Thanks for your answers,
Best,
Gildas
Brian Diggs a écrit :
> On 7/22/2010 5:01 AM, Allan Engelhardt wrote:
>> There are so many ways.... Here is one:
>>
>> aggregate(v ~ u, data=X, function(...) length(unique(...)))
>> # u v
>> # 1 T1 2
>> # 2 T2 1
>>
>> Hope this helps
>
> Here is one other way, using the plyr package (which is very good for
> taking a data structure (data.frame, list, array), pulling it apart by
> some criteria, doing something on each of the parts, and putting the
> results back together):
>
> library("plyr")
> ddply(X, .(u), function(x) {length(unique(x$v))})
> # u V1
> #1 T1 2
> #2 T2 1
>
>
>> Allan.
>>
>> On 22/07/10 12:52, Gildas Mazo wrote:
>>> Dear R users,
>>>
>>> I want to aggregate data in the following way:
>>>
>>> ###
>>>
>>> X<- data.frame(u = c("T1","T1","T1","T2"), v=c("a","a","b","a"))
>>> X
>>> library(sqldf)
>>> sqlOut<- sqldf("select count(distinct(v)) from X group by u")
>>> sqlOut
>>>
>>> ###
>>>
>>> Now I want to get the same result without using SQL. How can I achieve
>>> that ?
>>>
>>> Thanks for your help,
>>>
>>> Gildas
>>>
>>> ______________________________________________
>>> 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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