[Numpy-discussion] compatibility for supporting more than 2 versions of numpy

[Bruce Southey]

On 01/17/2011 10:32 AM, josef.pktd at gmail.com wrote:
> On Mon, Jan 17, 2011 at 11:28 AM,<josef.pktd at gmail.com>  wrote:
>> On Sat, Jan 15, 2011 at 3:27 PM,<josef.pktd at gmail.com>  wrote:
>>> After upgrading to numpy 1.5.1 I got caught by some depreciated
>>> features. Given the depreciation policy of numpy, if we want to
>>> support more than two versions of numpy, then we need some conditional
>>> execution.
>>>
>>> Does anyone have any compatibility functions?
>>>
>>> I haven't looked at it carefully yet, but statsmodels might need
>>> things like the following if we want to support numpy 1.3
>>>
>>>     if np.__version__<  '1.5':
>>>         freq,hsupp = np.histogram(rvs, histsupp, new=True)
>>>     else:
>>>         freq,hsupp = np.histogram(rvs,histsupp)
>>>
>>> matplotlib says it supports numpy>=1.1 but I didn't see any
>>> compatibility code that I could "borrow".
>>> Or do I worry for nothing? The compatibility.py in statsmodels is
>>> still almost empty.
>>
>> for scipy.linalg, in numdifftools, I changed this in core (in my copy)
>>
>>         if numpy.__version__<  '1.5':
>>             [qromb,rromb] = linalg.qr(rmat, econ=True)
>>         else:
>>             [qromb,rromb] = linalg.qr(rmat, mode='economic')
>>
> which is of course silly, since this is for the scipy update
>>> Josef

Scipy release notes usually state the supported numpy version eg from 
the current 0.8.0 release notes
"This release requires Python 2.4 - 2.6 and NumPy 1.4.1 or greater."
Consequently if you want to support different numpy versions, then you 
will need to maintain your own branch with that type of patch. That can 
get rather complex to maintain.

It would be better that you change the code calling numpy/scipy 
functions rather than the functions themselves such as passing the 
appropriate *args and **kwargs to the function.

I would expect that a try/except block would be more general as well as 
numpy.__version__ being a str.


Bruce