四舍五入为10的幂 [英] Rounding to a power of 10
问题描述
我有一个变量tauMax
,我想将向上舍入到最接近的10的幂(1,10,100,1000 ...).我正在使用以下表达式在tau数组中找到最接近最大值的整数.我正在寻找最大值,因为我正在尝试计算10的幂,这应该是x轴的截止值.因此,tauMax等于756,所以我想有一个输出1000或3(对于10 ^ 3)的表达式.
I have a variable, tauMax
, that I want to round up to the nearest power of ten(1, 10, 100, 1000...). I am using the below expression to find the closest integer to the max value in the tau array. I am finding the max value because I am trying to calculate the power of ten that should be the x axis cutoff. In this cause, tauMax is equal to 756, so I want to have an expression that outputs either 1000, or 3(for 10^3).
tauMax = round(max(tau));
非常感谢您的帮助!
推荐答案
由于您使用的是以10为底的数字,因此只需使用log10
即可获取位数.
Since you're talking base 10, you could just use log10
to get the number of digits.
怎么样:
>> ceil(log10(756))
ans =
3
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