A little more: decimal_portion
Seymore4Head at Hotmail.invalid
Sat Oct 4 18:11:41 CEST 2014
On Sun, 05 Oct 2014 01:46:03 +1000, Steven D'Aprano
<steve+comp.lang.python at pearwood.info> wrote:
>> A little more: decimal_portion
>> Write a function that takes two number parameters and returns a float
>> that is the decimal portion of the result of dividing the first
>> parameter by the second. (For example, if the parameters are 5 and 2,
>> the result of 5/2 is 2.5, so the return value would be 0.5)
>> def decimal_portion(a,b):
>> return float((b/a)-((b//a)))
>> print (decimal_portion(5,2))
>> I get 0.4 and the answer is supposed to be 0.5.
>Hint: given arguments a=5, b=2, you want 5/2. What do you calculate? Look at
>the order of the arguments a and b in the function def line, and in the
>calculations you perform.
>Once you fix that, I can suggest a more efficient way of calculating the
>answer: use the modulo operator %
>Given arguments 5 and 2, you want 0.5 == 1/2, and 5%2 returns 1.
>Given arguments 6 and 2, you want 0.0 == 0/2, and 6%2 returns 0.
>Given arguments 8 and 3, you want 0.6666... == 2/3, and 8%3 returns 2.
>P.S. the usual name for this function is "fraction part", sometimes "fp".
Yeah, I caught my mistake. Thanks for the suggestions.
I was expecting a lesson in rounding errors, so I had ruled out the
error was on my part. (Something I should never do)
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