從值列表中近似兩個目標值

Question

我有一個值列表（這些是指權重）和 2 個目標值（它們是這些權重選擇的總和）。 應將權重分配給兩個目標值之一，以便權重之和接近其目標值。 應該同時最大化兩個目標值的近似值。

例如，這是一個權重列表

重量
4.528
4.773
4.253
4.688
4.21
3.841
4.005
4.545
3.825
5.123
4.757

並且有兩個目標值：

價值觀
22.08
21.37

權重選擇的總和可能不會完全等於目標值，所以我需要一個近似值。

據我所知，Excel Solver 可以一次處理一個目標值，但我需要它一次處理多個目標值。 有誰知道如何解決這個問題？ 最好在 R 但 Python 或 Excel 也可以。

Answer 1

這是一個替代方案：

 n= length(Weight)
 
 # all combinations 2 to n-1
 cms = lapply(2:(n-1), function(x) combn(n,x))
 # sum weights selected and not selected and compute square errors
 res = lapply(cms, function(cm)
+ apply(cm,2,function(sel){
+   sum1 = sum(Weight[sel])
+   sum2 = sum(Weight[-sel])
+   sum((Values-c(sum1,sum2))^2)
+ }))
 
 # find the min square error 
 mins = sapply(res,function(re){
+   wm = which.min(re)
+   minv= re[wm]
+   c(wm=wm,minv=minv)})
 mins
#         [,1]    [,2]      [,3]     [,4]      [,5]      [,6]     [,7]     [,8]    [,9]
#wm    18.0000  81.000 169.00000 300.0000 163.00000 162.00000  85.0000  38.0000   2.000
#minv 447.1174 212.036  68.92069  14.0989  12.99698  54.91097 184.7123 406.2839 719.574
 which.min(mins[2,])
#[1] 5
 mins[,which.min(mins[2,])]
#       wm      minv 
#163.00000  12.99698 
 (sel = cms[[5]][,163]) # group 1
#[1] 1 3 5 6 7 9
 (1:n)[-sel] # group 2
#[1]  2  4  8 10 11
 c(sum(Weight[sel]), sum(Weight[-sel])) # sums
#[1] 24.662 23.886
 Values
#[1] 22.08 21.37

Answer 2

library(tidyverse)

weight<-c(4.528
,4.773
,4.253
,4.688
,4.21
,3.841
,4.005
,4.545
,3.825
,5.123
,4.757)



values <- c(
22.08,21.37)

# a heuristic that says I'm not going to add more than "these" number of entries to hit a target
(most_to_sum <- ceiling(max(values)/min(weight)))

#make combs
combs_to_do <- expand_grid(
  set_1 = seq_len(most_to_sum),
  set_2 = seq_len(most_to_sum)
) |> rowwise() |> mutate(rsum=sum(set_1,set_2)) |> filter(rsum<=length(weight)) |> ungroup()

unique_sets <- map(seq_len(most_to_sum),
                   ~combn(weight,.x,simplify=FALSE)) |> flatten()

unique_sets_evals <- map_dbl(unique_sets,
        ~abs(values[[1]] - sum(.x)))

# opportunity here to trade accuracy for speed/memoirt
(set1_reduced <- quantile(unique_sets_evals,1)) # use 1 for exhaustive search; though I got the correct result reducing to 0.01  and even good approximations with less  

set1_reduced_sets <- unique_sets[which(unique_sets_evals<=set1_reduced)]

do_second_set <- function(first_sets,size_of_second_set,weight,values){

  second_sets <- map(first_sets,~{
    available <- setdiff(weight,.x)
    if(length(available) < size_of_second_set) return(Inf)
    combn(available,size_of_second_set,simplify = FALSE)
  })
  coeval <- map2(first_sets,second_sets,
                 ~{
                   x <- .x
                   y <- .y 
                   xsum <- sum(x)
                   ysums <- map(y,sum)
                   evals <- map(ysums,
                                  ~sqrt((values[[1]]-xsum)^2+(values[[2]]-.x)^2))
                   best_y <- y[which.min(evals)]
                  list(best_second = best_y,
                       eval=evals[[which.min(evals)]])
                 })
  map2(first_sets,coeval,~c(list(set1=.x),.y))
}

almost_ <- map(seq_len(most_to_sum),
    ~do_second_set(set1_reduced_sets,.x,weight=weight,values=values)) |> flatten()

all_evals  <- map_dbl(almost_,~.x$eval)
(best_eval_num <- which.min(all_evals))

almost_[[best_eval_num]]

> almost_[[best_eval_num]]
$set1
[1] 4.528 4.773 4.253 4.688 3.825

$best_second
$best_second[[1]]
[1] 4.210 3.841 4.005 4.545 4.757


$eval
[1] 0.01769181

> sum( 4.528, 4.773, 4.253,4.688 ,3.825)
[1] 22.067
> sum(4.210, 3.841, 4.005,4.545, 4.757)
[1] 21.358