Message134203
> it may be worth looking at what numpy does here.
... or it may not. NumPy just uses (approximation to 1/log(2)) * log(x) when log2 doesn't already exist. And indeed, on Windows:
Python 2.7.1 |EPD 7.0-2 (64-bit)| (r271:86832, Dec 2 2010, 10:23:25) [MSC v.1500 64 bit (AMD64)] on win32
Type "help", "copyright", "credits" or "license" for more information.
>>> import numpy
>>> numpy.log2(8.0)
2.9999999999999996
I think we should be able to do better than this. :-) |
|
Date |
User |
Action |
Args |
2011-04-21 09:16:31 | mark.dickinson | set | recipients:
+ mark.dickinson, rhettinger, jcea, vstinner |
2011-04-21 09:16:31 | mark.dickinson | set | messageid: <1303377391.11.0.971336768502.issue11888@psf.upfronthosting.co.za> |
2011-04-21 09:16:30 | mark.dickinson | link | issue11888 messages |
2011-04-21 09:16:30 | mark.dickinson | create | |
|