Ian Kelly <ian.g.ke...@gmail.com> writes: > Everybody should know the generic algorithm, though: > from itertools import chain ...
For n>0, assuming you just want the converted digits and not a string.
String conversion and minus sign for n<0 left as exercise. Note this
returns a generator that you can convert to a list with list(...).
def convert(n, base):
a,b = divmod(n,base)
if a > 0:
for e in convert(a,base):
yield e
yield b
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