Round down and Round up?
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Hi everyone,
I have two numbers and want to round up and round down them.
minval = 0.4410 and maxval=0.8450
I want to round down the first number and it will be 0.4. The 2nd number after being rounded up will be 0.9.
I tried with floor(minval) it returns 0 and ceil(maxval) it returns 1. Those numbers are not the result I want.
Could someone give me some advice?
I'm using MATLAB R2014a.
Thank you so much.
1 件のコメント
shaadinama
2023 年 5 月 8 日
編集済み: DGM
2023 年 5 月 11 日
In MATLAB, the functions "floor" and "ceil" can be used to round down and round up, respectively.
The "floor" function rounds a given input value down to the nearest integer. For example, "floor(3.7)" would return 3, and "floor(-2.3)" would return -3.
The "ceil" function, on the other hand, rounds a given input value up to the nearest integer. For example, "ceil(3.2)" would return 4, and "ceil(-2.9)" would return -2.
Both functions can be useful in various mathematical and programming contexts where it is necessary to round numbers to the nearest whole integer.
採用された回答
Star Strider
2014 年 12 月 30 日
編集済み: Star Strider
2014 年 12 月 30 日
In R2014b, the round function now supports rounding to a specific number of digits:
minval = 0.4410;
maxval = 0.8450;
r1_minval = round(minval, 1);
r1_maxval = round(maxval, 1);
If you don’t have R2014b, this works:
diground = @(x,d) round(x*10^d)/10^d;
r2_minval = diground(minval,1);
r2_maxval = diground(maxval,1);
7 件のコメント
marchammer
2022 年 2 月 25 日
編集済み: marchammer
2022 年 2 月 25 日
If the requirement is based on the 2nd decimal as it could be inferred in the problem statement, given a common input and no prompt as to whether to round up or down, round(round(val,2),1) will work. Otherwise the first answer with floor() and ceil() multiplying by a power of 10 and dividing again after rounding is an easy and common way to do this.
It is interesting that the accepted answer doesn't round the second value up as requested.
Stephen23
2022 年 2 月 26 日
編集済み: Stephen23
2023 年 4 月 27 日
"It is interesting that the accepted answer doesn't round the second value up as requested. "
It is simply because the actual value is less than 5 at that digit (not equal to 5 as it seems when displayed to a much lower precision):
minval = 0.4410
maxval = 0.8450
fprintf('%.40f\n',minval,maxval)
So ROUND is working correctly, it is another reminder to always be aware of the behaviors of floating point numbers.
Given hints in the the OP's question, it seems like the correct answer was given by Rajesh here:
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