Excluding 0.5 from rounding

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?

Answer 1

[John D'Errico]

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

Answer 2

[Max Heimann]

if mod(x,1) ~= 0.5
x = round(x)
end

[John D'Errico]

Good idea. But while that would work for scalar x, it is not vectorized, and it will fail for vectors and arrays.

[Max Heimann]

How about this for vectors and matrices:

% Matrix with test values
x = [0 -4.5 -4.4; 3.3 0.5 1];
% Code
indices = mod(x,1) ~= 0.5;
x(indices) = round(x(indices))

[John D'Errico]

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.