On Monday, July 18, 2016 at 3:45:26 PM UTC+5:30, Chris Angelico wrote:
> On Mon, Jul 18, 2016 at 8:00 PM, Marko Rauhamaa wrote:
> > Python programmers (among others) frequently run into issues with
> > surprising results in floating-point arithmetics. For better or worse,
> > Scheme has tried to abstract the concept. You don't need to explain the
> > ideas of IEEE 64-bit floating-point numbers or tie the hands of the
> > implementation. Instead, what you have is "reliable" arithmetics and
> > "best-effort" arithmetics, a bit like TCP is "reliable" and UDP is
> > "best-effort".
>
> The problem with that is that failing to explain IEEE floating point
> and just calling it "inexact" scares people off unnecessarily. I've
> seen a lot of very intelligent people who think that you should never
> compare floats with the == operator, because floats randomly introduce
> "inaccuracy". And then you get these sorts of functions:
>
> EPSILON = 0.000001 # Adjust to control numeric accuracy
> def is_equal(f1, f2, epsilon=EPSILON):
> if abs(f1) > abs(f2):
> f1, f2 = f2, f1
> return abs(f2-f1) < f1*epsilon
>
> and interminable debates about how to pick an epsilon, whether it
> should be relative to the smaller value (as here) or the larger (use
> f2 instead), or maybe should be an absolute value, or maybe it should
> be relative to the largest/smallest value that was ever involved in
> the calculation, or........
>
I dont know what point you are trying to make
Here is behavior. Should one use == ??
Python 2.7.11+ (default, Apr 17 2016, 14:00:29)
[GCC 5.3.1 20160413] on linux2
Type "help", "copyright", "credits" or "license" for more information.
>>> .1+.1+.1 == .3
False
EPSILON = 0.000001 # Adjust to control numeric accuracy
def is_equal(f1, f2, epsilon=EPSILON):
if abs(f1) > abs(f2):
f1, f2 = f2, f1
return abs(f2-f1) < f1*epsilon
>>> ... ... ... ...
>>> is_equal(.1+.1+.1, .3)
True
>>>
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