Reputation: 301
Here is Matlab code to form the matrix of logical values of '0' and '1'
A=[1 2 3 4 5 6 7 8 9 10 ];
N = numel(A);
step = 2; % Set this to however many zeros you want to add each column
index = N:-step:1;
val = (1:N+step).' <= index;
Which result in
val=
1 1 1 1 1
1 1 1 1 1
1 1 1 1 0
1 1 1 1 0
1 1 1 0 0
1 1 1 0 0
1 1 0 0 0
1 1 0 0 0
1 0 0 0 0
1 0 0 0 0
0 0 0 0 0
0 0 0 0 0
How to do same task in r ,particularly val = (1:N+step).' <= index
this step?
Upvotes: 1
Views: 149
Reputation: 887511
One option is
i <- seq_len(ncol(m1))
sapply(rev(i), function(.i) {
m1[,.i][sequence(.i *2)] <- 1
m1[,.i]
})
# [,1] [,2] [,3] [,4] [,5]
# [1,] 1 1 1 1 1
# [2,] 1 1 1 1 1
# [3,] 1 1 1 1 0
# [4,] 1 1 1 1 0
# [5,] 1 1 1 0 0
# [6,] 1 1 1 0 0
# [7,] 1 1 0 0 0
# [8,] 1 1 0 0 0
# [9,] 1 0 0 0 0
#[10,] 1 0 0 0 0
#[11,] 0 0 0 0 0
#[12,] 0 0 0 0 0
Or vectorize it
i1 <- rep(i, rev(2*i))
m1[cbind(ave(i1, i1, FUN = seq_along), i1)] <- 1
m1
# [,1] [,2] [,3] [,4] [,5]
# [1,] 1 1 1 1 1
# [2,] 1 1 1 1 1
# [3,] 1 1 1 1 0
# [4,] 1 1 1 1 0
# [5,] 1 1 1 0 0
# [6,] 1 1 1 0 0
# [7,] 1 1 0 0 0
# [8,] 1 1 0 0 0
# [9,] 1 0 0 0 0
#[10,] 1 0 0 0 0
#[11,] 0 0 0 0 0
#[12,] 0 0 0 0 0
Or another option without creating a matrix
beforehand
n <- 5
i1 <- seq(10, 2, by = -2)
r1 <- c(rbind(i1, rev(i1)))
matrix(rep(rep(c(1, 0), n), r1), ncol = n)
# [,1] [,2] [,3] [,4] [,5]
# [1,] 1 1 1 1 1
# [2,] 1 1 1 1 1
# [3,] 1 1 1 1 0
# [4,] 1 1 1 1 0
# [5,] 1 1 1 0 0
# [6,] 1 1 1 0 0
# [7,] 1 1 0 0 0
# [8,] 1 1 0 0 0
# [9,] 1 0 0 0 0
#[10,] 1 0 0 0 0
#[11,] 0 0 0 0 0
#[12,] 0 0 0 0 0
m1 <- matrix(0, 12, 5)
Upvotes: 2