> Subject: Re: [Tutor] flow problem with a exercise > From: evert....@gmail.com > Date: Sat, 21 Aug 2010 12:39:05 +0200 > CC: tutor@python.org > To: rwob...@hotmail.com > > > In [39]: t = 3 > > > > In [40]: round((t-32)/1.8) > > Out[40]: -16.0 > > > > In [41]: t = 3.0 > > > > In [42]: round((t-32)/1.8) > > Out[42]: -16.0 > > > > Works fine for me. > > > > Correct, > > But I see one wierd thing. > > > > round ((42-32)/1.8) gives a output -16.0 but (42-32)/1.8) gives also -16.0 > > I was expectting that round will give 16 as output because round (32.0) is > > the same as round (32.0, 0) so there will be 0 decimals. > > And I see one decimal. > > The rounding doesn't influence how it's printed. > From help(round): > > """ > round(...) > round(number[, ndigits]) -> floating point number > > Round a number to a given precision in decimal digits (default 0 digits). > This always returns a floating point number. Precision may be negative. > """ > > It *rounds* the number to ndigits, not prints. It always returns a floating > point number though (not an integer). > Since floats are normally represented with at least one trailing digit after > the decimal dot, you see a zero. > Eg, > >>> float(1) > 1.0 > > If you want to get rid of the trailing .0, convert it to an integer: > >>> int(round(1.1)) > 1 > > Also consider: > >>> round(1.15, 0) > 1.0 > >>> round(1.15, 1) > 1.2 > >>> round(1.15, 2) > 1.1499999999999999 > > > (Note again how the last rounded 1.15 is represented. It's just a > representation though, and showing the imprecision you intrinsicaly get when > dealing with floating point numbers.) > > If you really want different behaviour when printing (or using) floating > point numbers, perhaps have a look at the decimal module: > http://docs.python.org/library/decimal.html > > Oke, Thank you. Learned another thing about Python. Roelof
_______________________________________________ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
