Conditional removal of rows in R-CodePudding

Consider the following matrix:

testMat <- matrix(c(1,2,1,3,
                 3,2,3,3,
                 1,3,1,3,
                 2,3,3,3,
                 3,3,3,3,
                 3,2,3,1), byrow = T, ncol = 4)

and following condition:

cond <- c(0,0) # binary

Problem: If cond[1] is 0 and there is a 1 in either the first or third column, the corresponding rows will be removed. Similarly if cond[2] is 0 and there is a 1 in either the second or fourth column, the corresponding rows will be removed. For example the new matrix will be:

newMat <- testMat[-c(1,3,6),] # for cond <- c(0,0)
newMat <- testMat[-c(1,3),] # for cond <- c(0,1)
newMat <- testMat[-c(6),] # for cond <- c(1,0)
newMat <- testMat # for cond <- c(1,1)

I tried in following way which is both wrong and clumsy.

newMat <- testMat[-(cond[1] == 0 & testMat[,c(1,3)] == 1),]
newMat <- newMat[-(cond[2] == 0 & newMat[,c(2,4)] == 1),]

Can you help to find a base R solution?

CodePudding user response：

This is ugly, but seems to work:
(generalized for length of cond, assuming that the matrix has length(cond)*2 columns)

keepRows <- apply(testMat, 1,
                  function(rw, cond) {
                    logicrow <- vector(length=2)
                    for (b in 1:length(cond)) {
                      logicrow[b] <- ifelse(cond[b]==0, all(rw[b]!=1) & all(rw[length(cond) b]!=1), TRUE)
                    }
                    all(logicrow)
                  }, cond = cond)

newMat <- testMat[keepRows, ]

(edited according to comment)

CodePudding user response：

Assuming 1) cond can be of arbitrary length, 2) testMat has an even number of columns, and 3) the rule is to look at the i-th and (i 2)-th column of testMat

cond=c(0,0)

unlist(
  sapply(
    1:length(cond),
    function(i){
      j=rowSums(testMat[,c(i,i 2)]==1)
      if (cond[i]==0 & sum(j)>0) which(j>0) else nrow(testMat) 1
    }
  )
)
[1] 1 3 6

which returns the rows which satisfy your conditions, you can then remove these

testMat[-.Last.value,]