CODEWITHSUNDEEP
javafile-listing
nafas
nafas

Reputation: 5423

How to get (some) filenames within a directory efficiently

I have a directory with 20M+ files in it. If I try

folder.list();

It'll take me about 5~10min to get the file (sometimes even more). I don't need all file names, just a handful everytime.

In Linux if I try:

So I can list files quickly if I use ProcessBuilder and run something like ls -f | head -n 100

Is there a native way for listing a fixed number of files within a directory without requiring to using ProcessBuilder?

Upvotes: 2

Views: 187

Answers (2)

Tunaki
Tunaki

Reputation: 137084

Yes, there is way using Java NIO.2 and the class DirectoryStream. This class implements a lazy iterable over the entries of a directory. You can obtain an instance of DirectoryStream<Path> using Files.newDirectoryStream(path) (and you can obtain an instance of Path with the static factories Paths.get).

If you are using Java 8, an even simpler solution would be to use Files.list():

Return a lazily populated Stream, the elements of which are the entries in the directory. The listing is not recursive.

You could then have

List<String> fileNames = Files.list(path)
                              .map(Path::getFileName)
                              .map(Path::toString)
                              .limit(100)
                              .collect(Collectors.toList());

to retrieve the 100 file names of the given path.

Upvotes: 5

Kai Iskratsch
Kai Iskratsch

Reputation: 191

you could try if the java7 nio file operations are faster for you. they were a big improvement in one case where i used them: https://docs.oracle.com/javase/tutorial/essential/io/walk.html

Upvotes: 1

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