How do I sort file names by their extension?

By Wayan in Apache Commons, Commons IO 0 Comment

To sort file names by their extension we can use the ExtensionFileComparator class from the Apache Commons IO library. This class provide a couple instances of comparator such as:

Comparator Description
EXTENSION_COMPARATOR Case sensitive extension comparator
EXTENSION_REVERSE Reverse case sensitive extension comparator
EXTENSION_INSENSITIVE_COMPARATOR Case insensitive extension comparator
EXTENSION_INSENSITIVE_REVERSE Reverse case insensitive extension comparator
EXTENSION_SYSTEM_COMPARATOR System sensitive extension comparator
EXTENSION_SYSTEM_REVERSE Reverse system sensitive path comparator

The following snippet show you how to use the first two comparators listed above.

package org.kodejava.commons.io;

import org.apache.commons.io.FilenameUtils;

import static org.apache.commons.io.comparator.ExtensionFileComparator.*;

import java.io.File;
import java.util.Arrays;

public class FileSortByExtension {
    public static void main(String[] args) {
        File file = new File(".");

        // Excludes directory in the list
        File[] files = file.listFiles(File::isFile);

        if (files != null) {
            // Sort in ascending order.
            Arrays.sort(files, EXTENSION_COMPARATOR);
            FileSortByExtension.displayFileOrder(files);

            // Sort in descending order.
            Arrays.sort(files, EXTENSION_REVERSE);
            FileSortByExtension.displayFileOrder(files);
        }
    }

    private static void displayFileOrder(File[] files) {
        System.out.printf("%-20s | %s%n", "Name", "Ext");
        System.out.println("--------------------------------");
        for (File file : files) {
            System.out.printf("%-20s | %s%n", file.getName(),
                    FilenameUtils.getExtension(file.getName()));
        }
        System.out.println();
    }
}

The result of the code snippet:

Name                 | Ext
--------------------------------
README               | 
lipsum.doc           | doc
lipsum.docx          | docx
data.html            | html
contributors.txt     | txt
pom.xml              | xml

Name                 | Ext
--------------------------------
pom.xml              | xml
contributors.txt     | txt
data.html            | html
lipsum.docx          | docx
lipsum.doc           | doc
README               |

Maven Dependencies

<dependency>
    <groupId>commons-io</groupId>
    <artifactId>commons-io</artifactId>
    <version>2.11.0</version>
</dependency>

Maven Central

How do I sort files and directories based on their size?

By Wayan in Apache Commons, Commons IO 0 Comment

In this example you will learn how to sort files and directories based on their size. Using the Apache Commons IO we can utilize the SizeFileComparator class. This class provides some instances to sort file size such as:

Comparator Description
SIZE_COMPARATOR Size comparator instance – directories are treated as zero size
SIZE_REVERSE Reverse size comparator instance – directories are treated as zero size
SIZE_SUMDIR_COMPARATOR Size comparator instance which sums the size of a directory’s contents
SIZE_SUMDIR_REVERSE Reverse size comparator instance which sums the size of a directory’s contents

Now let’s jump to the code snippet below:

package org.kodejava.commons.io;

import org.apache.commons.io.FileUtils;

import java.io.File;
import java.util.Arrays;

import static org.apache.commons.io.comparator.SizeFileComparator.*;

public class FileSortBySize {
    public static void main(String[] args) {
        File dir = new File(".");
        File[] files = dir.listFiles();

        if (files != null) {
            // Sort files in ascending order based on file size.
            System.out.println("Ascending order.");
            Arrays.sort(files, SIZE_COMPARATOR);
            FileSortBySize.displayFileOrder(files, false);

            // Sort files in descending order based on file size
            System.out.println("Descending order.");
            Arrays.sort(files, SIZE_REVERSE);
            FileSortBySize.displayFileOrder(files, false);

            // Sort files in ascending order based on file / directory
            // size
            System.out.println("Ascending order with directories.");
            Arrays.sort(files, SIZE_SUMDIR_COMPARATOR);
            FileSortBySize.displayFileOrder(files, true);

            // Sort files in descending order based on file / directory
            // size
            System.out.println("Descending order with directories.");
            Arrays.sort(files, SIZE_SUMDIR_REVERSE);
            FileSortBySize.displayFileOrder(files, true);
        }
    }

    private static void displayFileOrder(File[] files, boolean displayDirectory) {
        for (File file : files) {
            if (!file.isDirectory()) {
                System.out.printf("%-25s - %s%n", file.getName(),
                        FileUtils.byteCountToDisplaySize(file.length()));
            } else if (displayDirectory) {
                long size = FileUtils.sizeOfDirectory(file);
                String friendlySize = FileUtils.byteCountToDisplaySize(size);
                System.out.printf("%-25s - %s%n", file.getName(),
                        friendlySize);
            }
        }
        System.out.println("------------------------------------");
    }
}

In the code snippet above we use a couple method from the FileUtils class such as the FileUtils.sizeOfDirectory() to calculate the size of a directory and FileUtils.byteCountToDisplaySize() to create human-readable file size.

The result of the code snippet:

Ascending order.
.editorconfig             - 389 bytes
kodejava.iml              - 868 bytes
pom.xml                   - 1 KB
------------------------------------
Descending order.
pom.xml                   - 1 KB
kodejava.iml              - 868 bytes
.editorconfig             - 389 bytes
------------------------------------
Ascending order with directories.
.editorconfig             - 389 bytes
src                       - 851 bytes
kodejava.iml              - 868 bytes
pom.xml                   - 1 KB
apache-commons-example    - 8 KB
hibernate-example         - 29 KB
.idea                     - 85 KB
------------------------------------
Descending order with directories.
.idea                     - 85 KB
hibernate-example         - 29 KB
apache-commons-example    - 8 KB
pom.xml                   - 1 KB
kodejava.iml              - 868 bytes
src                       - 851 bytes
.editorconfig             - 389 bytes

Maven Dependencies

<dependency>
    <groupId>commons-io</groupId>
    <artifactId>commons-io</artifactId>
    <version>2.11.0</version>
</dependency>

Maven Central

How do I sort strings data using CollationKey class?

By Wayan in Core API, Text Package 0 Comment

When the strings must be compared multiple times, for example when sorting a list of strings. It’s more efficient to use CollationKey class. Using CollationKey to compare strings is generally faster than using Collator.compare().

You can not create CollationKey directly. Rather, generate them by calling Collator.getCollationKey() method. You can only compare CollationKey generated from the same Collator object.

package org.kodejava.text;

import java.text.CollationKey;
import java.text.Collator;
import java.util.Arrays;

public class CollationKeyExample {
    public static void main(String[] args) {
        String[] countries = {
                "German",
                "United Kingdom",
                "United States",
                "French",
                "Japan",
                "Myanmar",
                "India"
        };

        System.out.println("original:");
        System.out.println(Arrays.toString(countries));

        // Gets Collator object of default locale
        Collator collator = Collator.getInstance();

        // Creates and initializes CollationKey array
        CollationKey[] keys = new CollationKey[countries.length];

        for (int i = 0; i < countries.length; i++) {
            // Generate CollationKey by calling
            // Collator.getCollationKey() method then assign into
            // keys which is an array of CollationKey.
            // The CollationKey for the given String based on the 
            // Collator's collation rules.
            keys[i] = collator.getCollationKey(countries[i]);
        }

        // Sort the keys array
        Arrays.sort(keys);

        // Print out the sorted array
        System.out.println("sorted result: ");
        StringBuilder sb = new StringBuilder();
        for (CollationKey key : keys) {
            sb.append(key.getSourceString()).append(",");
        }
        System.out.println(sb);
    }
}

Below is the result of the program:

original:
[German, United Kingdom, United States, French, Japan, Myanmar, India]
sorted result: 
French,German,India,Japan,Myanmar,United Kingdom,United States,

How do I sort string of numbers in ascending order?

By Wayan in Core API, Util Package 2 Comments

In the following example we are going to sort a string containing the following numbers "2, 5, 9, 1, 10, 7, 4, 8" in ascending order, so we will get the result of "1, 2, 4, 5, 7, 8, 9, 10".

package org.kodejava.util;

import java.util.Arrays;

public class SortStringNumber {
    public static void main(String[] args) {
        // We have some string numbers separated by comma. First we
        // need to split it, so we can get each individual number.
        String data = "2, 5, 9, 1, 10, 7, 4, 8";
        String[] numbers = data.split(",");

        // Convert the string numbers into Integer and placed it into
        // an array of Integer.
        Integer[] intValues = new Integer[numbers.length];
        for (int i = 0; i < numbers.length; i++) {
            intValues[i] = Integer.parseInt(numbers[i].trim());
        }

        // Sort the number in ascending order using the
        // Arrays.sort() method.
        Arrays.sort(intValues);

        // Convert back the sorted number into string using the
        // StringBuilder object. Prints the sorted string numbers.
        StringBuilder builder = new StringBuilder();
        for (int i = 0; i < intValues.length; i++) {
            Integer intValue = intValues[i];
            builder.append(intValue);
            if (i < intValues.length - 1) {
                builder.append(", ");
            }
        }
        System.out.println("Before = " + data);
        System.out.println("After  = " + builder);
    }
}

When we run the program we will get the following output:

Before = 2, 5, 9, 1, 10, 7, 4, 8
After  = 1, 2, 4, 5, 7, 8, 9, 10

How do I sort an java.util.Enumeration?

By Wayan in Core API, Util Package 1 Comment

In this code snippet you will see how to sort the content of an Enumeration object. We start by creating a random numbers and stored it in a Vector. We use these numbers and create a Enumeration object by calling Vector‘s elements() method. We convert it to java.util.List and then sort the content of the List using Collections.sort() method. Here is the complete code snippet.

package org.kodejava.util;

import java.util.*;

public class EnumerationSort {
    public static void main(String[] args) {
        // Creates random data for sorting source. Will use java.util.Vector
        // to store the random integer generated.
        Random random = new Random();
        Vector<Integer> data = new Vector<>();
        for (int i = 0; i < 10; i++) {
            data.add(Math.abs(random.nextInt()));
        }

        // Get the enumeration from the vector object and convert it into
        // a java.util.List. Finally, we sort the list using
        // Collections.sort() method.
        Enumeration<Integer> enumeration = data.elements();
        List<Integer> list = Collections.list(enumeration);
        Collections.sort(list);

        // Prints out all generated number after sorted.
        for (Integer number : list) {
            System.out.println("Number = " + number);
        }
    }
}

An example result of the code above is:

Number = 20742427
Number = 163885840
Number = 204704456
Number = 560032429
Number = 601762809
Number = 1300593322
Number = 1371678147
Number = 1786580321
Number = 1786731301
Number = 1856215303

How do I sort LinkedList elements?

By Wayan in Core API, Util Package 0 Comment

To sort the elements of LinkedList we can use the Collections.sort(List<T> list) static methods. The default order of the sorting is a descending order.

package org.kodejava.util;

import java.util.LinkedList;
import java.util.Collections;

public class LinkedListSort {
    public static void main(String[] args) {
        LinkedList<String> grades = new LinkedList<>();
        grades.add("E");
        grades.add("C");
        grades.add("A");
        grades.add("F");
        grades.add("B");
        grades.add("D");

        System.out.println("Before sorting:");
        System.out.println("===============");
        for (String grade : grades) {
            System.out.println("Grade = " + grade);
        }

        // Sort the elements of linked list based on its data
        // natural order.
        Collections.sort(grades);

        System.out.println("After sorting:");
        System.out.println("===============");
        for (String grade : grades) {
            System.out.println("Grade = " + grade);
        }
    }
}

The result of the program are:

Before sorting:
===============
Grade = E
Grade = C
Grade = A
Grade = F
Grade = B
Grade = D
After sorting:
===============
Grade = A
Grade = B
Grade = C
Grade = D
Grade = E
Grade = F

How do I search collection elements?

By Wayan in Core API, Util Package 0 Comment

This code example use the Collections.binarySearch() to search an specified object inside a specified collections. Prior to calling the binarySearch() method we need to sort the elements of the collection. If the object is not sorted according to their natural order the search result will be undefined.

package org.kodejava.util;

import java.util.Arrays;
import java.util.LinkedList;
import java.util.Collections;
import java.text.DateFormatSymbols;

public class CollectionSearch {
    public static void main(String[] args) {
        DateFormatSymbols dfs = new DateFormatSymbols();

        LinkedList<String> monthList =
                new LinkedList<>(Arrays.asList(dfs.getMonths()));

        // Sort the collection elements
        Collections.sort(monthList);
        System.out.println("Months = " + monthList);

        // Get the position of November inside the monthList. It returns a positive
        // value if the item found in the monthList.
        int index = Collections.binarySearch(monthList, "November");
        if (index > 0) {
            System.out.println("Found at index = " + index);
            System.out.println("Month = " + monthList.get(index));
        }
    }
}

The output of the code snippet above is below.

Months = [, April, August, December, February, January, July, June, March, May, November, October, September]
Found at index = 10
Month = November

How do I sort files based on their last modified date?

By Wayan in Apache Commons, Commons IO 0 Comment

This example demonstrates how to use Apache Commons IO LastModifiedFileComparator class to sort files based on their last modified date in ascending and descending order. There are two comparators defined in this class, the LASTMODIFIED_COMPARATOR and the LASTMODIFIED_REVERSE.

package org.kodejava.commons.io;

import static org.apache.commons.io.comparator.LastModifiedFileComparator.*;

import java.io.File;
import java.util.Arrays;

public class FileSortLastModified {
    public static void main(String[] args) {
        File dir = new File(System.getProperty("user.home"));
        File[] files = dir.listFiles();

        if (files != null) {
            // Sort files in ascending order based on file's last
            // modification date.
            System.out.println("Ascending order.");
            Arrays.sort(files, LASTMODIFIED_COMPARATOR);
            FileSortLastModified.displayFileOrder(files);

            System.out.println("------------------------------------");

            // Sort files in descending order based on file's last
            // modification date.
            System.out.println("Descending order.");
            Arrays.sort(files, LASTMODIFIED_REVERSE);
            FileSortLastModified.displayFileOrder(files);
        }
    }

    private static void displayFileOrder(File[] files) {
        for (File file : files) {
            System.out.printf("%2$td/%2$tm/%2$tY - %s%n", file.getName(),
                    file.lastModified());
        }
    }
}

Here are the example results produced by the code snippet:

Ascending order.
15/12/2020 - ntuser.dat.LOG1
15/12/2020 - ntuser.ini
15/12/2020 - .m2
18/12/2020 - Contacts
22/12/2020 - Videos
01/01/2021 - VirtualBox VMs
02/01/2021 - Desktop
02/01/2021 - Documents
------------------------------------------
Descending order.
02/01/2021 - Documents
02/01/2021 - Desktop
01/01/2021 - VirtualBox VMs
22/12/20202 - Videos
18/12/20202 - Contacts
15/12/20202 - .m2
15/12/20202 - ntuser.ini
15/12/20202 - ntuser.dat.LOG1

Maven Dependencies

<dependency>
    <groupId>commons-io</groupId>
    <artifactId>commons-io</artifactId>
    <version>2.11.0</version>
</dependency>

Maven Central

How do I sort strings using Collator class?

By Wayan in Core API, Text Package 0 Comment

In this example we demonstrate how to use the java.text.Collator class to sort strings in language-specific order. Using the java.text.Collator class makes the string not just sorted by the ASCII code of their characters, but it will follow the language natural order of the characters.

package org.kodejava.text;

import java.util.List;
import java.util.ArrayList;
import java.util.Locale;
import java.text.Collator;

public class StringShortWithCollator {
    public static void main(String[] args) {
        List<String> fruits = new ArrayList<>();
        fruits.add("Guava");
        fruits.add("Banana");
        fruits.add("Orange");
        fruits.add("Mango");
        fruits.add("Apple");

        // Define a collator for US English.
        Collator collator = Collator.getInstance(Locale.US);

        // Sort the list base on the collator
        fruits.sort(collator);

        for (String fruit : fruits) {
            System.out.println("Fruit = " + fruit);
        }
    }
}

The result of the code snippet above are:

Fruit = Apple
Fruit = Banana
Fruit = Guava
Fruit = Mango
Fruit = Orange

How do I sort items in a Set?

By Wayan in Core API, Util Package 0 Comment

The trick to sort a java.util.Set is to use the implementation of a java.util.SortedSet such as the java.util.TreeSet class. The example below shows you the result of using the java.util.TreeSet class, in which the items in it will be sorted based on the element’s natural order.

package org.kodejava.util;

import java.util.Set;
import java.util.TreeSet;

public class TreeSetDemo {
    public static void main(String[] args) {
        // The TreeSet class is an implementation of a SortedSet, this means
        // that when you are using the TreeSet to store you data collections
        // you'll get the items ordered base on its elements natural order.
        Set<String> set = new TreeSet<>();

        // In the example below we add some letters to the TreeSet, this mean
        // that the alphabets will be ordered based on the alphabet order
        // which is from A to Z.
        set.add("Z");
        set.add("A");
        set.add("F");
        set.add("B");
        set.add("H");
        set.add("X");
        set.add("N");

        for (String item : set) {
            System.out.print(item + " ");
        }
    }
}

This demo prints:

A B F H N X Z