Am Sonntag 23 August 2009 15:27:48 schrieb Steve: > On Sun, 2009-08-23 at 15:12 +0400, Eugene Kirpichov wrote: > > > There is *not* the same problem in Python: > > > $ python > > > Python 2.6.2 (r262:71600, Jul 9 2009, 23:16:53) > > > [GCC 4.4.0 20090506 (Red Hat 4.4.0-4)] on linux2 > > > Type "help", "copyright", "credits" or "license" for more information. > > > > > >>>> import math > > >>>> math.log10(1000) > > > > > > 3.0 > > > > > >>> import math > > >>> math.log(1000,10) > > > > 2.9999999999999996 > > That is surprising. I think its a Python bug - the results should be > consistent.
On the other hand, it is also desirable to have log(a,b) == log(a)/log(b). If you have a special implementation of log10, you must decide which consistency you want to break, log(a,10) == log10(a) or log(a,10) == log(a)/log(10). > > > > Recent work in Python 3 (and Python 2.6) has improved the handling of > > > floating point numbers, and addresses exactly the problem that Roberto > > > has raised. > > > > > > I see no reason why Haskell could not improve its handling of floating > > > point numbers by using similar techniques. > > > > You mean introducing a "log10" function into the definition of the > > Floating class ? That might be a proposal for Haskell Prime. > > I was not thinking of the log10 function. I was thinking of the changes > mentioned in > http://docs.python.org/3.1/whatsnew/3.1.html > where it says > "Python now uses David Gay’s algorithm for finding the shortest floating > point representation that doesn’t change its value. This should help > mitigate some of the confusion surrounding binary floating point > numbers." Note however that that only concerns the output, not the underlying value. It doesn't change the fact that log(1000)/log(10) (== log(1000,10)) < 3 in Python, too. > > Also, I had a problem using floating point in Python where > > >>> round(697.04157958254996, 10) > > gave > 697.04157958259998 > which is closer to > 697.0415795826 > than the desired result of > 697.0415795825 Well, 10^10*(697.04157958254996 :: Double) == 6970415795825.5, that gets rounded to 6970415795826 under both, round-half-up and round-half-to-even, and (6970415795826/10^10 :: Double) == 697.04157958259998 Though ghci displays it as 697.0415795826, unlike Python. > > Its been fixed in the latest versions of Python: > >>> round(697.04157958254996, 10) > > 697.0415795825 > > > I'm not sure what the equivalent in Haskell is. Is there a function for > rounding to a number of decimal digits ? > I came up with this: > > roundN :: Double -> Int -> Double > roundN n ndigits = fromIntegral (round $ n * m) / m > where m = 10 ^ ndigits > > ghci> roundN 697.04157958254996 10 > 697.0415795826 > > which is not the desired result. Prelude Numeric> let roundN :: Double -> Int -> Double; roundN x d = let { m = 10^(d+1); i = floor (m*x); r = i `mod` 10; j = if r < 5 then i-r else i-r+10 } in fromInteger j/m roundN :: Double -> Int -> Double roundN x d = fromInteger j / m where m = 10^(d+1) i = floor (m*x) r = i `mod` 10 j | r < 5 = i-r | otherwise = i+10-r ghci> roundN 697.04157958254996 10 697.0415795825 I haven't treated negative input, but, more importantly, floating point representation being what it is, there are cases where this doesn't do what you want/expect either. > > Steve _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
