Combinatorial perumutations in a time series using R -

imagine time series experiment each of 3 time points (a, b, c) replicated 3 times:

a1, a2, a3, b1, b2, b3, c1, c2, c3 

we concatenate data (generating time series of 3 cycles: abc,abc,abc) in possible permutations (using different time-point replicate each time). here example few realizations:

a1, b1, c1,  a1, b1, c2, a1, b1, c3 a1, b2, c1,  a1, b2, c2, a1, b2, c3 a1, b3, c1,  a1, b3, c2, a1, b3, c3 

could kindly suggest r script carry task? in advance, eran

using expand grid:

expand.grid(a = c("a1","a2","a3"), b = c("b1","b2","b3"), c = c("c1","c2","c3")) 

returns

     b  c 1  a1 b1 c1 2  a2 b1 c1 3  a3 b1 c1 ... 23 a2 b2 c3 24 a3 b2 c3 25 a1 b3 c3 26 a2 b3 c3 27 a3 b3 c3 

edit: based on comments below, there more subtle trick problem.

i'm there more efficient way this, can't seem think of way beyond testing permutations:

library(gtools) library(data.table)  dt <- data.table(expand.grid(a = c(1,2,3), b = c(1,2,3), c = c(1,2,3)))  perm <- data.frame(permutations(nrow(dt),3,1:nrow(dt))) 

perm data frame possible permutations of rows.

we can create list, each element being combination:

mylist <- apply(perm,1,function(x,y){y[as.numeric(x)]},dt) 

we can test see list elements have every column 3 unique values

mylist[as.logical(lapply(mylist,function(x){all(lapply(x,function(y){length(unique(y))}) == 3)}))] 

this doozy of formula, job. returns list of 3x3 data frames meet conditions above.

edit 2: nvm, simpler:

library(data.table) library(gtools)  dt <- data.table(permutations(n = 3, r = 3, v = 1:3)) perm <- data.frame(permutations(nrow(dt),3,1:nrow(dt), repeats.allowed = t)) mylist <- apply(perm,1,function(x,y){y[as.numeric(x)]},dt)  mylist <- lapply(mylist,t) mylist <- lapply(mylist,`colnames<-`,c("a","b","c"))