Hi again,
I was contemplating the solution using base 3:
set.seed(3)
mat2 <- matrix(sample(0:2, 15, replace=T), 5, 3)
Extracting the line numbers is simple:
bases <- c(3, 3, 3)^(2:0) # or just 3^(2:0)
colSums(apply(mat2, 1, function(x) x*bases)) + 1
[1] 7 23 25 8 1
The problem is somet
I think you are computing your bases in the wrong way. If the data
represents 3 columns with base 3,3,2, then the multiplier has to be
c(6,2,1) not c(9,3,1). I think this should compute it correctly:
# create a matrix of all combination of bases 3,3,2
mat1 <- expand.grid(0:1, 0:2, 0:2)[,3:1]
bas
On Sunday 21 January 2007 16:30, jim holtman wrote:
> I think you are computing your bases in the wrong way. If the data
> represents 3 columns with base 3,3,2, then the multiplier has to be
> c(6,2,1) not c(9,3,1). I think this should compute it correctly:
>
> # create a matrix of all combinatio
