Harm_Kirchhoff@[...].jp wrote:
>  I have the following problem:
>  
>  >>> float(8.7)
>  8.6999999999999993
>  >>> float(870)/100
>  8.6999999999999993
>  >>> 8.7 * 100
>  869.99999999999989
>  >>> round(8.7,2)
>  8.6999999999999993
>  >>> math.modf(8.7)
>  (0.69999999999999929, 8.0)
>  >>> int(float(8.7*100))
>  869
>  
>  especially anoying is:
>  >>> n = float(8.7*100) ; print n
>  870.0
>  >>> n
>  869.99999999999989
>  >>> 
>  
>  How can I ensure that 8.7 stays 8.7, so that
>  8.7 * 100 = 870 and not 869.9999...
>  
>  The way I came up with is:
>  >>> f = float(8.7*100)   ;  i = int(f)   ; i,f
>  (869, 869.99999999999989)
>  >>> if f>=(i+.5): i += 1 ;  i
>  
>  870
>  
>  I am writing a commercial sales database. To save
>  memory I use integers 
>  internally, rather than floats.
>  This means that an amount of 8.70 is converted to
>  870 internally.
>  if it is converted to 869 I loose 1/10.

This is annoying, but the only way short of using a
library to perform this simple task that I can thing
of is to convert the float into a string and then into
an integer. Here's a simple function that does that:

> >> convert = lambda x: int(str(x*100)[:-2])
> >> convert(8.7)
870

You need the [:-2] to truncate the last two letters
because otherwise it would be in the form '870.0',
which it can't handle.

Daniel Ehrenberg

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