Reputation: 15021

How to retrieve a list of directories QUICKLY in Java?

Suppose a very simple program that lists out all the subdirectories of a given directory. Sound simple enough? Except the only way to list all subdirectories in Java is to use FilenameFilter combined with File.list().

This works for the trivial case, but when the folder has say 150,000 files and 2 sub folders, it's silly waiting there for 45 seconds iterating through all the files and testing for file.isDirectory(). Is there a better way to list sub directories??

PS. Sorry, please save the lectures on having too many files in the same directory. Our live environment has this as part of the requirement.

Upvotes: 24

Answers (14)

user1324109

Reputation: 461

As of 2020, the DirectoryStream does seem to be faster than using File.listFiles() and checking each file with isDirectory().

I learned the answer from here:

https://www.baeldung.com/java-list-directory-files

I'm using Java 1.8 on Windows 10.

Upvotes: 0

Xtra Coder

Reputation: 3497

I came across similar question when debugging performance in a Java application enumerating plenty of files. It is using old approach

for (File f : new File("C:\\").listFiles()) {
    if (f.isDirectory()) {
        continue;
    }        
}

And it appears that each f.isDirectory() is the call into native FileSsystem which, at least on NTFS, is very slow. Java7 NIO has additional API, but not all methods are good there. I'll just provide JMH benchmark result here

Benchmark                  Mode  Cnt  Score    Error  Units
MyBenchmark.dir_listFiles  avgt    5  0.437 ?  0.064   s/op
MyBenchmark.path_find      avgt    5  0.046 ?  0.001   s/op
MyBenchmark.path_walkTree  avgt    5  1.702 ?  0.047   s/op

Number come from execution of this code:

java -jar target/benchmarks.jar -bm avgt -f 1 -wi 5 -i 5 -t 1

static final String testDir = "C:/Sdk/Ide/NetBeans/src/dev/src/";
static final int nCycles = 50;

public static class Counter {
    int countOfFiles;
    int countOfFolders;
}

@Benchmark
public List<File> dir_listFiles() {
    List<File> files = new ArrayList<>(1000);

    for( int i = 0; i < nCycles; i++ ) {
        File dir = new File(testDir);

        files.clear();
        for (File f : dir.listFiles()) {
            if (f.isDirectory()) {
                continue;
            }
            files.add(f);
        }
    }
    return files;
}

@Benchmark
public List<Path> path_walkTree() throws Exception {
    final List<Path> files = new ArrayList<>(1000);

    for( int i = 0; i < nCycles; i++ ) {
        Path dir = Paths.get(testDir);

        files.clear();
        Files.walkFileTree(dir, new SimpleFileVisitor<Path> () {
            @Override
            public FileVisitResult visitFile(Path path, BasicFileAttributes arg1) throws IOException {
                files.add(path);
                return FileVisitResult.CONTINUE;
            }

            @Override
            public FileVisitResult preVisitDirectory(Path path, BasicFileAttributes arg1) 
                    throws IOException {
                return path == dir ? FileVisitResult.CONTINUE : FileVisitResult.SKIP_SUBTREE;
            }
        });
    }

    return files;
}

@Benchmark
public List<Path> path_find() throws Exception {
    final List<Path> files = new ArrayList<>(1000);

    for( int i = 0; i < nCycles; i++ ) {
        Path dir = Paths.get(testDir);

        files.clear();
        files.addAll(Files.find(dir, 1, (path, attrs) 
                -> true /*!attrs.isDirectory()*/).collect(Collectors.toList()));
    }

    return files;
}

Upvotes: 6

dfa

Reputation: 116314

there is also a recursive parallel scanning at http://blogs.oracle.com/adventures/entry/fast_directory_scanning. Essentially siblings are processed in parallel. There also encouraging performance tests.

Upvotes: 1

Roman Zenka

Reputation: 3604

The key problem could be File.isDirectory() function called in a loop.

File.isDirectory() can be extremely slow. I saw NFS take 10 seconds to process 200 file directory.

If you can by all means prevent File.isDirectory() calls (e.g. test for extension, no extension == directory), you could improve the performance drastically.

Otherwise I would suggest doing JNA/JNI/writing a native script that does this for you.

The jCifs library lets you manipulate windows network shares more efficiently. I am not aware of a library that would do this for other network file systems.

Upvotes: 6

Emil H

Reputation: 40230

As has already been mentioned, this is basicly a hardware problem. Disk access is always slow, and most file systems aren't really designed to handle directories with that many files.

If you for some reason have to store all the files in the same directory, I think you'll have to maintain your own cache. This could be done using a local database such as sqlite, HeidiSQL or HSQL. If you want extreme performance, use a java TreeSet and cache it in memory. This means at the very least that you'll have to read the directory less often, and it could possibly be done in the background. You could reduce the need to refresh the list even further by using your systems native file update notification API (inotify on linux) to subscribe to changes to the directory.

This doesn't seem to be possible for you, but I once solved a similiar problem by "hashing" the files into subdirectories. In my case, the challenge was to store a couple of millions images with numeric ids. I constructed the directory structure as follows:

images/[id - (id % 1000000)]/[id - (id % 1000)]/[id].jpg

This has worked well for us, and it's the solution that I would recommend. You could do something similiar to alpha-numeric filenames by simply taking the first two letters of the filename, and then the next two letters. I've done this as well once, and it did the job as well.

Upvotes: 11

Brian Agnew

Reputation: 272237

Here's an off-the wall solution, and devoid of any testing at all. It's also dependent on having a filesystem that supports symbolic links. This isn't a Java solution. I suspect your problem is filesystem/OS-related, and not Java related.

Is it possible to create a parallel directory structure, with subdirectories based on initial letters of the filenames, and then symbolically link to the real files ? An illustration

/symlinks/a/b/cde

would link to

/realfiles/abcde

(where /realfiles is where your 150,000 files reside)

You'd have to create and maintain this directory structure, and I don't have enough info to determine if that's practical. But the above would create a fast(er) index into your non-hierarchical (and slow) directory.

Upvotes: 1

dfa

Reputation: 116314

if your OS is 'stable' give a try to JNA:

opendir/readdir on UNIX
FindFirstFile and related API on Windows
Java7 with NIO2

these are all "streaming API". They doesn't force you to allocate a 150k list/array before start searching. IMHO this is a great advantage in your scenario.

Upvotes: 2

akarnokd

Reputation: 69997

In that case you might try some JNA solution - a platform dependant directory traverser (FindFirst, FindNext on Windows) with the possibility of some iteration pattern. Also Java 7 will have much better file system support, worth checking out the specs (I don't remember any specifics).

Edit: An idea: one option is to hide the slowness of the directory listing from the user's eyes. In a client side app, you could use some animation while the listing is working to distract the user. Actually depends on what else your application does beside the listing.

Upvotes: 0

lavinio

Reputation: 24299

I don't know if the overhead of shelling out to cmd.exe would eat it up, but one possibility would be something like this:

...
Runtime r = Runtime.getRuntime();
Process p = r.exec("cmd.exe /k dir /s/b/ad C:\\folder");
BufferedReader br = new BufferedReader(new InputStreamReader(p.getInputStream()));
for (;;) {
    String d = br.readLine();
    if (d == null)
        break;
    System.out.println(d);
}
...

/s means search subdirectories
/ad means only return directories
/b means return the full pathname from the root

Upvotes: 5

DVK

Reputation: 129383

Do you know the finite list of possible subdirectory names? If so, use a loop over all possible names and check for directory's existence.

Otherwise, you can not get ONLY directory names in most underlying OSs (e.g. in Unix, the directory listing is merely reading contents of "directory" file, so there's no way to find "just directories" quickly without listing all the files).

However, in NIO.2 in Java7 (see http://java.sun.com/developer/technicalArticles/javase/nio/#3 ), there's a way to have a streaming directory list so you don't get a full array of file elements cluttering your memory/network.

Upvotes: 8

Yoni Roit

Reputation: 28686

Well, either JNI, or, if you say your deployment is constant, just run "dir" on Windows or "ls" on *nixes, with appropriate flags to list only directories (Runtime.exec())

Upvotes: 0

kdgregory

Reputation: 39606

There's actually a reason why you got the lectures: it's the correct answer to your problem. Here's the background, so that perhaps you can make some changes in your live environment.

First: directories are stored on the filesystem; think of them as files, because that's exactly what they are. When you iterate through the directory, you have to read those blocks from the disk. Each directory entry will require enough space to hold the filename, and permissions, and information on where that file is found on-disk.

Second: directories aren't stored with any internal ordering (at least, not in the filesystems where I've worked with directory files). If you have 150,000 entries and 2 sub-directories, those 2 sub-directory references could be anywhere within the 150,000. You have to iterate to find them, there's no way around that.

So, let's say that you can't avoid the big directory. Your only real option is to try to keep the blocks comprising the directory file in the in-memory cache, so that you're not hitting the disk every time you access them. You can achieve this by regularly iterating over the directory in a background thread -- but this is going to cause undue load on your disks, and interfere with other processes. Alternatively, you can scan once and keep track of the results.

The alternative is to create a tiered directory structure. If you look at commercial websites, you'll see URLs like /1/150/15023.html -- this is meant to keep the number of files per directory small. Think of it as a BTree index in a database.

Of course, you can hide that structure: you can create a filesystem abstraction layer that takes filenames and automatically generates the directory tree where those filenames can be found.

Upvotes: 7