将数字向量舍入为整数，同时保留它们的总和

Question

How to round floats to integers while preserving their sum?有以下answer用伪代码编写，它将向量四舍五入为整数值，从而使元素之和不变，并且舍入误差最小。我想在 R 中有效地实现这一点(即，如果可能的话，进行矢量化)。

例如，四舍五入这些数字会产生不同的总数:

set.seed(1)
(v <- 10 * runif(4))
# [1] 2.655087 3.721239 5.728534 9.082078
(v <- c(v, 25 - sum(v)))
# [1] 2.655087 3.721239 5.728534 9.082078 3.813063
sum(v)
# [1] 25
sum(round(v))
# [1] 26

从 answer 复制伪代码以供引用

// Temp array with same length as fn.
tempArr = Array(fn.length)

// Calculate the expected sum.
arraySum = sum(fn)

lowerSum = 0
-- Populate temp array.
for i = 1 to fn.lengthf
    tempArr[i] = { result: floor(fn[i]),              // Lower bound
                   difference: fn[i] - floor(fn[i]),  // Roundoff error
                   index: i }                         // Original index

    // Calculate the lower sum
    lowerSum = lowerSum + tempArr[i] + lowerBound
end for

// Sort the temp array on the roundoff error
sort(tempArr, "difference")

// Now arraySum - lowerSum gives us the difference between sums of these
// arrays. tempArr is ordered in such a way that the numbers closest to the
// next one are at the top.
difference = arraySum - lowerSum

// Add 1 to those most likely to round up to the next number so that
// the difference is nullified.
for i = (tempArr.length - difference + 1) to tempArr.length
    tempArr.result = tempArr.result + 1
end for

// Optionally sort the array based on the original index.
array(sort, "index")

Answer 1

以更简单的形式，我会说这个算法是:

从四舍五入开始

四舍五入具有最高小数部分的数字，直到达到所需的总和。

这可以通过以下方式在 R 中以矢量化方式实现:

用 floor 向下舍入

按小数部分排列的订单号(使用 order )

使用 tail获取具有 k 个最大小数部分的元素的索引，其中 k 是我们需要增加总和以达到我们的目标值

的数量。

将这些指数中的每一个的输出值增加 1

在代码中:

smart.round <- function(x) {
  y <- floor(x)
  indices <- tail(order(x-y), round(sum(x)) - sum(y))
  y[indices] <- y[indices] + 1
  y
}
v
# [1] 2.655087 3.721239 5.728534 9.082078 3.813063
sum(v)
# [1] 25
smart.round(v)
# [1] 2 4 6 9 4
sum(smart.round(v))
# [1] 25