Opening dataframe in R from it's name

Question

I load all .csv files from a given folder and store all the names in the filenames variable.

# Loading files
filenames <- gsub("\\.csv$", "", list.files(pattern = "\\.csv$"))

# Reading files
for(i in filenames)
  {
    assign(i, read.csv(paste(i, ".csv", sep = ""), sep = ","))
  }

Now I'd like to access any dataframe with name that is in filenames. How can I do that?

Does this answer your question? How to call an object with the character variable of the same name You can use get(i) within the loop or just pass the specific string. — caldwellst
– caldwellst, Commented Nov 23, 2021 at 10:59

Francesco Grossetti · Accepted Answer · 2021-11-23 12:57:13Z

2

I would not recommend using the function get() as this can have bad consequences. I'd suggest using a more structured approach by mean of a list containing all you csv files. I don't have any data here so I will build upon your code.

# Loading files
filenames <- gsub("\\.csv$", "", list.files(pattern = "\\.csv$"))

# create a list with NULL
file_list = vector("list", length(filenames))

# name each element according to filenames
names(file_list) = filenames

# Loop over the list
# Please note that the iterator "i" now loops over the elements
# and does not take the real values in filenames

for (i in seq_along(file_list)) {
  current_file = read.csv(paste0(i, ".csv"))
  file_list[[ i ]] = current_file
}

Now you can access each element either with the standard indexing ([[) or via the names with the $ operator.

UPDATE

For an example about why you should avoid get(), please refer to this SO post.

answered Nov 23, 2021 at 12:57

Francesco Grossetti

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1 Comment

user10917479 Over a year ago

And if you still really want each file as it's own dataframe object in the environment, you can always use list2env(file_list) later. Either way, this should be the approach to read, not using assign().

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