fill matrix with numbers in each row to create a parttern - r

Question

I'm trying to generate a trip parttern using a matrix.
Glastonbury Munich Venice Paris Ibiza Kamar-Taj
[1,] 1 2 3 4 5 6
[2,] 1 2 3 4 5 0
[3,] 1 2 3 4 0 0
[4,] 1 2 3 0 0 0
[5,] 1 2 0 0 0 0
[6,] 3 2 1 4 5 6
[7,] 0 2 3 4 1 0
[8,] 0 1 2 0 0 0
[9,] 5 1 3 2 0 0
each row represents a single trip, and the number in a cell represents the order they visited each locations. Zero means they didn't visit that place.
currently I'm creating it like this:
tripMatrix <- list()
tripMatrix[[ 1 ]] <- c(1, 2, 3, 4, 5, 6)
tripMatrix[[ 2 ]] <- c(1, 2, 3, 4, 5, 0)
tripMatrix[[ 3 ]] <- c(1, 2, 3, 4, 0, 0)
tripMatrix[[ 4 ]] <- c(1, 2, 3, 0, 0, 0)
tripMatrix[[ 5 ]] <- c(1, 2, 0, 0, 0, 0)
tripMatrix[[ 6 ]] <- c(3, 2, 1, 4, 5, 6)
tripMatrix[[ 7 ]] <- c(0, 2, 3, 4, 1, 0)
tripMatrix[[ 8 ]] <- c(4, 5, 3, 2, 1, 0)
tripMatrix[[ 8 ]] <- c(0, 1, 2, 0, 0, 0)
tripMatrix[[ 10 ]] <- c(5, 1, 3, 2, 0, 0)
trips <- matrix(unlist(tripMatrix), ncol = 6, byrow = TRUE)
I can only do this for a few rows, but I'll like to generate N number of rows with different conbinations of places visited during a trip.
Please I can I do this for N number of rows without having to manually create a list of trips?

Answer 1

Starting with a math approach by thinking about the theoretical number of trips. For example, out of the 6 places, you can choose 4 to visit (i.e. 6C4 = 6 choose 4) and among these 4 places, you can visit them in any order giving 4! (4 factorial) ways of visiting them in different sequence. The same applies to other number of places, giving a theoretical number of 6! + 6C5*5! + 6C4*4! + 6C3*3! + 6C2*2! + 6C1 = 1956 possible trips.
Here is an option using utils::combn and RcppAlgos::permuteGeneral to generate all 1,956 possibilities:
nc <- 6L
l <- lapply(1L:nc, function(n) combn(1L:nc, n,
function(x) {
if (length(x) > 1L) {
p <- RcppAlgos::permuteGeneral(x, length(x))
a <- matrix(0L, nrow=nrow(p), ncol=nc)
a[cbind(c(row(p)), c(p))] <- col(p)
} else {
a <- integer(nc)
a[x] <- 1L
}
a
}, simplify=FALSE))
m <- do.call(rbind, unlist(l, recursive=FALSE))
head(m, 20):
[,1] [,2] [,3] [,4] [,5] [,6]
[1,] 1 0 0 0 0 0
[2,] 0 1 0 0 0 0
[3,] 0 0 1 0 0 0
[4,] 0 0 0 1 0 0
[5,] 0 0 0 0 1 0
[6,] 0 0 0 0 0 1
[7,] 1 2 0 0 0 0
[8,] 2 1 0 0 0 0
[9,] 1 0 2 0 0 0
[10,] 2 0 1 0 0 0
[11,] 1 0 0 2 0 0
[12,] 2 0 0 1 0 0
[13,] 1 0 0 0 2 0
[14,] 2 0 0 0 1 0
[15,] 1 0 0 0 0 2
[16,] 2 0 0 0 0 1
[17,] 0 1 2 0 0 0
[18,] 0 2 1 0 0 0
[19,] 0 1 0 2 0 0
[20,] 0 2 0 1 0 0

Answer 2

You could use sample and replicate like :
N <- 10
unique_trip_locations <- 6
t(replicate(N, sample(0:unique_trip_locations, unique_trip_locations)))
# [,1] [,2] [,3] [,4] [,5] [,6]
# [1,] 2 3 1 5 0 6
# [2,] 5 3 6 2 0 1
# [3,] 2 4 1 3 0 5
# [4,] 1 6 5 3 0 2
# [5,] 6 3 1 4 2 0
# [6,] 1 2 0 6 5 4
# [7,] 5 1 3 0 4 6
# [8,] 0 4 6 2 3 1
# [9,] 0 5 1 3 2 6
#[10,] 5 6 0 1 3 2
However, this has got one limitation that 0 can occur max only once like all other numbers.

Answer 3

You can use brute force to do it, and in this case it is totally feasible.
I think mine is clearer than the previous answers:
vector<-c(1:6, rep(0,6)) # This is what you need as combinations
Data<-data.frame() # Initilizing Dataframe
i<-1 # Just to check total iterations
N<-1956 # Number of unique combinations to get
while(nrow(Data)<=N){
i<- i+1
Data<- rbind(Data,sample(vector, 6, replace = F))
Data<-Data[!duplicated(Data),]
}

Answer 4

Here is a base R solution without using any additional packages:
# define your own permutation function
perm <- function(x) {
if (length(x)==1) return(t(x))
subset(r <- do.call(expand.grid,replicate(length(x),x,simplify = F)), colSums(apply(r, 1, duplicated))==0)
}
# define function that give all permutations with given number of posistions
f <- function(x,vl) {
p <- perm(x)
unname(t(apply(p, 1, function(q) replace(rep(0,vl),q,seq_along(q)))))
}
# generate the desired output
v = 1:6
res <- do.call(rbind,
Map(f,
vl = length(v),
unlist(sapply(seq_along(v), function(k) combn(v,k,simplify = F)),recursive = F)))
such that
> head(res,25)
[,1] [,2] [,3] [,4] [,5] [,6]
[1,] 1 0 0 0 0 0
[2,] 0 1 0 0 0 0
[3,] 0 0 1 0 0 0
[4,] 0 0 0 1 0 0
[5,] 0 0 0 0 1 0
[6,] 0 0 0 0 0 1
[7,] 2 1 0 0 0 0
[8,] 1 2 0 0 0 0
[9,] 2 0 1 0 0 0
[10,] 1 0 2 0 0 0
[11,] 2 0 0 1 0 0
[12,] 1 0 0 2 0 0
[13,] 2 0 0 0 1 0
[14,] 1 0 0 0 2 0
[15,] 2 0 0 0 0 1
[16,] 1 0 0 0 0 2
[17,] 0 2 1 0 0 0
[18,] 0 1 2 0 0 0
[19,] 0 2 0 1 0 0
[20,] 0 1 0 2 0 0
[21,] 0 2 0 0 1 0
[22,] 0 1 0 0 2 0
[23,] 0 2 0 0 0 1
[24,] 0 1 0 0 0 2
[25,] 0 0 2 1 0 0