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Getting module path string from a class instance

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  Getting module path string from a class instance Some Developer <someukdeveloper@gmail.com> - 2012-11-13 06:38 +0000
    Re: Getting module path string from a class instance Steven D'Aprano <steve+comp.lang.python@pearwood.info> - 2012-11-13 07:19 +0000
      Re: Getting module path string from a class instance Some Developer <someukdeveloper@gmail.com> - 2012-11-13 07:54 +0000
        Re: Getting module path string from a class instance Steven D'Aprano <steve+comp.lang.python@pearwood.info> - 2012-11-13 08:49 +0000
          Re: Getting module path string from a class instance Some Developer <someukdeveloper@gmail.com> - 2012-11-13 16:15 +0000

#33222 — Getting module path string from a class instance

I'm trying to find a way to get a string of the module path of a class.

So for instance say I have class Foo and it is in a module called 
my.module. I want to be able to get a string that is equal to this: 
"my.module.Foo". I'm aware of the __repr__ method but it does not do 
what I want it to do in this case.

Can anyone offer any advice at all?

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#33224

On Tue, 13 Nov 2012 06:38:31 +0000, Some Developer wrote:

> I'm trying to find a way to get a string of the module path of a class.
> 
> So for instance say I have class Foo and it is in a module called
> my.module. I want to be able to get a string that is equal to this:
> "my.module.Foo". I'm aware of the __repr__ method but it does not do
> what I want it to do in this case.
> 
> Can anyone offer any advice at all?

py> from multiprocessing.pool import Pool
py> repr(Pool)
"<class 'multiprocessing.pool.Pool'>"

Seems pretty close to what you ask for. You can either pull that string 
apart:

py> s = repr(Pool)
py> start = s.find("'")
py> end = s.rfind("'")
py> s[start+1:end]
'multiprocessing.pool.Pool'

or you can construct it yourself:

py> Pool.__module__ + '.' + Pool.__name__
'multiprocessing.pool.Pool'


-- 
Steven

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#33226

On 13/11/2012 07:19, Steven D'Aprano wrote:
> On Tue, 13 Nov 2012 06:38:31 +0000, Some Developer wrote:
>
>> I'm trying to find a way to get a string of the module path of a class.
>>
>> So for instance say I have class Foo and it is in a module called
>> my.module. I want to be able to get a string that is equal to this:
>> "my.module.Foo". I'm aware of the __repr__ method but it does not do
>> what I want it to do in this case.
>>
>> Can anyone offer any advice at all?
> py> from multiprocessing.pool import Pool
> py> repr(Pool)
> "<class 'multiprocessing.pool.Pool'>"
>
> Seems pretty close to what you ask for. You can either pull that string
> apart:
>
> py> s = repr(Pool)
> py> start = s.find("'")
> py> end = s.rfind("'")
> py> s[start+1:end]
> 'multiprocessing.pool.Pool'
>
> or you can construct it yourself:
>
> py> Pool.__module__ + '.' + Pool.__name__
> 'multiprocessing.pool.Pool'
>
>
Yeah I considered doing it this way but was wary of that method because 
of possible changes to the implementation of the __repr__ method in the 
upstream code. If the Django developers don't consider the __repr__ 
method a public API then it could change in the future breaking my code.

Of course this might not happen but I was hoping that there was a more 
generic way of doing it that did not rely on a certain implementation 
being in existence.

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#33228

On Tue, 13 Nov 2012 07:54:32 +0000, Some Developer wrote:

> On 13/11/2012 07:19, Steven D'Aprano wrote:
>> On Tue, 13 Nov 2012 06:38:31 +0000, Some Developer wrote:
>>
>>> I'm trying to find a way to get a string of the module path of a
>>> class.
>>>
>>> So for instance say I have class Foo and it is in a module called
>>> my.module. I want to be able to get a string that is equal to this:
>>> "my.module.Foo". I'm aware of the __repr__ method but it does not do
>>> what I want it to do in this case.
>>>
>>> Can anyone offer any advice at all?
>> py> from multiprocessing.pool import Pool py> repr(Pool)
>> "<class 'multiprocessing.pool.Pool'>"
>>
>> Seems pretty close to what you ask for. You can either pull that string
>> apart:
>>
>> py> s = repr(Pool)
>> py> start = s.find("'")
>> py> end = s.rfind("'")
>> py> s[start+1:end]
>> 'multiprocessing.pool.Pool'
>>
>> or you can construct it yourself:
>>
>> py> Pool.__module__ + '.' + Pool.__name__ 'multiprocessing.pool.Pool'
>>
>>
> Yeah I considered doing it this way but was wary of that method because
> of possible changes to the implementation of the __repr__ method in the
> upstream code. If the Django developers don't consider the __repr__
> method a public API then it could change in the future breaking my code.

I didn't call SomeClass.__repr__. That is an implementation detail of 
SomeClass, and could change.

I called repr(SomeClass), which calls the *metaclass* __repr__. That is 
less likely to change, although not impossible.


If you're worried, just use the second way:

SomeClass.__module__ + '.' + SomeClass.__name__
 


> Of course this might not happen but I was hoping that there was a more 
> generic way of doing it that did not rely on a certain implementation 
> being in existence.

SomeClass.__name__ is the official way to get the name of a class; 
SomeClass.__module__ is the official way to get the name of the module or 
package it comes from.


-- 
Steven

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#33248

On 13/11/2012 08:49, Steven D'Aprano wrote:
> On Tue, 13 Nov 2012 07:54:32 +0000, Some Developer wrote:
>
>> On 13/11/2012 07:19, Steven D'Aprano wrote:
>>> On Tue, 13 Nov 2012 06:38:31 +0000, Some Developer wrote:
>>>
>>>> I'm trying to find a way to get a string of the module path of a
>>>> class.
>>>>
>>>> So for instance say I have class Foo and it is in a module called
>>>> my.module. I want to be able to get a string that is equal to this:
>>>> "my.module.Foo". I'm aware of the __repr__ method but it does not do
>>>> what I want it to do in this case.
>>>>
>>>> Can anyone offer any advice at all?
>>> py> from multiprocessing.pool import Pool py> repr(Pool)
>>> "<class 'multiprocessing.pool.Pool'>"
>>>
>>> Seems pretty close to what you ask for. You can either pull that string
>>> apart:
>>>
>>> py> s = repr(Pool)
>>> py> start = s.find("'")
>>> py> end = s.rfind("'")
>>> py> s[start+1:end]
>>> 'multiprocessing.pool.Pool'
>>>
>>> or you can construct it yourself:
>>>
>>> py> Pool.__module__ + '.' + Pool.__name__ 'multiprocessing.pool.Pool'
>>>
>>>
>> Yeah I considered doing it this way but was wary of that method because
>> of possible changes to the implementation of the __repr__ method in the
>> upstream code. If the Django developers don't consider the __repr__
>> method a public API then it could change in the future breaking my code.
> I didn't call SomeClass.__repr__. That is an implementation detail of
> SomeClass, and could change.
>
> I called repr(SomeClass), which calls the *metaclass* __repr__. That is
> less likely to change, although not impossible.
>
>
> If you're worried, just use the second way:
>
> SomeClass.__module__ + '.' + SomeClass.__name__
>   

Ah, my mistake. Thanks. That sounds exactly like what I want.

>> Of course this might not happen but I was hoping that there was a more
>> generic way of doing it that did not rely on a certain implementation
>> being in existence.
> SomeClass.__name__ is the official way to get the name of a class;
> SomeClass.__module__ is the official way to get the name of the module or
> package it comes from.

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