Read multiple CSV files into separate data frames

Thank you all for replying.

For completeness here is my final answer for loading any number of (tab) delimited files, in this case with 6 columns of data each where column 1 is characters, 2 is factor, and remainder numeric:

##Read files named xyz1111.csv, xyz2222.csv, etc.
filenames <- list.files(path="../Data/original_data",
    pattern="xyz+.*csv")

##Create list of data frame names without the ".csv" part 
names <-substr(filenames,1,7)

###Load all files
for(i in names){
    filepath <- file.path("../Data/original_data/",paste(i,".csv",sep=""))
    assign(i, read.delim(filepath,
    colClasses=c("character","factor",rep("numeric",4)),
    sep = "\t"))
}

This answer is intended as a more useful complement to Hadley's answer.

While the OP specifically wanted each file read into their R workspace as a separate object, many other people naively landing on this question may think that that's what they want to do, when in fact they'd be better off reading the files into a single list of data frames.

So for the record, here's how you might do that.

#If the path is different than your working directory
# you'll need to set full.names = TRUE to get the full
# paths.
my_files <- list.files("path/to/files")

#Further arguments to read.csv can be passed in ...
all_csv <- lapply(my_files,read.csv,...)

#Set the name of each list element to its
# respective file name. Note full.names = FALSE to
# get only the file names, not the full path.
names(all_csv) <- gsub(".csv","",
                       list.files("path/to/files",full.names = FALSE),
                       fixed = TRUE)

Now any of the files can be referred to by my_files[["filename"]], which really isn't much worse that just having separate filename variables in your workspace, and often it is much more convenient.

Use assign with a character variable containing the desired name of your data frame.

for(i in 1:100)
{
   oname = paste("file", i, sep="")
   assign(oname, read.csv(paste(oname, ".txt", sep="")))
}

Quick draft, untested:

1. Use list.files() aka dir() to dynamically generate your list of files.

2. This returns a vector, just run along the vector in a for loop.

3. Read the i-th file, then use assign() to place the content into a new variable file_i

That should do the trick for you.