On Wed, Jun 15, 2011 at 10:29 PM, Olivier LEMAIRE <m.olivier.lema...@gmail.com> wrote: > b = str(bin(number))[2:] > if len(b) !=size: > b = (size-len(b))*"0"+b
You don't need the str() there as bin() already returns a number. Here's a relatively trivial simplification - although it does make the code more cryptic: b = (size*"0"+bin(number)[2:])[-size:] After typing this up, I saw Daniel's post, which is rather better. Go with that one for zero-filling; mine's more general though, you can pad with other tokens. ChrisA -- http://mail.python.org/mailman/listinfo/python-list
