moonchild on Rounding Percentages

float

hm ...

  >>> def percent(progress, total):
  ...     return round(99 * progress / total + 0.5)
  ...
  >>> x = 9007199254740990.0
  >>> x - 1 < x
  True
  >>> percent(x - 1, x)
  100

tavianatorOP2y

You're cheating by using floating-point arguments; they should be integers:

    >>> x = 9007199254740990
    >>> percent(x - 1, x)
    99

But then again,

    >>> percent(10 * x - 1, 10 * x)
    100

If your procedure has 2^53 steps, feel free to ignore my rounding recommendations, since I'll never see it get anywhere near 100% :)

Also if your username is named after the band, good taste :)

moonchild 2y

they should be integers

i am not cheating; you are cheating by trying to do integer arithmetic instead of float arithmetic. in particular: 99 * progress is a (potentially big) integer; then the quotient, from my understanding of python, is equal to the mathematical quantity rn(99 * progress / total), which is not trivial to compute. (although cpython does tend to do a particularly bad job of this sort of thing.) (compare with c or with my version of the python, where it would be rn(rn(99 * progress) / rn(total)), rounding twice, which is very easy to compute. i'm not saying the c semantics is better, mind.) when you scale up by 10x, the numerator is <1/2 'ulp' away from the denominator, and so the quotient rounds up to 99 exactly; there is still double rounding (would have been triple rounding in c and my python), because we got rn(rn(99 * progress / total) + 0.5) where what we wanted was rn(99 * progress / total + 0.5) (which is mathematically always the correct result)

i agree it is not common to have so many steps. but if i were providing a routine that was intended to be robust where others were not, i would try to be comprehensive. and i would not try to do int math with floats unless i could show it to be robust (i have sketched such a stunt! https://gist.github.com/moon-chilled/60bd2ba687dc197d93a9d22...). the integer routine is simpler and more honest anyway, and it is obvious that it works uniformly for the entire range

note also that, with x floating, the python expressions 10 * x and 10 * x - 1 are equivalent, meaning the error is on input to the percent function. (if we set progress to 10 * x - 8, the immediately preceding fp number, we do get 99, but there is no deep reason for this, and it differs for different values of 10.)

if your username is named after the band, good taste :)

my username comes from a book: the neverending story. i am more using the german version of the name these days but i do not feel like making a new account on this godforesaken website

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