Is ?split what you are after?
Emmanuel Levy wrote:
Dear Peter and Henrik, Thanks for your replies - this helps speed up a bit, but I thought there would be something much faster. What I mean is that I thought that a particular value of a level could be accessed instantly, similarly to a "hash" key. Since I've got about 6000 levels in that data frame, it means that making a list L of the form L[[1]] = values of name "1" L[[2]] = values of name "2" L[[3]] = values of name "3" ... would take ~1hour. Best, Emmanuel 2008/8/12 Henrik Bengtsson <[EMAIL PROTECTED]>:To simplify: n <- 2.7e6; x <- factor(c(rep("A", n/2), rep("B", n/2))); # Identify 'A':s t1 <- system.time(res <- which(x == "A")); # To compare a factor to a string, the factor is in practice # coerced to a character vector. t2 <- system.time(res <- which(as.character(x) == "A")); # Interestingly enough, this seems to be faster (repeated many times) # Don't know why. print(t2/t1); user system elapsed 0.632653 1.600000 0.754717 # Avoid coercing the factor, but instead coerce the level compared to t3 <- system.time(res <- which(x == match("A", levels(x)))); # ...but gives no speed up print(t3/t1); user system elapsed 1.041667 1.000000 1.018182 # But coercing the factor to integers does t4 <- system.time(res <- which(as.integer(x) == match("A", levels(x)))) print(t4/t1); user system elapsed 0.4166667 0.0000000 0.3636364 So, the latter seems to be the fastest way to identify those elements. My $.02 /Henrik On Tue, Aug 12, 2008 at 7:31 PM, Peter Cowan <[EMAIL PROTECTED]> wrote:Emmanuel, On Tue, Aug 12, 2008 at 4:35 PM, Emmanuel Levy <[EMAIL PROTECTED]> wrote:Dear All, I have a large data frame ( 2700000 lines and 14 columns), and I would like to extract the information in a particular way illustrated below: Given a data frame "df":col1=sample(c(0,1),10, rep=T) names = factor(c(rep("A",5),rep("B",5))) df = data.frame(names,col1) dfnames col1 1 A 1 2 A 0 3 A 1 4 A 0 5 A 1 6 B 0 7 B 0 8 B 1 9 B 0 10 B 0 I would like to tranform it in the form:index = c("A","B") col1[[1]]=df$col1[which(df$name=="A")] col1[[2]]=df$col1[which(df$name=="B")]I'm not sure I fully understand your problem, you example would not run for me. You could get a small speedup by omitting which(), you can subset by a logical vector also which give a small speedup.n <- 2700000 foo <- data.frame(+ one = sample(c(0,1), n, rep = T), + two = factor(c(rep("A", n/2 ),rep("B", n/2 ))) + )system.time(out <- which(foo$two=="A"))user system elapsed 0.566 0.146 0.761system.time(out <- foo$two=="A")user system elapsed 0.429 0.075 0.588 You might also find use for unstack(), though I didn't see a speedup.system.time(out <- unstack(foo))user system elapsed 1.068 0.697 2.004 HTH PeterMy problem is that the command: *** which(df$name=="A") *** takes about 1 second because df is so big. I was thinking that a "level" could maybe be accessed instantly but I am not sure about how to do it. I would be very grateful for any advice that would allow me to speed this up. Best wishes, Emmanuel______________________________________________ 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.______________________________________________ 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.
______________________________________________ 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.
