This works. But I wish I could write it without a lot of trial and
error. :-(
ha = matrix(nrow=7, ncol=24)
colnames(ha) = as.character(c(0:23))
rownames(ha) = rownames(maxrdf)
m = as.matrix(maxrdf)
for(j in 1:7) {
x = aggregate(m[j,], by=list(s), FUN=sum)
ha[j,] = x[[2]]
}
On 2/7/2010 1:57 PM, James Rome wrote:
On 2/7/2010 1:35 PM, David Winsemius wrote:But to answer your question:
> apply(d, 1, function(z) aggregate(z, by=list(s), FUN=sum) )
David,
That works, but I do not understand why I could not use aggregate
directly. And the answer comes out as a list, which thus far baffles me.
How do I get the answer as a matrix in my original code, which I
modified to use apply?
ha = matrix(nrow=7, ncol=24)
colnames(ha) = as.character(c(0:23))
rownames(ha) = rownames(maxrdf)
for(j in 1:7) {
x = apply(maxrdf[j,], 1, function(z) aggregate(z, by=list(s),
FUN=sum) )
ha[j,] = x[[1]][2]
}
Unfortunately, ha gets converted into a list, and then I can't use it
for my plots. And you can probably educate me on how to get what I am
aiming for (a matrix with the rows as the days, the columns as the
hours, and the content as the hourly sum of the 15-minute chunks)
without using the above for loop.
Thanks for the help,
Jim
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