R：尽可能均匀地分配金额II

We have have a certain amount e.g. 300 units. This amount should be as evenly as possible distributed over 40 "slots" or "bins". It would be easy if each slot would be the same - so it would be 7,5 at each slot. However, the slots vary in size and we cannot "fill in" there more than its "size" allows for e.g. if its only 4. What we cannot "fill in" more than 4. Hence, we have to distribute more over the other ones.

我们有一定数量，例如300个单位。该数量应尽可能均匀地分布在40“槽”或“箱”上。如果每个插槽都相同则很容易 - 因此每个插槽的数量为7.5。但是，插槽的尺寸不同，我们不能“填充”那里，而不是“尺寸”允许例如如果它只有4.我们不能“填写”超过4个。因此，我们必须分配更多其他的。

Lets assume that there is another limitation: A general filling in limit of e.g. 5. That would mean that even if we have enough size in the slot to fill in say 12 and enough units remaining say 11, we can only fill in 5. The value that is excess after all slots are filled should be placed in a seperated remainder slot. With each filling in process we should also get a number how much of the maximum-filling in capacity in percent is used. I.e. if we fill in 4 and 5 is the general filling limit. We used 80%.

让我们假设还有另一个限制：一般填写限制，例如： 5.这意味着即使我们在插槽中有足够的尺寸来填充说12并且剩余足够的单位说11，我们只能填写5.填充所有插槽后超出的值应该分开放置剩余时间。每次填充过程中，我们还应该得到一个数字，即使用百分比的最大填充量。即如果我们填写4和5是一般填充限制。我们使用了80％。

We discussed this already earlier in anohter question: Distributing an amount as evenly as possible

我们在anohter问题中已经讨论过这个问题：尽可能均匀地分配金额

I have some ideas how do develop this formula further, however partially it still stucks. Thanks for your help!

我有一些想法如何进一步发展这个公式，但部分仍然坚持。谢谢你的帮助！

# developing slots and their "size" 
a <- rnorm(40,10,4) 
sum(a) 

# overall sum to distribute 
b <- 300  
# general filling in limit
c <- 8

optimal.fill <- function(a, b) 
{ 
  stopifnot(sum(a) >= b) 

  d <- rep(0, length(a))
  info <- rep(0, length(a))  
  while(b > 0) { 
    has.room  <- a > 0 
    num.slots <- sum(has.room) 
    min.size  <- min(a[has.room]) 
    add.size  <- min(b / num.slots, min.size)
    #maximum limitation
    add.size[add.size>c]  <- c
    #percentage info
    info[has.room] <- add.size/c
    d[has.room] <- d[has.room] + add.size 
    a[has.room] <- a[has.room] - add.size 
    b <- b - num.slots * add.size 
    } 
  return(d) 
} 
optimal.fill(a,b)

1 个解决方案

#1

How about this

这个怎么样

optimal.fill <- function(a, b, generalLimit = 8){
  a <- pmax(0, pmin(a, generalLimit))
  if(sum(a) < b){
    stop("not enough room")
  }
  if(length(a) * min(a) <= b){
    result <- rep(min(a), length(a))
  } else {
    result <- rep(floor(b / length(a)), length(a))
  }
  while(floor((b - sum(result)) / sum(result < a)) >= 1){
    if(min(a[result < a]) * sum(result < a) <= b - sum(result)){
      result[result < a] <- 
        result[result < a] + rep(min(a[result < a]), sum(result < a))
    } else {
      result[result < a] <- 
        result[result < a] + 
        rep(floor((b - sum(result)) / sum(result < a)), sum(result < a))
    }
  }
  extra <- sample(which(result < a), (b - sum(result)), replace = FALSE)
  result[extra] <- result[extra] + 1
  return(cbind(result,  result / a))
}
optimal.fill(ceiling(rnorm(40,10,4)), 300, 8)