>>>>> Marius Hofert <m_hof...@web.de>
>>>>> on Wed, 22 Jun 2011 01:00:21 +0200 writes:
> Thanks a lot! Works like charm :-))) Cheers,
well,
I'd still *strongly* vote for Bill Dunlap's version.
Martin
> 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.
>>>>
>>>
______________________________________________
> 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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