in [Python]

From: David Robinow on 1 Apr 2010 23:08 On Thu, Apr 1, 2010 at 10:44 PM, Steven D'Aprano <steve (a)remove-this-cybersource.com.au> wrote:> On Thu, 01 Apr 2010 19:49:43 -0500, Tim Chase wrote: > >> David Robinow wrote: >>> $ python -c "print 1/2 * 1/2" >>> 0 >>> >>> But that's not what I learned in grade school. >>> (Maybe I should upgrade to 3.1?) >> >> That's because you need to promote one of them to a float so you get a >> floating-point result: >> >> >>> 1/2 * 1/2 >> 0 >> >>> 1/2 * 1/2.0 >> 0.0 >> >> Oh...wait ;-) > > Tim, I'm sure you know the answer to this, but for the benefit of the > Original Poster, the problem is that you need to promote *both* divisions > to floating point. Otherwise one of them will give int 0, which gives 0.0 > when multiplied by 0.5. > >>>> 1.0/2 * 1/2.0 > 0.25 > > > If you want an exact result when multiplying arbitrary fractions, you > need to avoid floats and decimals and use Fractions: > >>>> Fraction(1, 2)**2 > Fraction(1, 4) I should have known he wouldn't get it.
From: Tim Chase on 1 Apr 2010 23:34 Steven D'Aprano wrote: >> That's because you need to promote one of them to a float so you get a >> floating-point result: >> >> >>> 1/2 * 1/2 >> 0 >> >>> 1/2 * 1/2.0 >> 0.0 >> >> Oh...wait ;-) > > Tim, I'm sure you know the answer to this, but for the benefit of the > Original Poster, the problem is that you need to promote *both* divisions > to floating point. Otherwise one of them will give int 0, which gives 0.0 > when multiplied by 0.5. > >>>> 1.0/2 * 1/2.0 > 0.25 You can get away with just promoting one of them...you just have to promote the _correct_ one (one involved in the first division) so that its promotion-of-subresult-to-float carries into all subsequent operations/operators: >>> 1/2 * 1/2 # (((1/2)*1)/2)==(((0)*1)/2) in 2.x 0 >>> 1/2 * 1/2.0 # (((1/2)*1)/2.0)==(((0)*1)/2.0) in 2.x 0.0 >>> 1/2 * 1.0/2 # (((1/2)*1.0)/2)==(((0)*1.0)/2) in 2.x 0.0 >>> 1/2.0 * 1/2 # (((1/2.0)*1)/2) 0.25 >>> 1.0/2 * 1/2 # (((1.0/2)*1)/2) 0.25 I'd rather be explicit in *real* code that I'd write and explicitly float'ify constants or float() integer variables. The OP's question was both OT and pretty basic middle-school math that google would have nicely answered[1] so IMHO warranted a bit of fun. :) -tkc [1] http://www.google.com/search?q=1%2F2+*+1%2F2
From: Patrick Maupin on 2 Apr 2010 01:25 On Apr 1, 11:52 pm, Dennis Lee Bieber <wlfr... (a)ix.netcom.com> wrote:> On Thu, 01 Apr 2010 22:44:51 +0200, superpollo <ute... (a)esempio.net>> declaimed the following in gmane.comp.python.general: > > > how much is one half times one half? > > import math > print math.exp((math.log(1) - math.log(2)) > + (math.log(1) - math.log(2))) That's all well and good, but base 'e' is kind of complicated. Some of us were using base 10, and others took Tim's lead and were using base 2: >>> print math.exp(((math.log(1)/math.log(2) - math.log(2)/math.log(2)) + (math.log(1)/math.log(2) - math.log(2)/math.log(2)))*math.log(2)) 0.25
From: Steven D'Aprano on 2 Apr 2010 01:40 On Thu, 01 Apr 2010 22:34:46 -0500, Tim Chase wrote: >> Tim, I'm sure you know the answer to this, but for the benefit of the >> Original Poster, the problem is that you need to promote *both* >> divisions to floating point. Otherwise one of them will give int 0, >> which gives 0.0 when multiplied by 0.5. >> >>>>> 1.0/2 * 1/2.0 >> 0.25 > > You can get away with just promoting one of them...you just have to > promote the _correct_ one Doh! Of course you do. I knew that! -- Steven
From: Wanderer on 2 Apr 2010 10:06
On Apr 1, 7:34 pm, Patrick Maupin <pmau... (a)gmail.com> wrote:> On Apr 1, 4:42 pm, Tim Chase <python.l... (a)tim.thechases.com> wrote:> > Uh, did you try it at the python prompt? When I try it at the IPython prompt, I get Object 'how much is one half times one half' not found. |