另一种尝试:

apply(m[,-ncol(m)] >= m[,ncol(m)], 1, match, x=TRUE)
#[1] NA NA  1  2

或删除apply:

chk <- m[,-ncol(m)] >= m[,ncol(m)]
replace(max.col(chk,"first"), rowSums(chk)==0, NA)
#[1] NA NA  1  2

它实际上创建了除最后一个列之外的所有m列的逻辑矩阵，测试值是否为最后一个列值的>=.然后，使用match提取每行中第一个TRUE的位置.

It essentially creates a logical matrix of the all the m columns except the last, testing if the values are >= to the last column values. Then the position of the first TRUE in each row is extracted using match.

针对Ben的解决方案使用更大的矩阵进行测试的速度:

Testing speed using a bigger matrix against Ben's solution:

m<-matrix(sample(10,1.6e6,replace=TRUE),nrow=4e5)

replicate(5,
system.time(
 apply(m[,-ncol(m)] >= m[,ncol(m)], 1, match, x=TRUE)
))
#elapsed     0.7 0.77 0.77 0.76 0.93

replicate(5,
system.time({
 m2 <- sweep(m,1,m[,ncol(m)],">=")
 v <- apply(m2,1,function(x) which(x)[1])
 ifelse(v==ncol(m),NA,v)
}))
#elapsed    1.11 1.04 1.10 1.06 1.06