Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
379 views
in Technique[技术] by (71.8m points)

r - Round vector of numerics to integer while preserving their sum

How to round floats to integers while preserving their sum? has the below answer written in pseudocode, which rounds a vector to integer values such that the sum of the elements in unchanged and the roundoff error is minimized. I'd like to implement this efficiently (i.e. vectorized if possible) in R.

For example, rounding these numbers yields a different total:

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

Copying pseudocode from answer for reference

// 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")
See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

In an even simpler form, I would say this algorithm is:

  1. Start with everything rounded down
  2. Round up the numbers with the highest fractional parts until the desired sum is reached.

This can be implemented in a vectorized way in R by:

  1. Round down with floor
  2. Order numbers by their fractional parts (using order)
  3. Use tail to grab the indices of the elements with the k largest fractional parts, where k is the amount that we need to increase the sum to reach our target value
  4. Increment the output value in each of these indices by 1

In code:

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

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...