将包含具有相同名称的对象的多个.RData文件合并到一个.RData文件中

Question

I have many many .RData files containing one dataframe that I had saved in a previous analysis and the data frame has the same name for each file loaded. So for example using load(file1.RData) I get a data frame called 'df', then using load(file2.RData) I get a data frame with the same name 'df'. I was wondering if it is at all possible to combine all these .RData files into one big .RData file since I need to load them all at once, with the name of each df equal to the file name so I can then use the different data frames.

I can do this using the code below, but it is very intricate, there must be a simpler way to do this… Thank you for your suggestions.

Say I have 3 .RData files and want to save all in a file called "main.RData" with their specific name (now they all come out as 'df'):

all.files = c("/Users/fra/file1.RData", "/Users/fra/file2.RData", "/Users/fra/file3.RData")
assign(gsub("/Users/fra/", "", all.files[1]), local(get(load(all.files[1]))))
rm(list= ls()[!(ls() %in% (ls(pattern = "file")))])
save.image(file="main.RData")


all.files = all.files = c("/Users/fra/file1.RData", "/Users/fra/file2.RData", "/Users/fra/file3.RData")

for (f in all.files[-1]) {
  assign(gsub("/Users/fra/", "", f), local(get(load(f))))
  rm(list= ls()[!(ls() %in% (ls(pattern = "file")))])
  save.image(file="main.RData")
}

Answer 1

I think the best answer I saw was the code below, which I copied from an SO answer which I can't track down right now. Apologies to the original author.

resave <- function(..., list = character(), file) {
   previous  <- load(file)
   var.names <- c(list, as.character(substitute(list(...)))[-1L])
   for (var in var.names) assign(var, get(var, envir = parent.frame()))
   save(list = unique(c(previous, var.names)), file = file)
}
#I took advantage of the fact the load function 
#returns the name of the loaded variables, so 
#I could use the function's environment instead of creating one.
#And when using get, I was careful to only look in the 
#environment from which the function is called, i.e. parent.frame()

Answer 2

Here's an option that incorporates several existing posts

all.files = c("file1.RData", "file2.RData", "file3.RData")

Read multiple dataframes into a single named list ( How can I load an object into a variable name that I specify from an R data file? )

mylist<- lapply(all.files, function(x) {
  load(file = x)
  get(ls()[ls()!= "filename"])
})

names(mylist) <- all.files #Note, the names here don't have to match the filenames

You can save the list, or transfer the dataframes into the global environment prior to saving ( Unlist a list of dataframes )

list2env(mylist ,.GlobalEnv)

Alternatively, if the dataframes were identical and you wanted to create a single big dataframe, you could collapse the list and add a variable with names of contributing files ( Dataframes in a list; adding a new variable with name of dataframe ).

all <- do.call("rbind", mylist)
all$id <- rep(all.files, sapply(mylist, nrow))