Re: [R] SQL/R

2010-07-26 Thread Bart Joosen
see the manual from package sqldf: http://cran.r-project.org/web/packages/sqldf/sqldf.pdf Bart -- View this message in context: http://r.789695.n4.nabble.com/SQL-R-tp2298545p2301975.html Sent from the R help mailing list archive at Nabble.com. __ R-h

Re: [R] SQL/R

2010-07-23 Thread Gildas Mazo
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 othe

Re: [R] SQL/R

2010-07-22 Thread Brian Diggs
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

Re: [R] SQL/R

2010-07-22 Thread Allan Engelhardt
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 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",

[R] SQL/R

2010-07-22 Thread Gildas Mazo
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 t