We help IT Professionals succeed at work.

Check out our new AWS podcast with Certified Expert, Phil Phillips! Listen to "How to Execute a Seamless AWS Migration" on EE or on your favorite podcast platform. Listen Now

x

Arraylist elements Compare with HashMap keys

satyabrata25
satyabrata25 asked
on
Medium Priority
654 Views
Last Modified: 2013-12-29

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

superList.add("A");superList.add("B");
superList.add("C");superList.add("E");superList.add("D");


HashMap hm = new HashMap();

hm.put("B","1111");
hm.put("D","1112");
hm.put("A","1113");
hm.put("E","1114");

I have to copmapre superList elements with the keys of HashMap hm where the matching key is there

there i have to pull the corresponding value of it and put it in a List.

Like
ArrayList al = new ArrayList();
al sud contain {1113,1111,1114,1112}


ArrayList<String> superList = new ArrayList<String>();
 
superList.add("A");superList.add("B");
superList.add("C");superList.add("E");superList.add("D");
 
 
HashMap hm = new HashMap();
 
hm.put("B","1111");
hm.put("D","1112");
hm.put("A","1113");
hm.put("E","1114");
 
I have to copmapre superList elements with the keys of HashMap hm where the matching key is there
 
there i have to pull the corresponding value of it and put it in a List.
 
Like
ArrayList al = new ArrayList();
={1113,1111,1114,1112}

Open in new window

Comment
Watch Question

CERTIFIED EXPERT
Top Expert 2016

Commented:

Author

Commented:
ArrayList<String> superList = new ArrayList<String>();

            superList.add("A");superList.add("B");superList.add("C");superList.add("E");superList.add("D");

            HashMap hm = new HashMap();
            hm.put("B","1111");hm.put("D","1112");hm.put("A","1113");hm.put("E","1114");
            
            Set<String> set =new HashSet<String>();
        Iterator it = superList.iterator();
         while (it.hasNext())
         {
                // Get element
                String element = (String)it.next();
                set.add(element);
          }
         System.out.println("list in set"+set);
         
         Set common = new HashSet(set.keySet());
         common.retainAll( hm.keySet());
           System.out.println(common);
After converting my list to hashset the set.keySet() will not work right.
ur logic will not work in that case...how to get those values to fit into ur logic....
            
CERTIFIED EXPERT
Top Expert 2016
Commented:
Unlock this solution with a free trial preview.
(No credit card required)
Get Preview
CERTIFIED EXPERT
Top Expert 2016

Commented:
With generics:

Set<String> common = new HashSet<String>(superList);
CERTIFIED EXPERT
Top Expert 2016

Commented:
The rest of it would be as follows:
        common.retainAll(hm.keySet());
        List<String> valuesList = new ArrayList<String>(common.size());
        for (String key : common) {
            valuesList.add(hm.get(key));
        }
        System.out.println(valuesList);

Open in new window

CERTIFIED EXPERT
Top Expert 2016

Commented:
:-)

Author

Commented:

ITS FAILING>>>>
public class test5
{
      public ArrayList<String> getData(HashMap<String, ArrayList<String>> AMap,ArrayList<String> temp)
      {
            
            System.out.println(" Invloking getData ");
            Set<String> tempSet = new HashSet<String>(temp);
        System.out.println("getData "+tempSet);
        Set commonKeys = new HashSet(AMap.keySet());
        ArrayList<String> ruleGroupIds = new ArrayList<String>(commonKeys);
        commonKeys.retainAll( tempSet );
           System.out.println(commonKeys);
             for (Object key : commonKeys)
           {
                  ArrayList<String>  value = (ArrayList<String>) AMap.get(key);
                   ruleGroupIds.addAll(value);
             }
             System.out.println("result " + ruleGroupIds);
           return ruleGroupIds;
      }
      public static void main(String args[])
      {
            ArrayList<String> compareList = new ArrayList<String>();
            compareList.add("A");      compareList.add("B");      compareList.add("D");
            ArrayList<String> superList = new ArrayList<String>();
            ArrayList<String> ruleGroupIds = new ArrayList<String>();
            ArrayList<String> ruleGroupIdsss = new ArrayList<String>();
            ruleGroupIdsss.add("1234");ruleGroupIdsss.add("12345");
            superList.add("A");superList.add("B");superList.add("C");superList.add("E");//superList.add("D");

            HashMap<String, ArrayList<String>> hm = new HashMap<String, ArrayList<String>>();
            hm.put("B",ruleGroupIdsss);hm.put("D",ruleGroupIdsss);hm.put("A",ruleGroupIdsss);hm.put("E",ruleGroupIdsss);
            hm.put("C",ruleGroupIdsss);
            
            test5 t = new test5();
            t.getData(hm, compareList);
      }
}


ITS FAILING.The result is
result [C, B, A, E, D, 1234, 12345, 1234, 12345, 1234, 12345] which is wrong.
It sud pull corresponding arraylist instead of keeping all keys and values..
can u check..
CERTIFIED EXPERT
Top Expert 2016

Commented:
Haven't really got time to look in detail but i would put more debug printouts in to see what you're starting and ending with

Author

Commented:
i modified the logic its working...no need to proceed...
Unlock the solution to this question.
Thanks for using Experts Exchange.

Please provide your email to receive a free trial preview!

*This site is protected by reCAPTCHA and the Google Privacy Policy and Terms of Service apply.

OR

Please enter a first name

Please enter a last name

8+ characters (letters, numbers, and a symbol)

By clicking, you agree to the Terms of Use and Privacy Policy.