从 0.5 向上舍入 [英] Round up from .5

问题描述

是的，我知道为什么我们总是在两个数字的正中间(即 2.5 变成 2)时四舍五入到最接近的偶数.但是当我想为某些人评估数据时，他们不想要这种行为.获得此信息的最简单方法是什么:

Yes I know why we always round to the nearest even number if we are in the exact middle (i.e. 2.5 becomes 2) of two numbers. But when I want to evaluate data for some people they don't want this behaviour. What is the simplest method to get this:

x <- seq(0.5,9.5,by=1)
round(x)

是 1,2,3,...,10 而不是 0,2,2,4,4,...,10.

to be 1,2,3,...,10 and not 0,2,2,4,4,...,10.

清除:1.4999 在四舍五入后应为 1.(我认为这很明显)

To clearify: 1.4999 should be 1 after rounding. (I thought this would be obvious)

推荐答案

这不是我自己的功能，不幸的是，我现在找不到我在哪里得到它(最初发现为统计显着博客)，但它应该有助于满足您的需求.

This is not my own function, and unfortunately, I can't find where I got it at the moment (originally found as an anonymous comment at the Statistically Significant blog), but it should help with what you need.

round2 = function(x, n) {
  posneg = sign(x)
  z = abs(x)*10^n
  z = z + 0.5 + sqrt(.Machine$double.eps)
  z = trunc(z)
  z = z/10^n
  z*posneg
}

x 是要舍入的对象，n 是要舍入的位数.

x is the object you want to round, and n is the number of digits you are rounding to.

一个例子

x = c(1.85, 1.54, 1.65, 1.85, 1.84)
round(x, 1)
# [1] 1.8 1.5 1.6 1.8 1.8
round2(x, 1)
# [1] 1.9 1.5 1.7 1.9 1.8

(感谢@Gregor 添加 + sqrt(.Machine$double.eps).)

(Thanks @Gregor for the addition of + sqrt(.Machine$double.eps).)

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