从值列表中近似两个目标值

Question

I have a list of values (these refer to weights) and 2 target values (which are the sum of a selection of these weights). The weights should be assigned to one of the two target values, so that the sum of the weights approximates its target value. The approximation should be maximised for both target values simultaneously.

For example, this is a list of weights

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

and there are two target values:

Values
22.08
21.37

The sum of a selection of weights will probably not be exactly equal to the target value, so I need an approximation.

Excel Solver can do this for one target value at a time, as far as I know, but I need it to handle multiple target values at a time. Does anyone have any idea how to tackle this? Preferably in R but Python or Excel are also fine.

Answer 1

Here is an alternative:

 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