[R] Tricky (?) conversion from data.frame to matrix where not all pairs exist

[Marius Hofert]

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 <djmuser at 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_hofert at 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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>>