Hi Ivan, thank you very much for your detailed explanations and response! Please excuse, that I did not include a reproducible example.
Yes, what I am looking for is such a function - but I suspect, that it does not exist already, so I have to write it on my own. Best, Friedericksen On 11/11/2010 10:23 AM, Ivan Calandra wrote:
Hi, A reproducible example would have been nice, but a correct code even more (you forgot some commas)! So if you meant this: x <- list( list( list(df1,df2), list(df3, list(df4,df5)), list(df6,df7))) check str(x): (I removes the details of each df) List of 1 $ :List of 3 ..$ :List of 2 .. ..$ :'data.frame': 10 obs. of 2 variables: ## df1 .. ..$ :'data.frame': 10 obs. of 2 variables: ## df2 ..$ :List of 2 .. ..$ :'data.frame': 10 obs. of 2 variables: ## df3 .. ..$ :List of 2 .. .. ..$ :'data.frame': 10 obs. of 2 variables: ## df4 .. .. ..$ :'data.frame': 10 obs. of 2 variables: ## df5 ..$ :List of 2 .. ..$ :'data.frame': 10 obs. of 2 variables: ## df6 .. ..$ :'data.frame': 10 obs. of 2 variables: ## df7 And check x: (I removed the details again) [[1]] [[1]][[1]] [[1]][[1]][[1]] df1 [[1]][[1]][[2]] df2 [[1]][[2]] [[1]][[2]][[1]] df3 [[1]][[2]][[2]] [[1]][[2]][[2]][[1]] df4 [[1]][[2]][[2]][[2]] df5 [[1]][[3]] [[1]][[3]][[1]] df6 [[1]][[3]][[2]] df7 If you want to access them, you need to follow the structure of your list, e.g. for df1: x[[1]][[1]][[1]] for df5: x[[1]][[2]][[2]][[2]] You need to access them with indexes because the list elements are unnamed. Now if you do: x <- list(uppermost=list( medium1=list(df1=df1,df2=df2), medium2=list(df3=df3, lowermost=list(df4=df4,df5=df5)), medium3=list(df6=df6,df7=df7))) You can access them with names (but you still need to follow the structure, i.e. x[[df5]] cannot work): For df1: x$uppermost$medium1$df1 or x[["uppermost"]][["medium1"]][["df1"]] for df5: x$uppermost$medium2$lowermost$df5 or x[["uppermost"]][["medium2"]][["lowermost"]][["df5"]] But I guess you can write a function that would "search" through the structure of x for a given df (that has to be named then) and return it. Maybe you have your reasons to have such a complicated list, but if it were me, I would make it easier to understand and to access (i.e. less nested, max 2 levels) HTH, Ivan Le 11/11/2010 09:05, Friedericksen Hope a écrit :Hello, I have a nested named list structure, like the following: x <- list( list( list(df1,df2) list(df3, list(df4,df5)) list(df6,df7))) with df1...d7 as data frames. Every data frame is named. Is there a way to get a specific named element in x? so, for example, x[[c("df5")]] gives me the data frame 5? Thank you in advance! Best, Friedericksen ______________________________________________ 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.
