Hi Alfredo,
Wilfrieds code is without doubt the fastest, but not memory safe!
calc calls the function for each pixel, so "x" is a single pixel (if "test" is
a single layer), I think this makes calc slow.
If the solution below is still to slow, you could think using more cores if you
machine has some.
cheers, Matteo
##############################
###
test <- raster(system.file("external/test.grd", package="raster"))
funNA <- function(x) {x[is.na(x)] <- 0; return(x)}
a<- proc.time()[[3]]
test <- calc(test,funNA)
proc.time()[[3]]-a
###
# write to new file, memory save (you can write directly back to test if you
like!)
test <- raster(system.file("external/test.grd", package="raster"))
a <-proc.time()[[3]]
tr <- blockSize(test)
out <- raster(test)
out <- writeStart(out,filename="result.grd",overwrite=T)
for (l in 1:tr$n){
val <- getValues(test,tr$row[l],tr$nrows[l])
val[is.na(val)] <- 0 # in such a case you fun it acts on a
vector!
out <- writeValues(out,val,tr$row[l])
}
out <- writeStop(out)
proc.time()[[3]]-a
###
a<- proc.time()[[3]]
test[is.na(test[])] <- 0
proc.time()[[3]]-a
>>> Alfredo Alessandrini <alfreal...@gmail.com> 10.05.2011 16:21 >>>
Hi,
I've to replace NA value with zero in some raster.
I use the raster package.
Is there a faster method than this:
funNA <- function(x) {x[is.na(x)] <- 0; return(x)}
test <- calc(test,funNA)
Alfredo
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