Reputation: 1420
I have a 7000 X 7000 matrix in R. For example purpose I will use a smaller matrix as following:-
a <- matrix(c(0:-9, 1:-8, 2:-7, 3:-6, 4:-5, 5:-4, 6:-3, 7:-2, 8:-1, 9:0),
byrow = TRUE, ncol = 10, nrow = 10)
I want to create a new matrix which has values equal to 1 where the absolute values in matrix a
are between the closed interval of 2 and 5. And rest all other values equal to zero.
This would make the following matrix:-
b <- matrix(c(0,0,1,1,1,1,0,0,0,0
0,0,0,1,1,1,1,0,0,0
1,0,0,0,1,1,1,1,0,0
1,1,0,0,0,1,1,1,1,0
1,1,1,0,0,0,1,1,1,1
1,1,1,1,0,0,0,1,1,1
0,1,1,1,1,0,0,0,1,1
0,0,1,1,1,1,0,0,0,1
0,0,0,1,1,1,1,0,0,0
0,0,0,0,1,1,1,1,0,0),
byrow = TRUE, ncol = 10, nrow = 10)
I can do this using for loop, but I just want to know if there is a much better and effcient solution to do this.
Thanks in advance.
Upvotes: 2
Views: 58
Reputation: 102221
Perhaps you can try
> +((abs(a) - 2) * (abs(a) - 5) <= 0)
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,] 0 0 1 1 1 1 0 0 0 0
[2,] 0 0 0 1 1 1 1 0 0 0
[3,] 1 0 0 0 1 1 1 1 0 0
[4,] 1 1 0 0 0 1 1 1 1 0
[5,] 1 1 1 0 0 0 1 1 1 1
[6,] 1 1 1 1 0 0 0 1 1 1
[7,] 0 1 1 1 1 0 0 0 1 1
[8,] 0 0 1 1 1 1 0 0 0 1
[9,] 0 0 0 1 1 1 1 0 0 0
[10,] 0 0 0 0 1 1 1 1 0 0
Upvotes: 0
Reputation: 132854
You can just write down the comparison. It gives you a logical matrix and you can then use unary +
to turn the result into an integer matrix.
+(abs(a) >= 2 & abs(a) <= 5)
# [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
# [1,] 0 0 1 1 1 1 0 0 0 0
# [2,] 0 0 0 1 1 1 1 0 0 0
# [3,] 1 0 0 0 1 1 1 1 0 0
# [4,] 1 1 0 0 0 1 1 1 1 0
# [5,] 1 1 1 0 0 0 1 1 1 1
# [6,] 1 1 1 1 0 0 0 1 1 1
# [7,] 0 1 1 1 1 0 0 0 1 1
# [8,] 0 0 1 1 1 1 0 0 0 1
# [9,] 0 0 0 1 1 1 1 0 0 0
#[10,] 0 0 0 0 1 1 1 1 0 0
Upvotes: 2