I have thousands of files, but I only want to load some of them, those I want to load all are all named "xxxx5429.spe" e.i the all have the same ending, is there a easy way to do it?
答案 0 :(得分:3)
Guess those x
's are meant to be wildcards. Also guessing they are meant to be exactly 4 characters wide. If my guesses are both correct then probably can be handled witha regex pattern like:
patt= ".{4}5429[.]spe"
The list.files
function can handle a regex pattern to its second argument. The first argument needs to be a path. Or you can do this just in the working directory if you name the argument:
my_files <- list.files(patt= ".{4}5429[.]spe")
It's not exactly clear what you mean by "load files". The term load
in R generally applies to a package or a .Rdata file. If these are text files then one of the read*
functions would be used: the readLines function if the text is unstructured, or one of the variants of the read.table function if the files are rectangular and delimited files.
This would create a list object containing character vectors with the contents of such files (from the working directory) using readLines
:
my_files <- lapply( my_files, readLines)