Hi,
You could use:
as.matrix(expand.grid(vec1x3,vec1x3))
#or
as.matrix(expand.grid(rep(list(vec1x3),2)))
#or
library(gtools)
permutations(3, 2, vec1x3, repeats.allowed=TRUE)
A.K.
On Tuesday, October 15, 2013 7:14 PM, Stock Beaver <stockbea...@ymail.com>
wrote:
# I understand that a good way to build a vector from a sequence of integers,
# is to use syntax like this:
myvec = c(1:99)
# Here is the 'short' version of my question:
# I want to understand a 'good' way to build a matrix from a sequence of
integers.
# If that question is not clear, here is a longer version:
# Here is what I did for a 1D-matrix:
# I pick the sequence 1:3
# I build a vector:
vec1x3 = c(1:3)
vec1x3
# I transform it into a 1 x 3 matrix:
m1x3 = matrix(vec1x3, c(length(vec1x3),1))
m1x3
# [,1]
# [1,] 1
# [2,] 2
# [3,] 3
# >
# That was easy.
# Next I want to expand from a 1 x 3 matrix to a 2 x 9 matrix
# which contains all combinations of 1:3
# So the first 4 rows would look like this:
# 1 1
# 1 2
# 1 3 I call this a rowvec
# 2 1
# My first idea is write a loop like this:
for (i in 1:3) {
for(j in 1:3) {
rowvec = c(i,j)
# Place rowvec in matrix
}
}
# I'm curious if a skilled R-person would do it differently?
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