[Tutor] flow problem with a exercise

[Roelof Wobben]

> Subject: Re: [Tutor] flow problem with a exercise
> From: evert.rol at gmail.com
> Date: Sat, 21 Aug 2010 12:39:05 +0200
> CC: tutor at python.org
> To: rwobben at hotmail.com
> 
> > In [39]: t = 3
> > 
> > In [40]: round((t-32)/1.8)
> > Out[40]: -16.0
> > 
> > In [41]: t = 3.0
> > 
> > In [42]: round((t-32)/1.8)
> > Out[42]: -16.0
> > 
> > Works fine for me.
> > 
> > Correct,
> > But I see one wierd thing.
> > 
> > round ((42-32)/1.8) gives a output -16.0 but (42-32)/1.8) gives also -16.0
> > I was expectting that round will give 16 as output because round (32.0) is the same as round (32.0, 0) so there will be 0 decimals.
> > And I see one decimal.
> 
> The rounding doesn't influence how it's printed.
> From help(round):
> 
> """
> round(...)
> round(number[, ndigits]) -> floating point number
> 
> Round a number to a given precision in decimal digits (default 0 digits).
> This always returns a floating point number. Precision may be negative.
> """
> 
> It *rounds* the number to ndigits, not prints. It always returns a floating point number though (not an integer).
> Since floats are normally represented with at least one trailing digit after the decimal dot, you see a zero.
> Eg,
> >>> float(1)
> 1.0
> 
> If you want to get rid of the trailing .0, convert it to an integer:
> >>> int(round(1.1))
> 1
> 
> Also consider:
> >>> round(1.15, 0)
> 1.0
> >>> round(1.15, 1)
> 1.2
> >>> round(1.15, 2)
> 1.1499999999999999
> 
> 
> (Note again how the last rounded 1.15 is represented. It's just a representation though, and showing the imprecision you intrinsicaly get when dealing with floating point numbers.)
> 
> If you really want different behaviour when printing (or using) floating point numbers, perhaps have a look at the decimal module: http://docs.python.org/library/decimal.html
> 
> 


Oke, 

 

Thank you. Learned another thing about Python.

 

Roelof

 
 		 	   		  
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