Conditional removal of rows in R - r

Question

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?

Answer 1

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)

Answer 2

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,]