Rounding towards zero or from zero

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Nick Austinos
Nick Austinos 2022 年 9 月 26 日
回答済み: Image Analyst 2022 年 9 月 26 日
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")

採用された回答

Les Beckham
Les Beckham 2022 年 9 月 26 日
編集済み: Les Beckham 2022 年 9 月 26 日
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 件のコメント
Les Beckham
Les Beckham 2022 年 9 月 26 日
You are quite welcome.

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その他の回答 (3 件)

Sam Chak
Sam Chak 2022 年 9 月 26 日
Does this work for you?
x = 3.6;
y = floor(x)
y = 3

Steven Lord
Steven Lord 2022 年 9 月 26 日
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 2022 年 9 月 26 日
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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