finding out the precision of floats
B.Ogryczak at gmail.com
Thu Mar 1 11:33:23 CET 2007
On Feb 28, 10:29 pm, "John Machin" <sjmac... at lexicon.net> wrote:
> On Mar 1, 4:19 am, "BartOgryczak" <B.Ogryc... at gmail.com> wrote:
> > On Feb 28, 3:53 pm, "John Machin" <sjmac... at lexicon.net> wrote:
> > > On Feb 28, 10:38 pm, "BartOgryczak" <B.Ogryc... at gmail.com> wrote:
> > > >  eg. consider calculating interests rate, which often is defined as
> > > > math.pow(anualRate,days/365.0).
> > > More importantly, the formula you give is dead wrong. The correct
> > > formula for converting an annual rate of interest to the rate of
> > > interest to be used for n days (on the basis of 365 days per year) is:
> > > (1 + annual_rate) ** (n / 365.0) - 1.0
> > > or
> > > math.pow(1 + annual_rate, n / 365.0) - 1.0
> > > if you prefer.
> > YPB? Anyone with half a brain knows, that you can either express rate
> > as 0.07 and do all those ridiculous conversions above, or express it
> > as 1.07 and apply it directly.
> A conversion involving an exponentiation is necessary. "All those"?? I
> see only two.
> Please re-read your original post, and note that there are *TWO* plus-
> or-minus 1.0 differences between your formula and mine. For an annual
> rate of 10%, yours would calculate the rate for 6 months (expressed as
> 182.5 days) as:
> math.pow(0.10, 0.5) = 0.316... i.e. 31.6%
You're assuming that I'd store annual rate as 0.1, while actually it'd
stored as 1.1.
Which is logical and applicable directly.
More information about the Python-list