CODEWITHSUNDEEP
javaandroidarraysandroid-activityarraylist
user3375371
user3375371

Reputation:

How can I check if an ArrayList<String> contains any elements from an array of Strings?

Within Android, I'd like to perform an if statement to check whether an ArrayList contains any element from an array of Strings? e.g.

Check whether any of the elements from singingGroup are also containined in Winners[]

String Winners[] = {"Jennifer", "Steven", "Peter", "Parker"};

ArrayList<String> singingGroup  = new ArrayList<String>();

singingGroup.add("Patrick");

singingGroup.add("Jane");

singingGroup.add("Joe");

singingGroup.add("Susan");

singingGroup.add("Amy");

How can I do this? as I know how to check if one item is contained as in another array as below. But not if any from one, exist in another.

if (Arrays.asList(Winners).contains(singingGroup)) {

Upvotes: 3

Views: 6271

Answers (5)

Gorilla
Gorilla

Reputation: 102

//for loop would be perfect to check if element i = element i

int i =0;
int loopCount = 0;
while(loopCount < Winners.lenght)
{
  for(int i =0; i < singingGroup.length; i++)
  {
    if(Winners[loopCount] == singingGroup[i])
    {
    System.out.println(Winners[loopCount] + "is apart of the winners");
    }//end of comparing if

    if(i == singing.Group.length)
    { 
    loopCount ++;
    } //end of i == singingGroup

  }//end of for loop

}//end of while  loop

This is not the most optimal code but if you need it in a hurry this will work

Upvotes: 0

D. Pereira
D. Pereira

Reputation: 116

You can use the CollectionUtils class provided by Apache Commons.

Using the intersection method (useful if you want to do something with the common elements):

Collection<String> intersection = CollectionUtils.intersection(singingGroup, Arrays.asList(Winners));

if (intersection.size() > 0){

    // At least one element contained in the intersection
}

Or, using the containsAny method:

if (CollectionUtils.containsAny(singingGroup, Arrays.asList(Winners))){

    // True if at least one common element exists in both lists  
}

Upvotes: 0

Bacteria
Bacteria

Reputation: 8606

Collections.disjoint is one way to archive this but You can also use retainAll() method.

Retains only the elements in this list that are contained in the specified collection (optional operation). In other words, removes from this list all of its elements that are not contained in the specified collection.

Case I :elements from singingGroup are not containined in Winners[]

String Winners[] = {"Jennifer", "Steven", "Peter", "Parker"};
ArrayList<String> singingGroup  = new ArrayList<String>();

singingGroup.add("Patrick");
singingGroup.add("Jane");
singingGroup.add("Joe");
singingGroup.add("Susan");
singingGroup.add("Amy");

List<String> WinnerList = new ArrayList<>(Arrays.asList(Winners));
WinnerList.retainAll(singingGroup);
System.out.println("retainList = " + WinnerList);

Output 

list1 = []

Case II:elements from singingGroup are also containined in Winners[]

String Winners[] = {"Jennifer", "Steven", "Peter", "Parker"};
ArrayList<String> singingGroup  = new ArrayList<String>();

singingGroup.add("Steven");
singingGroup.add("Jane");
singingGroup.add("Joe");
singingGroup.add("Susan");
singingGroup.add("Jennifer");

List<String> WinnerList = new ArrayList<>(Arrays.asList(Winners));
WinnerList.retainAll(singingGroup);
System.out.println("retainList = " + WinnerList);

Output 

retainList = [Jennifer, Steven]

Upvotes: 2

Sonu Gupta
Sonu Gupta

Reputation: 367

you can also check like this:

String Winners[] = {"Jennifer", "Patrick", "Peter", "Parker"};

        ArrayList<String> singingGroup  = new ArrayList<String>();

        singingGroup.add("Patrick");

        singingGroup.add("Jane");

        singingGroup.add("Joe");

        singingGroup.add("Susan");

        singingGroup.add("Amy");

        for(int i=0; i< Winners.length;i++)
        {
            if(singingGroup.contains(Winners[i]))
            {
                System.out.println("duplicate");
            }
        }

Upvotes: 0

fabian
fabian

Reputation: 82461

You can use

Collections.disjoint(singingGroup, Arrays.asList(Winners));

to test, is the 2 arguments have no common element(s) in common. (see also javadoc)

The negation of the result seems to be what you're looking for.

Upvotes: 6

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