[r] How to convert a huge list-of-vector to a matrix more efficiently?

I have a list of length 130,000 where each element is a character vector of length 110. I would like to convert this list to a matrix with dimension 1,430,000*10. How can I do it more efficiently?\ My code is :

output=NULL
for(i in 1:length(z)) {
 output=rbind(output,
              matrix(z[[i]],ncol=10,byrow=TRUE))
}

you can use as.matrix as below:

output <- as.matrix(z)

I think you want

output <- do.call(rbind,lapply(z,matrix,ncol=10,byrow=TRUE))

i.e. combining @BlueMagister's use of do.call(rbind,...) with an lapply statement to convert the individual list elements into 11*10 matrices ...

Benchmarks (showing @flodel's unlist solution is 5x faster than mine, and 230x faster than the original approach ...)

n <- 1000
z <- replicate(n,matrix(1:110,ncol=10,byrow=TRUE),simplify=FALSE)
library(rbenchmark)
origfn <- function(z) {
    output <- NULL 
    for(i in 1:length(z))
        output<- rbind(output,matrix(z[[i]],ncol=10,byrow=TRUE))
}
rbindfn <- function(z) do.call(rbind,lapply(z,matrix,ncol=10,byrow=TRUE))
unlistfn <- function(z) matrix(unlist(z), ncol = 10, byrow = TRUE)

##          test replications elapsed relative user.self sys.self 
## 1   origfn(z)          100  36.467  230.804    34.834    1.540  
## 2  rbindfn(z)          100   0.713    4.513     0.708    0.012 
## 3 unlistfn(z)          100   0.158    1.000     0.144    0.008 

If this scales appropriately (i.e. you don't run into memory problems), the full problem would take about 130*0.2 seconds = 26 seconds on a comparable machine (I did this on a 2-year-old MacBook Pro).

You can also use,

output <- as.matrix(as.data.frame(z))

The memory usage is very similar to

output <- matrix(unlist(z), ncol = 10, byrow = TRUE)

Which can be verified, with mem_changed() from library(pryr).

It would help to have sample information about your output. Recursively using rbind on bigger and bigger things is not recommended. My first guess at something that would help you:

z <- list(1:3,4:6,7:9)
do.call(rbind,z)

See a related question for more efficiency, if needed.