How to find complementary rows in matrix in R

Question

I have this matrix:

      [,1] [,2] [,3] [,4]
 [1,]    1    0    0    0
 [2,]    0    1    0    0
 [3,]    0    0    1    0
 [4,]    0    0    0    1
 [5,]    1    1    0    0
 [6,]    0    0    1    1
 [7,]    1    0    1    0
 [8,]    0    1    0    1
 [9,]    1    1    1    1

So, there are some rows that are complementary. In this matrix these are:

[5,]    1    1    0    0
[6,]    0    0    1    1

and

[7,]    1    0    1    0
[8,]    0    1    0    1

What I want to do is to find these complementary rows and keep just the first one of them. The expected output should be this:

      [,1] [,2] [,3] [,4]
 [1,]    1    0    0    0
 [2,]    0    1    0    0
 [3,]    0    0    1    0
 [4,]    0    0    0    1
 [5,]    1    1    0    0
 [6,]    1    0    1    0
 [7,]    1    1    1    1

Is there a way to do this in R?

Answer 1

If your matrix is called m:

# find duplicate rows
dists <- as.matrix(dist(m, method = "manhattan"))
equals <- which(dists == ncol(m), arr.ind = TRUE, useNames = FALSE)

# remove symmetry (5,6 == 6,5)
equals <- equals[equals[,1] < equals[,2],]
to_drop <- equals[,2]

m <- m[-to_drop,]

This uses the Manhattan distance to find rows for which the sum of the differences equals the number of columns, hence all elements are different.

Answer 2

In base-R is all that is needed to run this code.

Sample data:

mydata<- matrix(c(1,0,0,0,1,0,1,0,1,0,1,0,0,1,0,0,1,1,0,0,1,0,0,1,1,0,1,0,0,0,1,0,1,0,1,1),ncol=4)

Code

i=1
while(i <= nrow(mydata)){
  test <- matrix(rep(mydata[i,],nrow(mydata)),nrow=nrow(mydata),byrow=T)+mydata
  RowsToRemove <- grep(1,sapply(1:nrow(mydata),function(x) prod(test[x,]==1)))
  if(length(RowsToRemove)!=0){
    mydata <- mydata[-RowsToRemove,]
  }
  i=i+1
}

Output

> mydata
     [,1] [,2] [,3] [,4]
[1,]    1    0    0    0
[2,]    0    1    0    0
[3,]    0    0    1    0
[4,]    0    0    0    1
[5,]    1    1    0    0
[6,]    1    0    1    0
[7,]    1    1    1    1