I have a column as below. Only for non-null elements, I want to get a matrix such as below. 6th column represent the actual value.
1 0 0 0 0 1 0 1 0 0 0 2 0 0 0 1 0 5 Any hint what is the efficient way to do this? which commands should I use? I am thinking of writing a if loop within for loop, but don't think it will be very efficient :(
abc=c('1','2','null','5','null') Assuming there is an error in your example, this is just a dummy variable coding essentially:
abc <- c('1','2','null','5','null') abc <- factor(abc,levels=1:5) cbind(model.matrix(~abc+0),orig=na.omit(abc)) # abc1 abc2 abc3 abc4 abc5 orig #1 1 0 0 0 0 1 #2 0 1 0 0 0 2 #4 0 0 0 0 1 5 If you want to automatically calculate the range of possible factors, try:
abc <- c('1','2','null','5','null') rng <- range(as.numeric(abc),na.rm=TRUE) abc <- factor(abc,levels=seq(rng[1],rng[2])) cbind(model.matrix(~abc+0),orig=na.omit(abc)) # abc1 abc2 abc3 abc4 abc5 orig #1 1 0 0 0 0 1 #2 0 1 0 0 0 2 #4 0 0 0 0 1 5 
rng <- range(as.numeric(abc),na.rm=TRUE) and use that to determine the levels of the factor: factor(abc,levels=seq(rng[1],rng[2]))as.numeric just forces non-numbers to NA, that is all.It's not clear why that matrix is six elements wide but if it is length(abc)+1 then just substitute that expression for my use of 6.
> abcn <- as.numeric(abc) > zero <- matrix(0,nrow=length(abcn[!is.na(abcn)]), ncol=6) > zero[ cbind(1:3, which( !is.na(abcn)) ) ] <- 1 > zero[ , 6] <- abcn[!is.na(abcn)] > zero [,1] [,2] [,3] [,4] [,5] [,6] [1,] 1 0 0 0 0 1 [2,] 0 1 0 0 0 2 [3,] 0 0 0 1 0 5 You can index teh [<- function for matrices with a two coulmn matrix and that's what I'm doing in the third line. The rest of it is ordinary matrix indexing.
