R: sum elements in a matrix up to threshold - r

Question

I have a matrix of values with thousands of rows and a couple dozen columns. For a given row, $$R_0$$, I'd like to find all other complementary rows. A complementary row is defined as:
if given row has a non-zero value for a column, then the complement must have a zero value for that column
the sum of the elements of a given row and its complements must be less than 1.0
To illustrate, here is a toy matrix:
[,1] [,2] [,3] [,4] [,5] [,6]
[1,] 0 0 0 0.1816416 0 0.1796779
[2,] 0.1889351 0 0 0 0 0
[3,] 0 0 0.1539683 0 0 0.1983812
[4,] 0 0.155489 0.1869410 0 0 0
[5,] 0 0 0 0 0.1739382 0
For row 1, there are values for columns 4 and 6. A complementary row must have "0" for columns 4 and 6.
I don't know what data structure my desired output should be. But I know the output should tell me:
row 1 has the following complementary rows: 2, 3, 5
row 2 has the following complementary rows: 1, 3, 4, 5
row 3 has the following complementary rows: 2, 5
row 4 has the following complementary rows: 1, 2, 5
row 5 has the following complementary rows: 1, 2, 3, 4
Perhaps a list of lists? I.e.:
[1: 2, 3, 5;
2: 1, 3, 4, 5;
3: 2, 5;
4: 1, 2, 5;
5: 1, 2, 3, 4]
But I'm open to other data structures.
The following code generates the toy matrix above.
set.seed(1)
a = runif(n=30, min=0, max=0.2)
a[a<0.15] = 0
A = matrix(a, # the data elements
nrow=5, # number of rows
ncol=6, # number of columns
byrow = TRUE) # fill matrix by rows
Is there a package or clever way to approach this problem?

Answer 1

We can create a function to check if the combination of two rows is a compliment
check_compliment <- function(x, y) {
all(A[y, A[x,] != 0] == 0) & sum(c(A[x, ], A[y, ])) < 1
}
Here, we subset row y for columns where x is not 0 and check if all of them are 0. Also check if sum of x and y rows is less than 1.
Apply this function for every combination using outer
sapply(data.frame(outer(1:nrow(A), 1:nrow(A), Vectorize(check_compliment))), which)
#$X1
#[1] 2 4 5
#$X2
#[1] 1 3 4 5
#$X3
#[1] 2 5
#$X4
#[1] 1 2 5
#$X5
#[1] 1 2 3 4
outer step gives us TRUE/FALSE value for every combination of a row with every other row indicating if it is a compliment
outer(1:nrow(A), 1:nrow(A), Vectorize(check_compliment))
# [,1] [,2] [,3] [,4] [,5]
#[1,] FALSE TRUE FALSE TRUE TRUE
#[2,] TRUE FALSE TRUE TRUE TRUE
#[3,] FALSE TRUE FALSE FALSE TRUE
#[4,] TRUE TRUE FALSE FALSE TRUE
#[5,] TRUE TRUE TRUE TRUE FALSE
We convert this to data frame and use which to get indices for every column.