Dataframes in a list; adding a new variable with n

问题:

I have a list of dataframes which I eventually want to merge while maintaining a record of their original dataframe name or list index. This will allow me to subset etc across all the rows. To accomplish this I would like to add a new variable 'id' to every dataframe, which contains the name/index of the dataframe it belongs to.

Edit: "In my real code the dataframe variables are created from reading multiple files using the following code, so I don't have actual names only those in the 'files.to.read' list which I'm unsure if they will align with the dataframe order:

mylist <- llply(files.to.read, read.csv)

A few methods have been highlighted in several posts: Working-with-dataframes-in-a-list-drop-variables-add-new-ones and Using-lapply-with-changing-arguments

I have tried two similar methods, the first using the index list:

df1 <- data.frame(x=c(1:5),y=c(11:15))
df2 <- data.frame(x=c(1:5),y=c(11:15))
mylist <- list(df1,df2)

# Adds a new coloumn 'id' with a value of 5 to every row in every dataframe.
# I WANT to change the value based on the list index.
mylist1 <- lapply(mylist, 
    function(x){
        x$id <- 5
        return (x)
    }
)
#Example of what I WANT, instead of '5'.
#> mylist1
#[[1]]
  #x  y id
#1 1 11  1
#2 2 12  1
#3 3 13  1
#4 4 14  1
#5 5 15  1
#
#[[2]]
  #x  y id
#1 1 11  2
#2 2 12  2
#3 3 13  2
#4 4 14  2
#5 5 15  2

The second attempts to pass the names() of the list.

# I WANT it to add a new coloumn 'id' with the name of the respective dataframe
# to every row in every dataframe.
mylist2 <- lapply(names(mylist), 
    function(x){
        portfolio.results[[x]]$id <- "dataframe name here"
        return (portfolio.results[[x]])
    }
)
#Example of what I WANT, instead of 'dataframe name here'.
# mylist2
#[[1]]
  #x  y id
#1 1 11  df1
#2 2 12  df1
#3 3 13  df1
#4 4 14  df1
#5 5 15  df1
#
#[[2]]
  #x  y id
#1 1 11  df2
#2 2 12  df2
#3 3 13  df2
#4 4 14  df2
#5 5 15  df2

But the names() function doesn't work on a list of dataframes; it returns NULL. Could I use seq_along(mylist) in the first example.

Any ideas or better way to handle the whole "merge with source id"

Edit - Added Solution below: I've implemented a solution using Hadleys suggestion and Tommy’s nudge which looks something like this.

files.to.read <- list.files(datafolder, pattern="\\_D.csv$", full.names=FALSE)
mylist <- llply(files.to.read, read.csv)
all <- do.call("rbind", mylist)
all$id <- rep(files.to.read, sapply(mylist, nrow))

I used the files.to.read vector as the id for each dataframe

I also changed from using merge_recurse() as it was very slow for some reason.

 all <- merge_recurse(mylist)

Thanks everyone.

回答1:

Personally, I think it's easier to add the names after collapse:

df1 <- data.frame(x=c(1:5),y=c(11:15))
df2 <- data.frame(x=c(1:5),y=c(11:15))
mylist <- list(df1 = df1, df2 = df2)

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

回答2:

Your first attempt was very close. By using indices instead of values it will work. Your second attempt failed because you didn't name the elements in your list.

Both solutions below use the fact that lapply can pass extra parameters (mylist) to the function.

df1 <- data.frame(x=c(1:5),y=c(11:15))
df2 <- data.frame(x=c(1:5),y=c(11:15))
mylist <- list(df1=df1,df2=df2) # Name each data.frame!
# names(mylist) <- c("df1", "df2") # Alternative way of naming...

# Use indices - and pass in mylist
mylist1 <- lapply(seq_along(mylist), 
        function(i, x){
            x[[i]]$id <- i
            return (x[[i]])
        }, mylist
)

# Now the names work - but I pass in mylist instead of using portfolio.results.
mylist2 <- lapply(names(mylist), 
    function(n, x){
        x[[n]]$id <- n
        return (x[[n]])
    }, mylist
)

回答3:

names() could work it it had names, but you didn't give it any. It's an unnamed list. You will need ti use numeric indices:

> for(i in 1:length(mylist) ){ mylist[[i]] <- cbind(mylist[[i]], id=rep(i, nrow(mylist[[i]]) ) ) }
> mylist
[[1]]
  x  y id
1 1 11  1
2 2 12  1
3 3 13  1
4 4 14  1
5 5 15  1

[[2]]
  x  y id
1 1 11  2
2 2 12  2
3 3 13  2
4 4 14  2
5 5 15  2

回答4:

dlply function form plyr package could be an answer:

library('plyr')
df1 <- data.frame(x=c(1:5),y=c(11:15))
df2 <- data.frame(x=c(1:5),y=c(11:15))
mylist <- list(df1 = df1, df2 = df2)

all <- ldply(mylist)