Java Compare Two Lists

Question

I have two lists ( not java lists, you can say two columns)

For example

**List 1**            **Lists 2**
  milan                 hafil
  dingo                 iga
  iga                   dingo
  elpha                 binga
  hafil                 mike
  meat                  dingo
  milan
  elpha
  meat
  iga                   
  neeta.peeta    

I'd like a method that returns how many elements are same. For this example it should be 3 and it should return me similar values of both list and different values too.

Should I use hashmap if yes then what method to get my result?

Please help

PS: It is not a school assignment :) So if you just guide me it will be enough

Answer 1

EDIT

Here are two versions. One using ArrayList and other using HashSet

Compare them and create your own version from this, until you get what you need.

This should be enough to cover the:

PS: It is not a school assignment :) So if you just guide me it will be enough

part of your question.

continuing with the original answer:

You may use a java.util.Collection and/or java.util.ArrayList for that.

The retainAll method does the following:

Retains only the elements in this collection that are contained in the specified collection

see this sample:

import java.util.Collection;
import java.util.ArrayList;
import java.util.Arrays;

public class Repeated {
    public static void main( String  [] args ) {
        Collection listOne = new ArrayList(Arrays.asList("milan","dingo", "elpha", "hafil", "meat", "iga", "neeta.peeta"));
        Collection listTwo = new ArrayList(Arrays.asList("hafil", "iga", "binga", "mike", "dingo"));

        listOne.retainAll( listTwo );
        System.out.println( listOne );
    }
}

EDIT

For the second part ( similar values ) you may use the removeAll method:

Removes all of this collection's elements that are also contained in the specified collection.

This second version gives you also the similar values and handles repeated ( by discarding them).

This time the Collection could be a Set instead of a List ( the difference is, the Set doesn't allow repeated values )

import java.util.Collection;
import java.util.HashSet;
import java.util.Arrays;

class Repeated {
      public static void main( String  [] args ) {

          Collection<String> listOne = Arrays.asList("milan","iga",
                                                    "dingo","iga",
                                                    "elpha","iga",
                                                    "hafil","iga",
                                                    "meat","iga", 
                                                    "neeta.peeta","iga");

          Collection<String> listTwo = Arrays.asList("hafil",
                                                     "iga",
                                                     "binga", 
                                                     "mike", 
                                                     "dingo","dingo","dingo");

          Collection<String> similar = new HashSet<String>( listOne );
          Collection<String> different = new HashSet<String>();
          different.addAll( listOne );
          different.addAll( listTwo );

          similar.retainAll( listTwo );
          different.removeAll( similar );

          System.out.printf("One:%s%nTwo:%s%nSimilar:%s%nDifferent:%s%n", listOne, listTwo, similar, different);
      }
}

Output:

$ java Repeated
One:[milan, iga, dingo, iga, elpha, iga, hafil, iga, meat, iga, neeta.peeta, iga]

Two:[hafil, iga, binga, mike, dingo, dingo, dingo]

Similar:[dingo, iga, hafil]

Different:[mike, binga, milan, meat, elpha, neeta.peeta]

If it doesn't do exactly what you need, it gives you a good start so you can handle from here.

Question for the reader: How would you include all the repeated values?

Answer 2

You can try intersection() and subtract() methods from CollectionUtils .

intersection() method gives you a collection containing common elements and the subtract() method gives you all the uncommon ones.

They should also take care of similar elements

Answer 3

Are these really lists (ordered, with duplicates), or are they sets (unordered, no duplicates)?

Because if it's the latter, then you can use, say, a java.util.HashSet<E> and do this in expected linear time using the convenient retainAll .

    List<String> list1 = Arrays.asList(
        "milan", "milan", "iga", "dingo", "milan"
    );
    List<String> list2 = Arrays.asList(
        "hafil", "milan", "dingo", "meat"
    );

    // intersection as set
    Set<String> intersect = new HashSet<String>(list1);
    intersect.retainAll(list2);
    System.out.println(intersect.size()); // prints "2"
    System.out.println(intersect); // prints "[milan, dingo]"

    // intersection/union as list
    List<String> intersectList = new ArrayList<String>();
    intersectList.addAll(list1);
    intersectList.addAll(list2);
    intersectList.retainAll(intersect);
    System.out.println(intersectList);
    // prints "[milan, milan, dingo, milan, milan, dingo]"

    // original lists are structurally unmodified
    System.out.println(list1); // prints "[milan, milan, iga, dingo, milan]"
    System.out.println(list2); // prints "[hafil, milan, dingo, meat]"

Answer 4

Using java 8 removeIf

public int getSimilarItems(){
    List<String> one = Arrays.asList("milan", "dingo", "elpha", "hafil", "meat", "iga", "neeta.peeta");
    List<String> two = new ArrayList<>(Arrays.asList("hafil", "iga", "binga", "mike", "dingo")); //Cannot remove directly from array backed collection
    int initial = two.size();

    two.removeIf(one::contains);
    return initial - two.size();
}

Answer 5

Simple solution :-

    List<String> list = new ArrayList<String>(Arrays.asList("a", "b", "d", "c"));
    List<String> list2 = new ArrayList<String>(Arrays.asList("b", "f", "c"));

    list.retainAll(list2);
    list2.removeAll(list);
    System.out.println("similiar " + list);
    System.out.println("different " + list2);

Output :-

similiar [b, c]
different [f]

Answer 6

如果您正在寻找一种方便的方法来测试两个集合的相等性，则可以使用org.apache.commons.collections.CollectionUtils.isEqualCollection ，它比较两个集合而不考虑其顺序。

Answer 7

我在List Compare中找到了一个非常基本的List比较示例。该示例首先验证大小，然后检查另一个列表中某个列表的特定元素的可用性。

Answer 8

Assuming hash1 and hash2

List< String > sames = whatever
List< String > diffs = whatever

int count = 0;
for( String key : hash1.keySet() )
{
   if( hash2.containsKey( key ) ) 
   {
      sames.add( key );
   }
   else
   {
      diffs.add( key );
   }
}

//sames.size() contains the number of similar elements.

Answer 9

Of all the approaches, I find using org.apache.commons.collections.CollectionUtils#isEqualCollection is the best approach. Here are the reasons -

• I don't have to declare any additional list/set myself
• I am not mutating the input lists
• It's very efficient. It checks the equality in O(N) complexity.

If it's not possible to have apache.commons.collections as a dependency, I would recommend to implement the algorithm it follows to check equality of the list because of it's efficiency.

Answer 10

public static boolean compareList(List ls1, List ls2){
    return ls1.containsAll(ls2) && ls1.size() == ls2.size() ? true :false;
     }

public static void main(String[] args) {

    ArrayList<String> one = new ArrayList<String>();
    one.add("one");
    one.add("two");
    one.add("six");

    ArrayList<String> two = new ArrayList<String>();
    two.add("one");
    two.add("six");
    two.add("two");

    System.out.println("Output1 :: " + compareList(one, two));

    two.add("ten");

    System.out.println("Output2 :: " + compareList(one, two));
  }