Hi David,
thanks for the quick response. That's nice. Is there also a way without loading
an additional package? I'd prefer loading less packages if possible.
Cheers,
Marius
On 2011-06-22, at 15:38 , David Winsemius wrote:
>
> On Jun 22, 2011, at 9:19 AM, Marius Hofert wrote:
>
>> Hi,
>>
>> and what's the simplest way to obtain a *data.frame* with all years?
>> The matching seems more difficult here because the years can/will show up
>> several times...
>>
>> (df <- data.frame(year=c(2000, 2001, 2002, 2001, 2000, 2001),
>> block=c("a","a","a","b","c","c"), value=1:6))
>> (df. <- data.frame(year=rep(2000:2002, 3), block=rep(c("a", "b", "c"),
>> each=3), value=0))
>> # how to fill in the given values?
>
> These days I think most people would reach for melt() in either reshape or
> reshape2 packages:
>
> > require(reshape2)
> Loading required package: reshape2
> > melt(xtb)
> year block value
> 1 2000 a 1
> 2 2001 a 2
> 3 2002 a 3
> 4 2000 b 0
> 5 2001 b 4
> 6 2002 b 0
> 7 2000 c 5
> 8 2001 c 6
> 9 2002 c 0
>
> It seems to do a good job of guessing what you want whereas the reshape
> function in my hands is very failure prone (... yes, the failings are mine.)
> --
> David
>>
>> Cheers,
>>
>> Marius
>>
>>
>> On 2011-06-22, at 14:40 , Dennis Murphy wrote:
>>
>>> I saw it as an xtabs object - I didn't think to check whether it was
>>> also a matrix object. Thanks for the clarification, David.
>>>
>>> Dennis
>>>
>>> On Wed, Jun 22, 2011 at 4:59 AM, David Winsemius <dwinsem...@comcast.net>
>>> wrote:
>>>>
>>>> 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 <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
>>>>>
>>>>
>>>>
>>>> David Winsemius, MD
>>>> West Hartford, CT
>>>>
>>>>
>>
>> ______________________________________________
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>> 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.
>
> David Winsemius, MD
> West Hartford, CT
>
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