how to sort integer from listFiles?

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I want to sort the integer from listFiles which not included the string. The smallest number will display"The smallest number=" and the biggest number will display"The biggest number="

public static void main(String[] args) {
        // TODO Auto-generated method stub
        File folder = new File("input/");
        File [] files = folder.listFiles();         

        for(int i = 0; i < files.length-1;i++) {

            String fileFullName = files[i].getName();       
            String fileSimple= fileFullName.substring(2,fileFullName.length()-4);           
            long fileNumber =Long.parseLong(fileSimple);

            String fileFullName2 = files[i+1].getName();        
            String fileSimple2= fileFullName2.substring(2,fileFullName2.length()-4);            
            long fileNumber2 =Long.parseLong(fileSimple2);                  

The file name included HR20190405.txt, AR20290405.txt,RG20290805.txt,RK21290405.txt

There is suitable method summaryStatistics from IntStream. Using stream api it will look like:

IntSummaryStatistics stat =
        .mapToInt(file -> Integer.valueOf(file.getName().substring(2, file.getName().length() - 4)))


how to File.listFiles in alphabetical order?, The listFiles method, with or without a filter does not guarantee any order. It does, however, return an array, which you can sort with Arrays. sort() . File[] files = XMLDirectory. I am trying to take input from File1(.txt) and File2(.txt), combine and sort the Integers contained in both files and create a third File3(.txt), containing the sorted integers from files 1 and 2.

You were quite close. We will just have to store the maximum & minimum values in variables. Also, the code to fetch the number from file name can be a separate method:

public static void main(String[] args) {

    File folder = new File("input/");
    File [] files = folder.listFiles();      

    //Set first number as smallest and largest   
    long smallest = getNumberFromName(files[0]);
    long largest = smallest;

    //Get number for each file and assign smallest & largest values
    for(int i = 1; i < files.length-1;i++) {
        long nextNumber = getNumberFromName(files[i]);
        if (smallest > nextNumber)
            smallest = nextNumber
        if (largest < nextNumber)
            largest = nextNumber
    System.out.println("The smallest number="+smallest)
    System.out.println("The biggest number="+largest)

private static long getNumberFromName(File nextFile) {
    String fileFullName = nextFile.getName();       
    String fileSimple= fileFullName.substring(2,fileFullName.length()-4);           
    return Long.parseLong(fileSimple);

How to list files sorted by filename ascending or descending in java, Calling the Collections.sort with only with one argument which is list, the list is sorted by default in ascending order. However if a comparator is� Sorting integer data from file and calculate execution time Prerequisite: Selection Sort In this article, we are going to apply selection sort algorithm, in which the source of input is A FILE CONTAINING 10000 INTEGERS and output will be the total time taken to sort.

Actually based on a previous answer, you can go even further and refine the code to use both the Stream API as well as java.nio. This will save you plenty of time as you won't be needing to perform manual iterations over the array of files you have read.

Also a good trick is to avoid the various magic number during substringing actions. A regex would have been way more suitable to avoid unwanted runtime errors.

Based on the above, a sample code would look like:

private static Optional<IntSummaryStatistics> getMinMax(final String filePath) {
    try (Stream<Path> pathStream = Files.list(Paths.get(filePath))){
        return Optional.of(pathStream
        .map(s -> s.replaceAll("\\D+",""))
    } catch (IOException e) {
    return Optional.empty();

And calling this would as simple as doing this:

public static void main(String[] args) {
        stats -> System.out.println("Min is " + stats.getMin() + " Max is " + stats.getMax()),
        () -> System.out.println("Nothing has been found")

List files sorted numerically, -v natural sort of (version) numbers within text The GNU sort (as available on Linux,) has a "version sort" mode that interprets numberes inside non-numbers� The only way to get the list of files is to use File.listFiles () and the documentation states that this makes no guarantees about the order of the files returned. Therefore you need to write a Comparator that uses File.lastModified () and pass this, along with the array of files, to Arrays.sort ().

How do I sort filenames containing text and numbers in numerical , In reading the MATLAB documentation, I understand the dir function sorts strings in ASCII dictionary order. And since the files I'm using have no leading zeros,� How to display integer array list values on screen and sort using Collections.sort() and reverse() method. In this tutorial we are using Collections.sort() and Collections.reverse() methods to automatically convert normal array list into ascending and descending order.

Sort a directory listing, listFiles(); Arrays.sort( files, new Comparator() { public int compare(final Object o1 , final Object o2) { return new Long(((File)o1).lastModified()).compareTo (new� Well organized and easy to understand Web building tutorials with lots of examples of how to use HTML, CSS, JavaScript, SQL, PHP, Python, Bootstrap, Java and XML.

Sorting files 'numerically' instead of alphabetically in java � GitHub, listFiles(fileFilter); // Sort files by name Arrays.sort(fileList, new Comparator<File>( ) /*System.out.println(fName1); System.out.println(fName2);*/ int i1 = Integer. Sorting the List in Descending Order using list.sort() # Sort the List in Place (Descending Order) listOfNum.sort(reverse=True) It will sort the list itself. listOfNum is now a List Sorted in Descending Order [67, 50, 45, 45, 23, 21, 11, 5, 2, 1] Complete example is as follows,