Calling variable in the loop by using assign and paste0

Keep dataframes in a list using ?lapply, then it gets easier to carry out same transformations on multiple dataframes in a list, something like:

# set vars
watersheds <- c('ANE', 'SAL', 'CER')
cols_dont_want <- c('B1', 'B2', 'B3')

# result, all dataframes in one list
myList <- lapply(watersheds, function(i){
  # read the file
  x <- read.csv(paste0(i, "_mean.csv"))
  # exclude columns and return
  x[, !colnames(x) %in% cols_dont_want]
} )

replace

paste0(watersheds[2], '_mean')

with

eval(parse(text = paste0(watersheds[2], '_mean')))

and it should work. Your guess is correct, paste0 just gives you a string but you need to call the variable which is done using eval()

Or you can do it in a for loop (some find the syntax more understandable). It's equivalent to zx8754's solution, except it assigns names to each dataframe as per the OP. It's trivial to modify zx8754's solution do do the same.

watersheds <- c('ANE', 'SAL', 'CER')
cols_dont_want <- c('B1', 'B2', 'B3')
ws.list <- list()

for (i in 1:length(watersheds)) {
    ws.list[[i]] <- read.csv(paste0(watersheds[i], '_mean.csv'))
    names(ws.list)[i] <- paste0(watersheds[i], '_mean')
    ws.list[[i]] <- ws.list[[i]][!names(ws.list[[i]]) %in% cols_dont_want]
}
names(ws.list)
# "ANE_mean" "SAL_mean" "CER_mean"

# If you absolutely want to call the data.frames by their 
# individual names, you can do so after you attach() the list.
attach(ws.list)
ANE_mean