如何从数组中获取唯一值

Question

I have an Array from which i want to remove Duplicate items.

for(int data1=startpos;data1<=lastrow;data1++) {
    String movie_soundtrk=cells.getCell(data1,Mmovie_sndtrk_cl).getValue().toString();
    al.add(movie_soundtrk);
}

String commaSeparated=al.toString();
String [] items = commaSeparated.split(",");
String[] trimmedArray = new String[items.length];
for (int i = 0; i < items.length; i++) {
    trimmedArray[i] = items[i].trim();
}

Set<String> set = new HashSet<String>();
Collections.addAll(set, trimmedArray);

System.out.println(set);

But this is not giving me unique Values from Array.

My Array:- {English, French, Japanese, Russian, Chinese Subtitles,English, French, Japanese, Russian, Chinese Subtitles}

Out Put :- [Japanese, Russian, French, Chinese Subtitles], Chinese Subtitles, [English, English]

Answer 1

You can do it in one line in java 7:

String[] unique = new HashSet<String>(Arrays.asList(array)).toArray(new String[0]);

and shorter and simpler in java 8:

String[] unique = Arrays.stream(array).distinct().toArray(String[]::new);

Answer 2

HashSet will do the job.

You can try this:

List<String> newList = new ArrayList<String>(new HashSet<String>(oldList));

Answer 3

Using the Stream API of Java 8 this is a solution with a generic Array type:

public static <T> T[] makeUnique(T... values)
{
    return Arrays.stream(values).distinct().toArray(new IntFunction<T[]>()
    {

        @Override
        public T[] apply(int length)
        {
            return (T[]) Array.newInstance(values.getClass().getComponentType(), length);
        }

    });
}

It works for any Object type array, but not for primitive arrays.

For primitive arrays it looks like this:

public static int[] makeUnique(int... values)
{
    return Arrays.stream(values).distinct().toArray();
}

And finally here is a little unit test:

@Test
public void testMakeUnique()
{
    assertArrayEquals(new String[] { "a", "b", "c" }, makeUnique("a", "b", "c", "b", "a"));
    assertArrayEquals(new Object[] { "a", "b", "c" }, makeUnique(new Object[] { "a", "b", "c", "b", "a" }));
    assertArrayEquals(new Integer[] { 1, 2, 3, 4, 5 }, makeUnique(new Integer[] { 1, 2, 2, 3, 3, 3, 1, 4, 5, 5, 5, 1 }));
    assertArrayEquals(new int[] { 1, 2, 3, 4, 5 }, makeUnique(new int[] { 1, 2, 2, 3, 3, 3, 1, 4, 5, 5, 5, 1 }));
}

Answer 4

You could get two sets, one with all the subtitles, and the other with the duplicates

String[] trimmedArray = new String[items.length];
Set<String> subtitles = new HashSet<String>();
Set<String> duplicatedSubtitles = new HashSet<String>();

foreach(String subtitle : trimmedArray){
    subtitle = subtitle.trim();
    if(subtitles.contains(subtitle)){
        duplicatedSubtitles.add(subtitle);
    }
    subtitles.add(subtitle);
}

Answer 5

Try instead of this

Set<String> set = new HashSet<String>();

to call this

set.addAll(trimmedArray);

Answer 6

Why did you first add items into array and then convert it to string? Just iterate over tha array and copy them to Set.Then print new created set which holds unique values.

Set<String> set = new HashSet<String>();
for (int i = 0; i < al.length; i++) {
    set.add(al[i]);
}

for (String str : set) {
    System.out.println(str);
}

Answer 7

This code will calculates distinct elements from an array, then finds their occurrence. And calculates percentage and save it to hashmap.

int _occurrence = 0;
        String[] _fruits = new String[] {"apple","apple","banana","mango","orange","orange","mango","mango","banana","banana","banana","banana","banana"};
        List<String> _initialList = Arrays.asList(_fruits);
        Set<String> treesetList = new TreeSet<String>(_initialList);
        String[] _distinct =  (String[]) treesetList.toArray(new String[0]);

        HashMap<String,String> _map = new HashMap<String,String>();
        int _totalElement = _fruits.length;
        for(int x=0;x<_distinct.length;x++){
            for(int i=0;i<_fruits.length;i++){
                if(_distinct[x].equals(_fruits[i])){
                    ++_occurrence;
                }
            }
            double _calPercentage = Math.round((((double)_occurrence/(double)_totalElement)*100));
            _map.put(_distinct[x], String.valueOf(_calPercentage+"%"));
            _occurrence = 0;
        }
        System.out.println(_map);

Answer 8

Let's see how to find distinct values from an array.

  public class Distinct  {
        public static void main(String args[]) {
             int num[]={1,4,3,2,6,7,4,2,1,2,8,6,7};
                for(int i=0; i<num.length; i++){
                    boolean isdistinct = true;
                    for(int j=0; j<i; j++){
                        if(num[i] == num[j]){
                            isdistinct =false;
                            break;
                        }
                   }
                    if(isdistinct){
                        System.out.print(num[i]+" ");
                    }
               }
           }
     }

Answer 9

If you don't wanna use Hashset or the new method in Java8 mentioned above you can write this code you just need first to sort the array so similar values will then be next to each other and then count the number of distinct pairs in adjacent cells.

    public static int solution(int[] A) {
    int count = 1;
    Arrays.sort(A);
    for (int i = 1; i < A.length - 1; i++) {
        if (A[i] != A[i + 1]) {
            count++;
        }
    }
    return count;
}

Answer 10

In python you can use Set.

s = Set()
data_list = [1,2,3,4]
s.update(data_list)

Answer 11

int arr[] = {1,1,2,2,3,3,4,5,5,6}; 
for(int i=0; i<arr.length; i++) {
int count = 0;
    for(int j=0; j<arr.length; j++) {
        if ((arr[i] == arr[j]) && (i!=j)) {
            count++ ;
        }
    }
        if(count==0) {
            System.out.println(arr[i]);
        }
}

Answer 12

public static int[] findAndReturnUniqueIntArray(int[] arr) {
    int[] distinctElements = {};
    int newArrLength = distinctElements.length;
    for (int i = 0; i < arr.length; i++) {
        boolean exists = false;
        if (distinctElements.length == 0) {
            distinctElements = new int[1];
            distinctElements[newArrLength] = arr[i];
            newArrLength++;
        }

        else {
            for (int j = 0; j < distinctElements.length; j++) {
                if (arr[i] == distinctElements[j]) {
                    exists = true;
                    break;
                }
            }
            if (exists == false) {
                distinctElements = Arrays.copyOf(distinctElements, distinctElements.length+1);
                distinctElements[distinctElements.length-1] = arr[i];
                newArrLength++;
                exists = false;
            }
        }
    }
    return distinctElements;
}