Hi Ivan,
Now I read the other answers I also think I misunderstood the
question... The good thing is that one of use certainly gave the right
answer to the question ;-)
Alain
On 20-Aug-10 12:37, Ivan Calandra wrote:
> Hi,
>
> I thought you were looking for table(), but the other answers gave you
> something really different; I might have wrongly understood your question.
>
> HTH,
> Ivan
>
> Le 8/20/2010 12:32, Alain Guillet a écrit :
>> Hi,
>>
>> You can try sapply(levels(as.factor(dat1)),nchar)
>>
>> Alain
>>
>> On 20-Aug-10 12:01, Ron Michael wrote:
>>> Dear all, let suppose I have following vector:
>>>
>>>> dat1<- c(rep("asd", 5), rep("xyz", 12), rep("erd", 17))
>>>> dat1<- dat1[sample(1:length(dat1), length(dat1), replace=F)]
>>>> dat1
>>> [1] "erd" "xyz" "erd" "asd" "asd" "erd" "xyz" "asd" "erd" "erd"
>>> "asd" "xyz" "erd" "asd" "xyz" "xyz" "erd" "xyz" "erd"
>>> [20] "erd" "erd" "xyz" "xyz" "erd" "erd" "erd" "erd" "xyz" "xyz"
>>> "xyz" "erd" "xyz" "erd" "erd"
>>>
>>> Here I want to know the length of replications for each unique
>>> items viz "asd", "xyz", and "erd". Is there any R function available
>>> to directly implement that?
>>> Thanks,
>>>
>>>
>>> [[alternative HTML version deleted]]
>>>
>>>
>>>
>>> ______________________________________________
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>
>
> ______________________________________________
> [email protected] mailing list
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> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
--
Alain Guillet
Statistician and Computer Scientist
SMCS - IMMAQ - Université catholique de Louvain
Bureau c.316
Voie du Roman Pays, 20
B-1348 Louvain-la-Neuve
Belgium
tel: +32 10 47 30 50
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