[R] assign a cluster based on a variable

[Dominik P.H. Kalisch]

Oh, right. Now I see.
Thanks for your help.

Dominik

On Thu, 09 Jun 2011 15:33:43 +0200, Uwe Ligges
<ligges at statistik.tu-dortmund.de> wrote:
> On 09.06.2011 14:31, Dominik P.H. Kalisch wrote:
>> Hi Uwe,
>>
>> thanks for your help. I didn't know that function (mea culpa)...
>> With merge() it works what I want to do.
>> But I'm still interessted, for a learning purpose, why the loop didn't
>> work. So here are the str() results:
>>
>>> str(cluster)
>>   int [1:18, 1:2] 12051000 12052000 12053000 12054000 12060000 12061000
>> 12062000 12063000 12064000 12065000 ...
>>   - attr(*, "dimnames")=List of 2
>>    ..$ : chr [1:18] "1" "2" "3" "4" ...
>>    ..$ : chr [1:2] "Kreis" "Cluster"
>>
>>> str(KreisSA)
>>   num [1:2302, 1:2] 12069000 12072000 12067000 12060000 12070000 ...
>>   - attr(*, "dimnames")=List of 2
>>    ..$ : NULL
>>    ..$ : chr [1:2] "Kreis" "Cluster"
> 
> 
> So KreisSA is a 2 column matrix rather than a vector. Hence you also
> get 2 column matrix (logical) from the comparison KreisSA ==
> cluster[i,1] and so you are assigning into KreisSAa.
> 
> Uwe Ligges
> 
> 
> 
>> Thanks for your help.
>>
>>
>> On Tue, 07 Jun 2011 22:49:04 +0200, Uwe Ligges
>> <ligges at statistik.tu-dortmund.de>  wrote:
>>> On 07.06.2011 16:24, Dominik P.H. Kalisch wrote:
>>>> Hi,
>>>>
>>>> I have two matrices of the following form:
>>>>
>>>> cluster (n=18):
>>>> 12062 1
>>>> 12063 2
>>>> 12064 2
>>>> 12065 3
>>>> 12066 5
>>>>
>>>> KreisSA (n=2304)
>>>> 12062
>>>> 12062
>>>> 12067
>>>> 12065
>>>> 12063
>>>> 12067
>>>>
>>>> I try to assign the cluster[,2] to KreisSAa by the follwoing loop:
>>>>
>>>> n<- nrow(cluster)
>>>> KreisSAa<- numeric()
>>>>
>>>> for(i in 1:n){
>>>> KreisSAa[KreisSA == cluster[i,1]]<- cluster[i,2]
>>>> }
>>>>
>>>> The result is a vector of the length n=4608 where after the entry 2304
>>>> are just NA's
>>>> Has someone an idea what I do wrong?
>>>
>>> Since we do not know the structure of your objects, we cannot say
>>> easily. You may want to provide
>>>
>>> str(cluster)
>>> and
>>> str(KreisSA)
>>>
>>> Anyway: I'd just use merge()
>>>
>>> Uwe Ligges
>>>
>>>
>>>
>>>
>>>
>>>
>>>> Thanks for your help.
>>>> Dominik
>>>>
>>>> ______________________________________________
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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.
>>
>> ______________________________________________
>> R-help at 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.