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

Question

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))
}

Solution 1

This should be equivalent to your current code, only a lot faster:

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

Solution 2

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).

Solution 3

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.

Solution 4

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).

View more solutions

Comments

[Ben Bolker]

Bingo. This should be much faster than my solution too, but I couldn't think of it fast enough.

[user1787675]

That's magical! It takes about 20 seconds to do this on my one-year-old toshiba machine, which saves me a lot of time. And your function to show the run time is very interesting too.

[Joshua Ulrich]

+1, but I'd recommend setting USE.NAMES=FALSE in unlist in order to save time and memory.

[Johan Larsson]

It should be use.names (i.e. in lowercase).

[mikey]

Just to clarify, it should be output <- matrix(unlist(z), ncol = 10, byrow = TRUE, use.names=FALSE) to be the most efficient.

[Felix]

@mikey Almost. It should be: output <- matrix(unlist(z, use.names = FALSE), ncol = 10, byrow = TRUE)