NuclearVision hello... I was working on the decimal exercice. And came up with this: def dec(n):#trying to separate the integral part from the fraction part return float(n)-int(n) output: >>> dec(3.8) 0.7999999999999998 Any ideas?
Joe It has so much more than just a dedicated website. Floating number arithmetic is one of the oldest problem computers faced. The reason is simple: how do you represent 0.1 or 10/3 in binary? The website I link to is trying to provide a simple answer to this complex problem, that's all.
arithma When I face those issues, I usually decide beforehand the number of decimals I want to use. >>> print "%.2f" % (3.8-int(3.8)) 0.80
NuclearVision m0ei wroteYou can do something like this: def dec(n): return float(format(n - int(n), '.2f')) Thanks : )