R: How do I translate solve_LSAP into a dataframe? - r

Question

I'm trying to do a large assignment problem, and am using LSAP. It works, but I am trying to get the output into a dataframe so that I can do more with it. However, the documentation on the function says "An object of class "solve_LSAP" with the optimal assignment of rows to columns", with no further information on the data. I cannot seem to crack open the class to break the data out into a more usable form.
I've provided their example code.
x <- matrix(c(5, 1, 4, 3, 5, 2, 2, 4, 4), nrow = 3)
y <- solve_LSAP(x, maximum = FALSE)
y
Output:
Optimal assignment:
1 => 3, 2 => 1, 3 => 2
I have 200+ assignments, and the baseline output is simply not usable for me. How can I translate it into a dataframe, or at least a matrix that looks something like the below?
Row Column
1 3
2 1
3 2

Answer 1

The solve_LSAP returns column indices by each row, therefore comprising all you need for reconstruction:
len <- length(y)
parsedMat <- cbind(
1:len,
as.integer(y)
)
parsedMat
[,1] [,2]
[1,] 1 3
[2,] 2 1
[3,] 3 2
This can be turned into a solved matrix by:
solvedMat <- matrix(0, nrow = len, ncol = len)
solvedMat[parsedMat] <- 1
solvedMat
[,1] [,2] [,3]
[1,] 0 0 1
[2,] 1 0 0
[3,] 0 1 0
You can also turn this into a function that will return both outputs in form of a list, e.g.:
parseClueOutput <- function(x) {
len <- length(x)
parsedMat <- cbind(
1:len,
as.integer(x)
)
solvedMat <- matrix(0, nrow = len, ncol = len)
solvedMat[parsedMat] <- 1
return(
list(
parsedMat = parsedMat,
solvedMat = solvedMat
)
)
}
And use it as:
parseClueOutput(y)
$parsedMat
[,1] [,2]
[1,] 1 3
[2,] 2 1
[3,] 3 2
$solvedMat
[,1] [,2] [,3]
[1,] 0 0 1
[2,] 1 0 0
[3,] 0 1 0
As for the structure, the solve_LSAP is not really a complicated object, it's essentially a numeric vector which you can see with:
is.numeric(y)
[1] TRUE
Or:
str(y)
'solve_LSAP' num [1:3] 3 1 2
You can also turn the solvedMat or parsedMat into a dataframe easily - for example, the parsedMat:
setNames(as.data.frame(parsedMat), c('Row', 'Column'))
Row Column
1 1 3
2 2 1
3 3 2