CODEWITHSUNDEEP
rfunctionsolver
Daan
Daan

Reputation: 43

Replicate excel solver in R

I am wondering if there is a simple function to solve the following problem in R:

Suppose I have the following dataframe:

Now I know that the sum of various values in B equal the values in A, e.g. '3 + 4 + 3 = 10' and '17 + 18 = 35', which always balances out in the complete dataset.

Question

Is there a function that can sum these values in B, through trial and error I suppose, and match the correctly summed values with A? For example, the function tries to sum 3 + 4 + 18, which is 25 and retries this because 25 is not a value in A.

I have tried several solutions myself but one problem that I often encountered was the fact that A always has less observations than B.

I would be very thankful if someone can help me out here! If more info is needed please let me know.

Cheers,

Daan

Edit

This example is with simplified numbers. In reality, it is a large dataset, so I am looking for a scalable solution.

Thanks again!

Upvotes: 4

Views: 250

Answers (2)

jeremycg
jeremycg

Reputation: 24945

This is a problem know as the subset sum problem, and there are a ton of examples online of how to solve it using dynamic programming, or greedy algorithms.

To give you an answer that just works, the package adagio has an implementation:

library(adagio)

sums = c(10, 35, 90)
values = c(3, 4, 17, 18, 50, 40, 3)


for(i in sums){
  #we have to subset the values to be less than the value
  #otherwise the function errors:
  print(subsetsum(values[values < i], i))
}

The output for each sum is a list, with the val and the indices in the array, so you can tidy up the output depending on what you want from there.

Upvotes: 1

Chriss Paul
Chriss Paul

Reputation: 1101

You can try the following but I am affraid is not scalable. For the case of 3 summands you have

x <- expand.grid(c(3, 4, 17, 18, 50, 40, 3),#building a matrix of the possible combinations of summands
                 c(3, 4, 17, 18, 50, 40, 3),
                 c(3, 4, 17, 18, 50, 40, 3))
x$sums <-rowSums(x) #new column with possible sums
idx<- x$sums%in%c(10, 35, 90) #checking the sums are in the required total
x[idx,]

        Var1 Var2 Var3 sums
2      4    3    3   10
8      3    4    3   10
14     3    4    3   10
44     4    3    3   10
50     3    3    4   10
56     3    3    4   10
92     3    3    4   10
98     3    3    4   10
296    4    3    3   10
302    3    4    3   10
308    3    4    3   10
338    4    3    3   10

For the case of 2 summands

x <- expand.grid(c(3, 4, 17, 18, 50, 40, 3),
                 c(3, 4, 17, 18, 50, 40,3))
x$sums <-rowSums(x) 
idx<- x$sums%in%c(10, 35, 90)
#Results
x[idx,]
   Var1 Var2 sums
18   18   17   35
24   17   18   35
34   40   50   90
40   50   40   90

Upvotes: 0

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