how to avoid duplicates and count the list once stored in LinkedHashMap

Question

I have a scenario to count number of names which is stored in the LinkedHashMap , and names can be duplicated but i should not count the duplicate name.

Below is Sample code:

         LinkedHashMap<Long,MyApplicationDTO> myApps = (LinkedHashMap<Long,MyApplicationDTO>) request.getAttribute("data");
                  for (Map.Entry app : myApps.entrySet()) {                  
                  Long  ID = (Long)app.getKey() ;
                  MyApplicationDTO singleMyApp =  (MyApplicationDTO) app.getValue();

        LinkedHashMap<Long, MyDTO> myList  = singleMyApp.getMyList();
          String name = "";  
        for (Map.Entry details : myList.entrySet()) {      
            Long id1 = (Long)details.getKey() ;
            MyDTO myDetails =  (MyDTO) details.getValue();

            name = myDetails.getName(); // For first time it stores A
            //how to loop so that i can only get the count of names as 3 by avoiding duplicate names from the below shown list.
            //A B A B A B C

}

}

On the Screen i have something as below:

Name : A B A B A B C

I have to print the count of the name as 3(non repeating names)

http://stackoverflow.com/questions/4415803/java-code-to-prevent-duplicate-key-value-pairs-in-hashmap-hashtable — Ya Wang, Feb 10 '15 at 18:19
Do you don't want to save the duplicates in your Map? Or just want to skip the duplicates while counting? — Vivin, Feb 10 '15 at 18:33
No , i want to count the number of names using myDetails.getName() from LinkedHashMap but should not count the repeated names. @Vwin. — user222, Feb 10 '15 at 18:34
@Vwin , i want to skip the duplicates names while counting , so my output should give me name count as 3. thanks. — user222, Feb 10 '15 at 18:35

score 3 · Accepted Answer · answered Feb 10 '15 at 18:33

3

As you iterate over the entrySet, add all names to a Set<String>. Then output set.size().

The Set will not add duplicates when you add names by set.add(name), so the size of the set will be the count of uniqe names.

answered Feb 10 '15 at 18:33

duckstep

1,148
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Can you please provide the sample from my code if possible. Thanks.@duckstep – user222 Feb 10 '15 at 18:37
Yes this is the easiest way to do – Vivin Feb 10 '15 at 18:43

score 2 · Answer 2 · answered Feb 10 '15 at 18:40

LinkedHashMap<Long,MyApplicationDTO> myApps = (LinkedHashMap<Long,MyApplicationDTO>) request.getAttribute("data");
for (Map.Entry app : myApps.entrySet()) {                  
Long  ID = (Long)app.getKey() ;
MyApplicationDTO singleMyApp =  (MyApplicationDTO) app.getValue();

LinkedHashMap<Long, MyDTO> myList  = singleMyApp.getMyList();
String name = "";
Set<String> uniqueNames = new HashSet<String>();
for (Map.Entry details : myList.entrySet()) {      
     Long id1 = (Long)details.getKey() ;
     MyDTO myDetails =  (MyDTO) details.getValue();

     name = myDetails.getName(); // For first time it stores A
       //how to loop so that i can only get the count of names as 3 by avoiding duplicate names from the below shown list.
            //A B A B A B C
     uniqueNames.add(name);

}

}

To get size do =

uniqueNames.size();

score 1 · Answer 3 · answered Feb 10 '15 at 18:39

I'm not sure I'm understanding your question entirely, but if you're just looking to count the occurrences of unique values in the LinkedHashmap you can do something like this `

public static void main(String[] args) {

    LinkedHashMap<Long, String> myApps = new LinkedHashMap<Long, String>();
    myApps.put(4L, "A");
    myApps.put(14L, "B");
    myApps.put(44L, "A");
    myApps.put(54L, "B");
    myApps.put(46L, "A");
    myApps.put(543L, "B");
    myApps.put(144L, "C");

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

    for (Map.Entry app : myApps.entrySet()) {

        if (!(names.contains(app.getValue()))) {
            names.add(app.getValue().toString());
        }
    }

    System.out.println(names.size());
    for (String s : names ) {
        System.out.print(s + " ");
    }
}

Looks like @duckstep gave the idea, so credit to him. Here's a representation of it. Output : 3 A B C — tlw11591, Feb 10 '15 at 18:41

score 1 · Answer 4 · answered Feb 10 '15 at 18:43

1

One liner with JAVA 8 Stream API

LinkedHashMap<Long, MyDTO> myList  = singleMyApp.getMyList();

long disctinctNamesCount = myList.values().stream().map(MyDTO::getNAme).distinct().count();

answered Feb 10 '15 at 18:43

iamiddy

3,015
3
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Then a Set is your friend as others have suggested. – iamiddy Feb 10 '15 at 18:50

how to avoid duplicates and count the list once stored in LinkedHashMap

4 Answers4