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 <m_hof...@web.de> 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
>
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> and provide commented, minimal, self-contained, reproducible code.
>
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