Hello,
Using the technique suggested here (
https://stat.ethz.ch/pipermail/r-devel/2007-September/047028.html),
we can update a matrix in place (see example 1 below), with huge time and
memory savings.
Is it possible to tweak example 2 to update in place also? The only
difference is that the matrix is held within a new environment.
Thanks,
Simon
# example 1 - fast and doesn't create copies
rm(list=ls())
main <- function() {
matrix1 <- matrix(1.0, nrow=60000, ncol=200)
if (sum(matrix1) == 12000000) { print ("OK") } else { print ("NOT
OK") }
funcExample <- function() eval.parent(substitute({
print(paste("Mem1", memory.size()))
for (i in 1:30) {
eval(parse(text="temp <- matrix1[10,]"))
eval(parse(text="matrix1[10,] <- temp * 0"))
}
print(paste("Mem2", memory.size()))
}))
print (paste("Time:", system.time(funcExample())[1], "secs"))
if (sum(matrix1) == 11999800) { print ("OK") } else { print ("NOT
OK") }
}
main()
[1] "OK"
[1] "Mem1 99.32"
[1] "Mem2 99.46"
[1] "Time: 0 secs"
[1] "OK"
# example 2 - slow and does create copies
rm(list=ls())
main <- function() {
envData <- new.env()
envData$matrix1 <- matrix(1.0, nrow=60000, ncol=200)
if (sum(envData$matrix1) == 12000000) { print ("OK") } else {
print ("NOT OK") }
funcExample <- function() eval.parent(substitute({
print(paste("Mem1", memory.size()))
for (i in 1:30) {
eval(parse(text="temp <- envData$matrix1[10,]"))
eval(parse(text="envData$matrix1[10,] <- temp * 0"
))
}
print(paste("Mem2", memory.size()))
}))
print (paste("Time:", system.time(funcExample())[1], "secs"))
if (sum(envData$matrix1) == 11999800) { print ("OK") } else {
print ("NOT OK") }
}
main()
[1] "OK"
[1] "Mem1 99.32"
[1] "Mem2 190.88"
[1] "Time: 2.6 secs"
[1] "OK"
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