HashMap logic

[trupti nigam]

I have two hashmaps.

HasMap1 has keys 100 , 200
HashMap2 has keys 200,300,400,100

I need to remove the elements with keys 300 and 400 which are also in HashMap1 from HashMap2. How to achieve that? I thought of different ways but the problem is even if I separate the duplicates keys how to make sure I remove only those from HashMap2.


thanks,
Trupti

[Campbell Ritchie]

Draw a diagram on a piece of paper and the logic will become obvious.
You need to be specific. By “also in HashMap1” do you mean they have the same “V”s?

[Ashwini Kashyap]

Hi Trupti,

I think you should check out once the API for Hashmap here.

For reading keys of the map, use keySet(). Iterate through them and proceed like for instance:

if(hashMap1.get(key).equals(hashMap2.get(key)){
hashMap2.remove(key);
}

Thanks and Regards,
Ashwini Kashyap | [e-mail] | www.infocepts.com

[trupti nigam]

Campbell Ritchie wrote:Draw a diagram on a piece of paper and the logic will become obvious.
You need to be specific. By “also in HashMap1” do you mean they have the same “V”s?

Yes the values are same.

[trupti nigam]

Ashwini Kashyap wrote:Hi Trupti,

I think you should check out once the API for Hashmap here.

For reading keys of the map, use keySet(). Iterate through them and proceed like for instance:

if(hashMap1.get(key).equals(hashMap2.get(key)){
hashMap2.remove(key);
}

Thanks and Regards,
Ashwini Kashyap | [e-mail] | www.infocepts.com

I don't want to remove the duplicate key elements but the non dupe one.

[John Jai]

Alternatively instead of iterating, you can use the retainAll() method in Set interface. First get the key set of the second map and call retainAll() on it with the first map's keys. Check methods - keySet() and retainAll()

[Ashwini Kashyap]

Yet its simple. Make use of !equals() i.e. not equals method.

Thanks and Regards,
Ashwini Kashyap | [e-mail] | www.infocepts.com

[trupti nigam]

It was pretty simple. Here is the sample program.

HashMap h1 =   new HashMap();
      h1.put(100,"100");
      h1.put(200,"200");

        HashMap h2 =   new HashMap();
        h2.put(100,"100");
        h2.put(200,"200");
        h2.put(300,"300");

        Iterator itr2 = h2.keySet().iterator();
        while(itr2.hasNext())
        {
            System.out.println("Hashmap h2 has before the logic" + itr2.next() );
        }
        Iterator itr = h2.keySet().iterator();
        while(itr.hasNext())
        {
          Object a =  itr.next();
            if(!h1.keySet().contains(a))
            {
               itr.remove();
            }

        }

        Iterator itr1 = h2.keySet().iterator();
        while(itr1.hasNext())
        {
         System.out.println("Hashmap h2 has" + itr1.next() );
        }

[Paul Clapham]

I would prefer something more simple than that. Based on John Jai's comment, I would replace this:

Iterator itr = h2.keySet().iterator();  
        while(itr.hasNext())  
        {  
          Object a =  itr.next();  
            if(!h1.keySet().contains(a))  
            {  
               itr.remove();  
            }  
  
        }  

by this:

Set commonKeys = h2.keySet().retainAll(h1.keySet());