On Jun 21, 2011, at 6:51 PM, Dennis Murphy wrote:
Ahhh...you want a matrix. xtabs() doesn't easily allow coercion to a
matrix object, so try this instead:
What am I missing? A contingency table already inherits from matrix-
class and if you insisted on coercion it appears simple:
> xtb <- xtabs(value ~ year + block, data = df)
> is.matrix(xtb)
[1] TRUE
> as.matrix(xtb)
block
year a b c
2000 1 0 5
2001 2 4 6
2002 3 0 0
--
David.
library(reshape)
as.matrix(cast(df, year ~ block, fill = 0))
a b c
2000 1 0 5
2001 2 4 6
2002 3 0 0
Hopefully this is more helpful...
Dennis
On Tue, Jun 21, 2011 at 3:35 PM, Dennis Murphy <[email protected]>
wrote:
Hi:
xtabs(value ~ year + block, data = df)
block
year a b c
2000 1 0 5
2001 2 4 6
2002 3 0 0
HTH,
Dennis
On Tue, Jun 21, 2011 at 3:13 PM, Marius Hofert <[email protected]>
wrote:
Dear expeRts,
In the minimal example below, I have a data.frame containing three
"blocks" of years
(the years are subsets of 2000 to 2002). For each year and block a
certain "value" is given.
I would like to create a matrix that has row names given by all
years ("2000", "2001", "2002"),
and column names given by all blocks ("a", "b", "c"); the entries
are then given by the
corresponding value or zero if not year-block combination exists.
What's a short way to achieve this?
Of course one can setup a matrix and use for loops (see below)...
but that's not nice.
The problem is that the years are not running from 2000 to 2002
for all three "blocks"
(the second block only has year 2001, the third one has only 2000
and 2001).
In principle, table() nicely solves such a problem (see below) and
fills in zeros.
This is what I would like in the end, but all non-zero entries
should be given by df$value,
not (as table() does) by their counts.
Cheers,
Marius
(df <- data.frame(year=c(2000, 2001, 2002, 2001, 2000, 2001),
block=c("a","a","a","b","c","c"), value=1:6))
table(df[,1:2]) # complements the years and fills in 0
year <- c(2000, 2001, 2002)
block <- c("a", "b", "c")
res <- matrix(0, nrow=3, ncol=3, dimnames=list(year, block))
for(i in 1:3){ # year
for(j in 1:3){ # block
for(k in 1:nrow(df)){
if(df[k,"year"]==year[i] && df[k,"block"]==block[j])
res[i,j] <- df[k,"value"]
}
}
}
res # does the job; but seems complicated
David Winsemius, MD
West Hartford, CT
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