A little more: decimal_portion
Steven D'Aprano
steve+comp.lang.python at pearwood.info
Sat Oct 4 17:46:03 CEST 2014
Seymore4Head 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)
>
> http://imgur.com/a0Csi43
>
> 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".
--
Steven
More information about the Python-list
mailing list