Multiply values of each "path" in a given matrix for all columns

0
votes

Given an example of a matrix below:

index 1 2 3 4 5 6 7 8 9 10
1 0.1 0.1 0.1 0.2 0.2 0.2 0.7 0.7 0.4 0.7
2 0.6 0.6 0.6 0.1 0.1 0.1 0.1 0.1 0.5 0.1
3 0.3 0.3 0.3 0.7 0.7 0.7 0.2 0.2 0.1 0.2

I would like to multiply all possible combination, selecting a value from each column multiply it over the row eg:

0.1* 0.1* 0.1* 0.2* 0.2* 0.2* 0.7* 0.7* 0.4* 0.7

0.1* 0.6* 0.1* 0.2* 0.2* 0.2* 0.7* 0.7* 0.4* 0.7

0.1* 0.3* 0.1* 0.2* 0.2* 0.2* 0.7* 0.7* 0.4* 0.7

0.1* 0.1* 0.6* 0.2* 0.2* 0.2* 0.7* 0.7* 0.4* 0.7

0.1* 0.1* 0.3* 0.2* 0.2* 0.2* 0.7* 0.7* 0.4* 0.7

0.1* 0.1* 0.1* 0.1* 0.2* 0.2* 0.7* 0.7* 0.4* 0.7

0.1* 0.1* 0.1* 0.7* 0.2* 0.2* 0.7* 0.7* 0.4* 0.7

And so on... Aim is to find the maximum value and get the row index selected for each 10 columns

I thought of creating all possible combinations into a row and the perform row multiplication for each row(which would be a combination) then use maximum.

How do I create a matrix with all possible "path" into a row, but then it would be difficult to identify which "path" the max value be for.

1
riterationmatrix-multiplication
The logic is not clear to me. It looks like the fourth and first example are identical. What do you mean by "all possible combination"? How a combination is formed? Are you just changing the second term in the multiplication? With which logic? – nicola
Sorry I have added more examples. I am not just changing second term. Logic is to change each element of each column and multiplying 10 values. – Rahil Vora

1 Answers

0
votes

To make every permutations, where I let your data as dummy, use expand.grid

grid <- expand.grid(rep(list(1:3), 10))

Because grid is very huge, I'll use first five of grid

grid <- grid[c(1:5),]

Let's define a function to pull out grid rownumberd(?) values of dummy.

func <- function(df, idx) {
  vec <- c()
  for (i in 1:length(idx)){
    vec[i] <- df[as.numeric(idx[i]),i]
  }
  vec
}

Finally, get multiplications of those values like

apply(grid, 1, function(x) prod(func(dummy, x)))

          1           2           3           4           5 
1.09760e-06 6.58560e-06 3.29280e-06 6.58560e-06 3.95136e-05

If you want what maximize that is

grid[which.max(apply(grid, 1, function(x) prod(func(dummy, x)))),]
  Var1 Var2 Var3 Var4 Var5 Var6 Var7 Var8 Var9 Var10
5    2    2    1    1    1    1    1    1    1     1

This is Second way

HOWEVER I think this way is much more efficient to get maximum value and it's index. Because all elements are positive,

x <- apply(dummy, 2, function(x) which.max(x))
 X1  X2  X3  X4  X5  X6  X7  X8  X9 X10 
  2   2   2   3   3   3   1   1   2   1 

prod(func(dummy,x))

[1] 0.01270609

Large size example for second way

microbenchmark::microbenchmark(test ={x <- runif(250000,0,1)
                               y <- matrix(x, nrow = 5)
                               xx <- apply(y, 2, function(x) which.max(x))
                               
                               prod(func(y,xx))}
)
Unit: milliseconds
 expr     min       lq     mean  median       uq      max neval
 test 86.0542 91.41575 111.9721 96.9571 140.7912 171.8394   100