How do I make a list of data frames?
If you have a large number of sequentially named data frames you can create a list of the desired subset of data frames like this:
d1 <- data.frame(y1=c(1,2,3), y2=c(4,5,6))
d2 <- data.frame(y1=c(3,2,1), y2=c(6,5,4))
d3 <- data.frame(y1=c(6,5,4), y2=c(3,2,1))
d4 <- data.frame(y1=c(9,9,9), y2=c(8,8,8))
my.list <- list(d1, d2, d3, d4)
my.list
my.list2 <- lapply(paste('d', seq(2,4,1), sep=''), get)
my.list2
where my.list2 returns a list containing the 2nd, 3rd and 4th data frames.
[[1]]
y1 y2
1 3 6
2 2 5
3 1 4
[[2]]
y1 y2
1 6 3
2 5 2
3 4 1
[[3]]
y1 y2
1 9 8
2 9 8
3 9 8
Note, however, that the data frames in the above list are no longer named. If you want to create a list containing a subset of data frames and want to preserve their names you can try this:
list.function <- function() {
d1 <- data.frame(y1=c(1,2,3), y2=c(4,5,6))
d2 <- data.frame(y1=c(3,2,1), y2=c(6,5,4))
d3 <- data.frame(y1=c(6,5,4), y2=c(3,2,1))
d4 <- data.frame(y1=c(9,9,9), y2=c(8,8,8))
sapply(paste('d', seq(2,4,1), sep=''), get, environment(), simplify = FALSE)
}
my.list3 <- list.function()
my.list3
which returns:
> my.list3
$d2
y1 y2
1 3 6
2 2 5
3 1 4
$d3
y1 y2
1 6 3
2 5 2
3 4 1
$d4
y1 y2
1 9 8
2 9 8
3 9 8
> str(my.list3)
List of 3
$ d2:'data.frame': 3 obs. of 2 variables:
..$ y1: num [1:3] 3 2 1
..$ y2: num [1:3] 6 5 4
$ d3:'data.frame': 3 obs. of 2 variables:
..$ y1: num [1:3] 6 5 4
..$ y2: num [1:3] 3 2 1
$ d4:'data.frame': 3 obs. of 2 variables:
..$ y1: num [1:3] 9 9 9
..$ y2: num [1:3] 8 8 8
> my.list3[[1]]
y1 y2
1 3 6
2 2 5
3 1 4
> my.list3$d4
y1 y2
1 9 8
2 9 8
3 9 8
The other answers show you how to make a list of data.frames when you already have a bunch of data.frames, e.g., d1, d2, .... Having sequentially named data frames is a problem, and putting them in a list is a good fix, but best practice is to avoid having a bunch of data.frames not in a list in the first place.
The other answers give plenty of detail of how to assign data frames to list elements, access them, etc. We'll cover that a little here too, but the Main Point is to say don't wait until you have a bunch of a data.frames to add them to a list. Start with the list.
The rest of the this answer will cover some common cases where you might be tempted to create sequential variables, and show you how to go straight to lists. If you're new to lists in R, you might want to also read What's the difference between [[ and [ in accessing elements of a list?.
Lists from the start
Don't ever create d1 d2 d3, ..., dn in the first place. Create a list d with n elements.
Reading multiple files into a list of data frames
This is done pretty easily when reading in files. Maybe you've got files data1.csv, data2.csv, ... in a directory. Your goal is a list of data.frames called mydata. The first thing you need is a vector with all the file names. You can construct this with paste (e.g., my_files = paste0("data", 1:5, ".csv")), but it's probably easier to use list.files to grab all the appropriate files: my_files <- list.files(pattern = "\\.csv$"). You can use regular expressions to match the files, read more about regular expressions in other questions if you need help there. This way you can grab all CSV files even if they don't follow a nice naming scheme. Or you can use a fancier regex pattern if you need to pick certain CSV files out from a bunch of them.
At this point, most R beginners will use a for loop, and there's nothing wrong with that, it works just fine.
my_data <- list()
for (i in seq_along(my_files)) {
my_data[[i]] <- read.csv(file = my_files[i])
}
A more R-like way to do it is with lapply, which is a shortcut for the above
my_data <- lapply(my_files, read.csv)
Of course, substitute other data import function for read.csv as appropriate. readr::read_csv or data.table::fread will be faster, or you may also need a different function for a different file type.
Either way, it's handy to name the list elements to match the files
names(my_data) <- gsub("\\.csv$", "", my_files)
# or, if you prefer the consistent syntax of stringr
names(my_data) <- stringr::str_replace(my_files, pattern = ".csv", replacement = "")
Splitting a data frame into a list of data frames
This is super-easy, the base function split() does it for you. You can split by a column (or columns) of the data, or by anything else you want
mt_list = split(mtcars, f = mtcars$cyl)
# This gives a list of three data frames, one for each value of cyl
This is also a nice way to break a data frame into pieces for cross-validation. Maybe you want to split mtcars into training, test, and validation pieces.
groups = sample(c("train", "test", "validate"),
size = nrow(mtcars), replace = TRUE)
mt_split = split(mtcars, f = groups)
# and mt_split has appropriate names already!
Simulating a list of data frames
Maybe you're simulating data, something like this:
my_sim_data = data.frame(x = rnorm(50), y = rnorm(50))
But who does only one simulation? You want to do this 100 times, 1000 times, more! But you don't want 10,000 data frames in your workspace. Use replicate and put them in a list:
sim_list = replicate(n = 10,
expr = {data.frame(x = rnorm(50), y = rnorm(50))},
simplify = F)
In this case especially, you should also consider whether you really need separate data frames, or would a single data frame with a "group" column work just as well? Using data.table or dplyr it's quite easy to do things "by group" to a data frame.
I didn't put my data in a list :( I will next time, but what can I do now?
If they're an odd assortment (which is unusual), you can simply assign them:
mylist <- list()
mylist[[1]] <- mtcars
mylist[[2]] <- data.frame(a = rnorm(50), b = runif(50))
...
If you have data frames named in a pattern, e.g., df1, df2, df3, and you want them in a list, you can get them if you can write a regular expression to match the names. Something like
df_list = mget(ls(pattern = "df[0-9]"))
# this would match any object with "df" followed by a digit in its name
# you can test what objects will be got by just running the
ls(pattern = "df[0-9]")
# part and adjusting the pattern until it gets the right objects.
Generally, mget is used to get multiple objects and return them in a named list. Its counterpart get is used to get a single object and return it (not in a list).
Combining a list of data frames into a single data frame
A common task is combining a list of data frames into one big data frame. If you want to stack them on top of each other, you would use rbind for a pair of them, but for a list of data frames here are three good choices:
# base option - slower but not extra dependencies
big_data = do.call(what = rbind, args = df_list)
# data table and dplyr have nice functions for this that
# - are much faster
# - add id columns to identify the source
# - fill in missing values if some data frames have more columns than others
# see their help pages for details
big_data = data.table::rbindlist(df_list)
big_data = dplyr::bind_rows(df_list)
(Similarly using cbind or dplyr::bind_cols for columns.)
To merge (join) a list of data frames, you can see these answers. Often, the idea is to use Reduce with merge (or some other joining function) to get them together.
But I really need sequentially named variables
They can be a pain to work with, and almost always you don't actually need them, but if you do, do everything you can in a list for ease, and then you can use list2env() to put all the list items into an environment, such as your .GlobalEnv.
Why put the data in a list?
Put similar data in lists because you want to do similar things to each data frame, and functions like lapply, sapply do.call, the purrr package, and the old plyr l*ply functions make it easy to do that. Examples of people easily doing things with lists are all over SO.
Even if you use a lowly for loop, it's much easier to loop over the elements of a list than it is to construct variable names with paste and access the objects with get. Easier to debug, too.
Think of scalability. If you really only need three variables, it's fine to use d1, d2, d3. But then if it turns out you really need 6, that's a lot more typing. And next time, when you need 10 or 20, you find yourself copying and pasting lines of code, maybe using find/replace to change d14 to d15, and you're thinking this isn't how programming should be. If you use a list, the difference between 3 cases, 30 cases, and 300 cases is at most one line of code---no change at all if your number of cases is automatically detected by, e.g., how many .csv files are in your directory.
You can name the elements of a list, in case you want to use something other than numeric indices to access your data frames (and you can use both, this isn't an XOR choice).
Overall, using lists will lead you to write cleaner, easier-to-read code, which will result in fewer bugs and less confusion.
This isn't related to your question, but you want to use = and not <- within the function call. If you use <-, you'll end up creating variables y1 and y2 in whatever environment you're working in:
d1 <- data.frame(y1 <- c(1, 2, 3), y2 <- c(4, 5, 6))
y1
# [1] 1 2 3
y2
# [1] 4 5 6
This won't have the seemingly desired effect of creating column names in the data frame:
d1
# y1....c.1..2..3. y2....c.4..5..6.
# 1 1 4
# 2 2 5
# 3 3 6
The = operator, on the other hand, will associate your vectors with arguments to data.frame.
As for your question, making a list of data frames is easy:
d1 <- data.frame(y1 = c(1, 2, 3), y2 = c(4, 5, 6))
d2 <- data.frame(y1 = c(3, 2, 1), y2 = c(6, 5, 4))
my.list <- list(d1, d2)
You access the data frames just like you would access any other list element:
my.list[[1]]
# y1 y2
# 1 1 4
# 2 2 5
# 3 3 6
You can also access specific columns and values in each list element with [ and [[. Here are a couple of examples. First, we can access only the first column of each data frame in the list with lapply(ldf, "[", 1), where 1 signifies the column number.
ldf <- list(d1 = d1, d2 = d2) ## create a named list of your data frames
lapply(ldf, "[", 1)
# $d1
# y1
# 1 1
# 2 2
# 3 3
#
# $d2
# y1
# 1 3
# 2 2
# 3 1
Similarly, we can access the first value in the second column with
lapply(ldf, "[", 1, 2)
# $d1
# [1] 4
#
# $d2
# [1] 6
Then we can also access the column values directly, as a vector, with [[
lapply(ldf, "[[", 1)
# $d1
# [1] 1 2 3
#
# $d2
# [1] 3 2 1