Thanks a lot! Works like charm :-)))
Cheers,
Marius
On 2011-06-22, at 24:51 , Dennis Murphy wrote:
> Ahhh...you want a matrix. xtabs() doesn't easily allow coercion to a
> matrix object, so try this instead:
>
> 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 <djmu...@gmail.com> 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 <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
>>>
>>> ______________________________________________
>>> R-help@r-project.org mailing list
>>> https://stat.ethz.ch/mailman/listinfo/r-help
>>> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
>>> and provide commented, minimal, self-contained, reproducible code.
>>>
>>
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