[Python-Dev] [Tutor] nice()

Josiah Carlson jcarlson at uci.edu
Mon Feb 13 01:14:50 CET 2006


"Alan Gauld" <alan.gauld at freenet.co.uk> wrote:
> However I do dislike the name nice() - there is already a nice() in the
> os module with a fairly well understood function. But I'm sure some
> time with a thesaurus can overcome that single mild objection. :-)

Presumably it would be located somewhere like the math module.

 - Josiah


> Alan G
> Author of the learn to program web tutor
> http://www.freenetpages.co.uk/hp/alan.gauld
> 
> 
> 
> ----- Original Message ----- 
> From: "Smith" <smiles at worksmail.net>
> To: <tutor at python.org>
> Cc: <edu-sig at python.org>; <python-dev at python.org>
> Sent: Sunday, February 12, 2006 6:44 PM
> Subject: [Tutor] nice()
> 
> 
> I've been thinking about a function that was recently proposed at python-dev 
> named 'areclose'. It is a function that is meant to tell whether two (or 
> possible more) numbers are close to each other. It is a function similar to 
> one that exists in Numeric. One such implementation is
> 
> def areclose(x,y,abs_tol=1e-8,rel_tol=1e-5):
>     diff = abs(x-y)
>     return diff <= ans_tol or diff <= rel_tol*max(abs(x),abs(y))
> 
> (This is the form given by Scott Daniels on python-dev.)
> 
> Anyway, one of the rationales for including such a function was:
> 
>   When teaching some programming to total newbies, a common frustration
>   is how to explain why a==b is False when a and b are floats computed
>   by different routes which ``should'' give the same results (if
>   arithmetic had infinite precision).  Decimals can help, but another
>   approach I've found useful is embodied in Numeric.allclose(a,b) --
>   which returns True if all items of the arrays are ``close'' (equal to
>   within certain absolute and relative tolerances)
> The problem with the above function, however, is that it *itself* has a 
> comparison between floats and it will give undesired result for something 
> like the following test:
> 
> ###
> >>> print areclose(2, 2.1, .1, 0) #see if 2 and 2.1 are within 0.1 of each 
> >>> other
> False
> >>>
> ###
> 
> Here is an alternative that might be a nice companion to the repr() and 
> round() functions: nice(). It is a combination of Tim Peter's delightful 
> 'case closed' presentation in the thread, "Rounding to n significant 
> digits?" [1] and the hidden magic of "prints" simplification of floating 
> point numbers when being asked to show them.
> 
> It's default behavior is to return a number in the form that the number 
> would have when being printed. An optional argument, however, allows the 
> user to specify the number of digits to round the number to as counted from 
> the most significant digit. (An alternative name, then, could be 'lround' 
> but I think there is less baggage for the new user to think about if the 
> name is something like nice()--a function that makes the floating point 
> numbers "play nice." And I also think the name...sounds nice.)
> 
> Here it is in action:
> 
> ###
> >>> 3*1.1==3.3
> False
> >>> nice(3*1.1)==nice(3.3)
> True
> >>> x=3.21/0.65; print x
> 4.93846153846
> >>> print nice(x,2)
> 4.9
> >>> x=x*1e5; print nice(x,2)
> 490000.0
> ###
> 
> Here's the function:
> ###
> def nice(x,leadingDigits=0):
>  """Return x either as 'print' would show it (the default) or rounded to the
>  specified digit as counted from the leftmost non-zero digit of the number,
> 
>  e.g. nice(0.00326,2) --> 0.0033"""
>  assert leadingDigits>=0
>  if leadingDigits==0:
>   return float(str(x)) #just give it back like 'print' would give it
>  leadingDigits=int(leadingDigits)
>  return float('%.*e' % (leadingDigits,x)) #give it back as rounded by the %e 
> format
> ###
> 
> Might something like this be useful? For new users, no arguments are needed 
> other than x and floating points suddenly seem to behave in tests made using 
> nice() values. It's also useful for those computing who want to show a 
> physically meaningful value that has been rounded to the appropriate digit 
> as counted from the most significant digit rather than from the decimal 
> point.
> 
> Some time back I had worked on the significant digit problem and had several 
> math calls to figure out what the exponent was. The beauty of Tim's solution 
> is that you just use built in string formatting to do the work. Nice.
> 
> /c
> 
> [1] http://mail.python.org/pipermail/tutor/2004-July/030324.html 
> 
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