I have a dataset:
time delta
0.47 0
0.01 1
0.30 1
0.07 0
0.38 0
0.68 1
0.13 0
0.09 1
0.08 1
0.04 0
0.13 0
0.41 1
0.22 0
0.11 0
0.85 0
0.26 0
I'm using R and I need to compare this matrix with itself. I want to generate a new matrix 16*16 with values:
1 time_i > time_j & delta_i= delta_j != 0;
0 otherwise.
where i, j = 1,..., 16.
I tried to use the sapply() function, but it is useful only if I want to compare with respect one condition.
Could someone help me? Thank you in advance.
You can use outer to apply a function to every pair of elements in two vectors, so you could do one outer for each of the two logical comparisons, combine them with a logical AND, then convert to numeric. Here I am assuming your matrix is called m:
1*(outer(m[,1], m[,1], `>`) & outer(m[,2], m[,2], function(x, y) x == y & x != 0))
This gives the following output:
#> [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13] [,14]
#> [1,] 0 0 0 0 0 0 0 0 0 0 0 0 0 0
#> [2,] 0 0 0 0 0 0 0 0 0 0 0 0 0 0
#> [3,] 0 1 0 0 0 0 0 1 1 0 0 0 0 0
#> [4,] 0 0 0 0 0 0 0 0 0 0 0 0 0 0
#> [5,] 0 0 0 0 0 0 0 0 0 0 0 0 0 0
#> [6,] 0 1 1 0 0 0 0 1 1 0 0 1 0 0
#> [7,] 0 0 0 0 0 0 0 0 0 0 0 0 0 0
#> [8,] 0 1 0 0 0 0 0 0 1 0 0 0 0 0
#> [9,] 0 1 0 0 0 0 0 0 0 0 0 0 0 0
#>[10,] 0 0 0 0 0 0 0 0 0 0 0 0 0 0
#>[11,] 0 0 0 0 0 0 0 0 0 0 0 0 0 0
#>[12,] 0 1 1 0 0 0 0 1 1 0 0 0 0 0
#>[13,] 0 0 0 0 0 0 0 0 0 0 0 0 0 0
#>[14,] 0 0 0 0 0 0 0 0 0 0 0 0 0 0
#>[15,] 0 0 0 0 0 0 0 0 0 0 0 0 0 0
#>[16,] 0 0 0 0 0 0 0 0 0 0 0 0 0 0
#> [,15] [,16]
#> [1,] 0 0
#> [2,] 0 0
#> [3,] 0 0
#> [4,] 0 0
#> [5,] 0 0
#> [6,] 0 0
#> [7,] 0 0
#> [8,] 0 0
#> [9,] 0 0
#>[10,] 0 0
#>[11,] 0 0
#>[12,] 0 0
#>[13,] 0 0
#>[14,] 0 0
#>[15,] 0 0
#>[16,] 0 0
You can more easily check that the elements of the matrix are in the correct position by making a matrix of the rows and columns where a 1 is to be found:
which(res == 1, arr.ind = TRUE)
#> row col
#> [1,] 3 2
#> [2,] 6 2
#> [3,] 8 2
#> [4,] 9 2
#> [5,] 12 2
#> [6,] 6 3
#> [7,] 12 3
#> [8,] 3 8
#> [9,] 6 8
#> [10,] 12 8
#> [11,] 3 9
#> [12,] 6 9
#> [13,] 8 9
#> [14,] 12 9
#> [15,] 6 12
The first entry in this table tells us that the criteria were met for row 3 of the original matrix when compared to row 2 of the original matrix. It is easy to confirm that this is indeed the case.