it to find the real module. But in general, there is nothing to start
with.
I'm eagerly waiting for someone to come and say I'm wrong...

--
Gabriel Genellina

Carl Banks

unread,
Jan 30, 2008, 12:02:52 PM1/30/08
to
On Jan 30, 4:31 am, Peter Schuller <peter.schul...@infidyne.com>
wrote:

> I don't know Python internals enough to state of believe with any
> authority wither, let's say, stomping __module__ and hacking
> sys.modules would be enough to *truly* do it correctly in a proper way
> such that it is entirely transparent. This is why I care about whether
> it truly changes the real identity of the class; it's not about
> satisfying my particular list of examples (because they *were* just
> examples).

Well, all I will say is that many people on this list, myself
included, do know Python internals, and we use the method we've been
suggesting here, without problems.

I think you're slipping to a level of paranoia that's more harmful
that helpful now.


The ironic thing is, breaking the one-to-one module-to-file
relationship is more likely to have "unintended consequences", by a
very large margin. Python has always been one-to-one module-to-file
(excepting modules built into the interpretter), and many codes and
tools have come to depend on it.


Carl Banks

Peter Schuller

unread,
Jan 30, 2008, 6:02:21 PM1/30/08
to
> Well, all I will say is that many people on this list, myself
> included, do know Python internals, and we use the method we've been
> suggesting here, without problems.

Ok. That is useful to know (that it is being done in practice without
problems).

Thanks!

Peter Schuller

unread,
Jan 30, 2008, 6:04:47 PM1/30/08
to
> It what sense will it not be? Why do you care so much about where the
> source code for Monkey is defined? If you actually want to read the
> source, you might need to follow the chain from "animal", see that Monkey
> is imported from "monkey", and go look at that. But the rest of the time,
> why would you care?
>
> There is a very good reason to care *in practice*: if there is code out
> there that assumes that the source code from Monkey is in the file it was
> found in. In practice, you might be stuck needing to work around that.
> But that's not a good reason to care *in principle*. In principle, the
> actual location of the source code should be an implementation detail of
> which we care nothing. It's possible that the source for Monkey doesn't

Exactly. I *DON'T* want anything to depend on the physical location on disk.
That was exactly what I was after from the beginning; a total separation of
location on disk from the location in the module hiearachy. As you say, the
location of the source should be an implementation detail. That is exactly
what I am after.

I'll have a closer look at the suggested practice of modifying __module__.

For this particular use case we probably won't end up doing that, but it
may come to be useful in the future.

Ben Finney

unread,
Jan 30, 2008, 8:31:29 PM1/30/08
to
Peter Schuller <peter.s...@infidyne.com> writes:

> I *DON'T* want anything to depend on the physical location on disk.

Importing the code in the first place will — unavoidably, it seems to
me — depend on the file location from which to load the module.

After that, nothing depends on the physical location on disk, unless
it's buggy. Imported modules are available from 'sys.modules', and
that's where subsequent 'import' statements will find them, with no
reference to file locations.

> That was exactly what I was after from the beginning; a total
> separation of location on disk from the location in the module
> hiearachy. As you say, the location of the source should be an
> implementation detail. That is exactly what I am after.

It *is* an implementation detail, once the module is loaded from disk.

--
\ "Philosophy is questions that may never be answered. Religion |
`\ is answers that may never be questioned." —anonymous |
_o__) |
Ben Finney

NickC

unread,
Feb 4, 2008, 5:46:13 PM2/4/08
to
On Jan 31, 12:27 am, Gabriel Genellina <gagsl-...@yahoo.com.ar> wrote:
> Python stores filename and line number information in code objects
> (only). If you have a reference to any code object (a method, a
> function, a traceback...) inspect can use it to retrieve that
> information.

Aside from general concerns about increasing the size of class objects
(and most programs don't contain enough of those to make a big
difference) I don't immediately see anything that would prevent the
interpreter being modified to add file and line number information to
class objects as well. While the information wouldn't always be
present (types implemented in C, types created by calling the
metaclass constructor directly), it would help address the inspect
module bugs Steven illustrated.

I would agree with Carl that modifying __module__ in the way he
suggests is legitimate - if it breaks the inspect module, then it is
the inspect module that needs fixing (and/or better support from the
interpreter to help find the real source code).

Cheers,
Nick.

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