Your code is doing too many needless things.
The following takes about one second on my slow Vista laptop.
n <- 500
mat <- matrix(1:(n*n), n)
v <- n:1
z <- 2*1:(n/2)
w <- c(z, rev(z))
for(i in seq_len(n)){
for(j in seq_len(n)){
if(v[j] + w[i] <= n)(mat[i,j] <- NA)
}
}
rownames(mat) <- v
colnames(mat) <- w
str(mat)
You end up with matrix, but if you really want a data.frame
with duplicate names, that's easy to get. Do you actually
want those row/col names or are they just used to identify
the cells that get NA?
Depending on what you really need, the following may be
good enough; takes about 0.1 seconds.
n <- 500
mat <- matrix(1:(n*n), n)
for(i in 1:(n/2)){mat[i, -(1:(2*i))] <- mat[n+1-i, -(1:(2*i))] <- NA}
-Peter Ehlers
Etienne Stockhausen wrote:
Hey everybody,
I have a small problem with a routine, which prepares some data for
plotting.
I've made a small example:
c=10
mat=data.frame(matrix(1:(c*c),c,c))
row.names(mat)=seq(c,1,length=c)
names(mat)=c(seq(2,c,length=c/2),seq(c,2,length=c/2))
v=as.numeric(row.names(mat))
w=as.numeric(names(mat))
for(i in 1:c)
{ for(j in 1:c)
{
if(v[j]+w[i]<=c)(mat[i,j]=NA)
}}
This produces exactly the data I need to go on, but if I increase the
constant c ,to for instance 500 , it takes a very long time to set the
NA's.
I've heard there is a much faster way to set the NA's using the command
apply( ), but I don't know how.
I'm looking forward for any ideas or hints, that might help me.
Best regards
Etienne
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--
Peter Ehlers
University of Calgary
403.202.3921
______________________________________________
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.