convert floats to their 4 byte representation

[Tim Chase]

> 	s = "%"+str(size) + "X"
> 	return (s % number).replace(' ', '0')

While I don't have a fast and easy way to represent floats, you 
may want to tweak this to be

	return ("%0*X" % (size,number))

which will zero-pad the number in hex to "size" number of places 
in a single step.  It also helps prevent problems where there might

> but I haven't been able to find anything for floats.  Any help
> would be great.

My first stab at such an attempt:

 >>> from struct import pack, unpack
 >>> s = pack("d", 3.14)
 >>> i = unpack("q", s)
 >>> "%X"%i
'40091EB851EB851F'
 >>> def floatAsHex(f, size):
...     return "%0*X" % (size, unpack("q", pack("d", f))[0])
...
 >>> floatAsHex(3.14, 20)
'000040091EB851EB851F'

It's ugly, it's hackish, it's likely architecture-dependant, but 
it seems to do what you're describing.

-tkc