Excluding 0.5 from rounding

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How can I exclude the 0.5 fraction from rounding such that the fractions less than or greater than 0.5 are only to be rounded?

Accepted Answer

John D'Errico
John D'Errico on 13 Jan 2022
You cannot do this. That is, there are only a few specific classes of rounds you can do, embodied in round, fix, floor, and ceil. (I think I listed them all.) There are no flags you can set that will control rounding.
You want to round down, for non-integer parts that are strictly less than 1/2, and round up for non-integer parts greater than 1/2, but leave those values that are exactly at 1/2 alone?
I suppose with some code, and some small effort, do what you want.
x = [1.5;rand(8,1)*10 - 5]
x = 9×1
1.5000 -4.6270 0.9246 -4.6491 1.7999 2.9374 -3.1723 -4.4182 3.1633
xr = strangeround(x)
xr = 9×1
1.5000 -5.0000 1.0000 -5.0000 2.0000 3.0000 -3.0000 -4.0000 3.0000
Does that do as required?
function xround = strangeround(x)
xint = floor(x);
xfrac = x - xint;
xfrac(xfrac < 1/2) = 0;
xfrac(xfrac > 1/2) = 1;
xround = xint + xfrac;
end

More Answers (1)

Max Heimann
Max Heimann on 13 Jan 2022
Edited: Max Heimann on 13 Jan 2022
if mod(x,1) ~= 0.5
x = round(x)
end
  3 Comments
John D'Errico
John D'Errico on 13 Jan 2022
Yes. That will work. And since 0.5 is exactly representable in floating point arithmetic as a double, the exact test for equality is sufficient.

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