Rounding towards zero or from zero

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Nick Austinos
Nick Austinos el 26 de Sept. de 2022
Respondida: Image Analyst el 26 de Sept. de 2022
Hi all; I want to round 3.6 to 3 but funny enough all the 'tozero' and 'fromzero' tiebreakers are returning the same answer i.e 4 which i dont want. How can this issue be fixed?
x=3.6
y=round(x,"TieBreaker","tozero")
z=round(x,"TieBreaker","fromzero")

Respuesta aceptada

Les Beckham
Les Beckham el 26 de Sept. de 2022
Editada: Les Beckham el 26 de Sept. de 2022
There is no tie involved in rounding 3.6. If you wish, you can use floor instead:
floor(3.6)
ans = 3
If x was 3.5 instead, that would be a tie:
x = 3.5;
y=round(x,"TieBreaker","tozero")
y = 3
z=round(x,"TieBreaker","fromzero")
z = 4
  2 comentarios
Nick Austinos
Nick Austinos el 26 de Sept. de 2022
I works thanks;
Les Beckham
Les Beckham el 26 de Sept. de 2022
You are quite welcome.

Iniciar sesión para comentar.

Más respuestas (3)

Sam Chak
Sam Chak el 26 de Sept. de 2022
Does this work for you?
x = 3.6;
y = floor(x)
y = 3

Steven Lord
Steven Lord el 26 de Sept. de 2022
The tiebreaker methods only apply when the quantity to be rounded is halfway between the two numbers to which it could be rounded. So if you had 3.5 that's halfway between 3 and 4 and the tiebreaker would determine to which of those numbers 3.5 gets rounded.
round(3.5, 'TieBreaker', 'tozero')
ans = 3
round(3.5, 'TieBreaker', 'fromzero')
ans = 4
There's no tie to be broken if you're rounding 3.6.
There are other rounding functions that you may want to use instead of round. See their help or documentation pages for more information on each function's specific behavior.
[fix(3.6), floor(3.6), round(3.6), ceil(3.6)]
ans = 1×4
3 3 4 4

Image Analyst
Image Analyst el 26 de Sept. de 2022
Try fix it rounds towards zero regardless if it's positive or negative, unlike floor which rounds towards negative infinity.
v = fix(3.6)
v = 3
v = fix(-9.4)
v = -9
v = floor(-9.4)
v = -10

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