Excluding 0.5 from rounding

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Israa Ahmed
Israa Ahmed el 13 de En. de 2022
Comentada: John D'Errico el 13 de En. de 2022
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?

Respuesta aceptada

John D'Errico
John D'Errico el 13 de En. de 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

Más respuestas (1)

Max Heimann
Max Heimann el 13 de En. de 2022
Editada: Max Heimann el 13 de En. de 2022
if mod(x,1) ~= 0.5
x = round(x)
end
  3 comentarios
Max Heimann
Max Heimann el 13 de En. de 2022
Editada: Max Heimann el 13 de En. de 2022
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
John D'Errico el 13 de En. de 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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