More than a sorting with dupplicate elimination

[Lucian Botezatul]

My Problem:

Have Quizz and a Part of the Programm show the User some Feedback.

I built some "dummy" Names, where the Number of answered Questions is updatet randomly, so the User think that some other users are around

Now my Problem i want something like this after sorting:

14 Questions: Mike, Anna
12 Questions: Michael, John
11 Questions: Andreas, Nick

But i get something like this:

14 Questions: Mike
14 Questions: Anna
12 Questions: Michael
12 Questions: John
....

Tried with hashSet and TreeMap but nothing worked ... hope


My Code:

         public static Object[] list_RF_Rnk(int a, String nam){  // ---------------------------------------- 
        int n1=1 , n2=2, n3=3, n4=4, n5=5, n6=6, n7=7, n8=8, n9=9, n10=10;               // the INT VARIABLE for THE DUMMY USER
        int RF_Rnk=0;   
        n1 = n1 + Experiment_RF_Rnk.n1a;                             
        n2 = n2 + Experiment_RF_Rnk.n2a;
        n3 = n3 + Experiment_RF_Rnk.n3a;                  // HERE INCREASE RANDOMLY THE VARIABLES
        n4 = n4 + Experiment_RF_Rnk.n4a;
        n5 = n5 + Experiment_RF_Rnk.n5a;
        n6 = n6 + Experiment_RF_Rnk.n6a;
        n7 = n7 + Experiment_RF_Rnk.n7a;
        n8 = n8 + Experiment_RF_Rnk.n8a;
        n9 = n9 + Experiment_RF_Rnk.n9a;
        n10 = n10 + Experiment_RF_Rnk.n10a;
        
        Object result [];                              
        String name = "--> "+ nam +" <--";           // nam the User Name, rest dummy names
        if (a ==RF_Rnk) a11 = name + ", ";
        else if (a ==n1) a1 = name + ", ";
        else if (a ==n2) a2 = name + ", ";
        else if (a ==n3) a3 = name+ ", ";
        else if (a ==n4) a4 = name+ ", ";
        else if (a ==n5)a5 =name+ ", ";
        else if (a ==n6)a6 =name+ ", ";
        else if (a ==n7)a7 =name+ ", ";
        else if (a ==n8)a8 =name+ ", ";
        else if (a ==n9)a9 =name+ ", ";
        else if (a ==n10)a10 =name+ ", ";        
       
        else{
            neuu = "\n  " + a + " Questions: \t" + name;
        }
        
               
        
            String[] coll = {//"\n " + RF_Rnk + " Questions: \t" + a11 + "",                                       // all the dummy names
                "\n" + n10 + " Questions: \t" + a10 + "Frank, Lisa03, Mark Schanz, TAbea",
                "\n" + n9 + " Questions: \t" + a9 + "David",
                "\n" + n8 + " Questions: \t" + a8 + "Christian, Julian",
                "\n" + n7 + " Questions: \t" + a7 + "Anka",
                "\n" + n6 + " Questions: \t" + a6 + "Ben",
                "\n" + n5 + " Questions: \t" + a5 + "Michaela",
                "\n" + n4 + " Questions: \t" + a4 + "Mustafa",
                "\n" + n3 +" Questions: \t" + a3 + "Jerry",
                "\n" + n2 + " Questions: \t" + a2 + "Uwe",
                "\n" + n1 + " Questions: \t" + a1 + "Jahn"                
                
       };
         Arrays.sort(coll);   // sorted array list
       ArrayList<String> liste = new ArrayList<String>() ;
       for (int i = 0; i<coll.length; i++){
           liste.add(coll[i]);
       }
       
       
        // new Line in the List
        if(a1.equals("")&&a2.equals("")&&a3.equals("")&&a4.equals("")&&a5.equals("")&&a6.equals("")&&a7.equals("")&&a8.equals("")&&a9.equals("")&&a10.equals("")&&a11.equals("")){
             if(a<10) liste.add(neuu);
             if(a>10) liste.add(neuu);
        } 
        result = liste.toArray();
        a1="";
        a2="";
        a3="";
        a4="";
        a5="";
        a6="";
        a7="";
        a8="";
        a9="";
        a10="";
        a11="";
        neuo = "";
        neuu = "";
        return result;    
        
        }]

I try first to sort array against array with hashSet:

 ArrayList arrayList2 = new ArrayList(collSet) ;
       Collections.sort(arrayList2);
       for (Object item : arrayList2)
            System.out.println(item)

did not work

Then use a TreeMap:

TreeMap<Object[], List<String>>NewSorted_list = new TreeMap<Object[],List<String>> ();  //---------------- TreeMap sort
        NewSorted_list(result,liste);  
        System.out.print(NewSorted_list);

Then took put the Numbers n1 ... n10 and names in a TreeMap:

TreeMap<String,String>NewSorted_list= new TreeMap<String,String> (); 
        NewSorted_list.put(c1, "Mike ");
        NewSorted_list.put(c1, " Anna");
         NewSorted_list.put(c2, " Michael");
       NewSorted_list.put(c2, " John");
.......

Problem no dupplicates but lost names:

instead of:

14 Questions: Mike, Anna
12 Questions: Michael, John

i got:

14 Questions: Anna
12 Questions: John

.. no dupplicates, but lost names ...

If any ideas ... BIG THANKS!!! ... if hard to understand what my Problem(s) are, i will try to post once more

[Jesper de Jong]

Welcome to JavaRanch.

Lucian Botezatul wrote:Tried with hashSet and TreeMap but nothing worked ... hope

Using a Map was a good idea.

Lucian Botezatul wrote:
Then use a TreeMap:

TreeMap<Object[], List<String>>NewSorted_list = new TreeMap<Object[],List<String>> ();  //---------------- TreeMap sort
        NewSorted_list(result,liste);  
        System.out.print(NewSorted_list);

Then took put the Numbers n1 ... n10 and names in a TreeMap:

TreeMap<String,String>NewSorted_list= new TreeMap<String,String> (); 
        NewSorted_list.put(c1, "Mike ");
        NewSorted_list.put(c1, " Anna");
         NewSorted_list.put(c2, " Michael");
       NewSorted_list.put(c2, " John");
.......

In the first try, why did you use Object[] (an array of objects) as the key in the map?

In the second try: That was almost working, but the reason it did not is this: When you call put() on a Map, and the Map already contains an element for the key that you are using, then you are replacing the element in the Map. So if there is already an entry 14 -> Mike in the map and you put Anna under key 14, then Mike will be replaced by Anna, so you'll not have Mike in the map anymore.

Instead of just putting Anna in there, you will want to get what is in the map first, and if it is not null, then add Anna to it. Like this:

String s = NewSorted_list.get(c1);
if (s == null) {
	s = "Anna";  // new entry
}
else {
	s = s + ", Anna";  // use whatever was there and add Anna to it
}
NewSorted_list.put(c1, s);

[Lucian Botezatul]

@Jesper, Thank You !

That works very fine, one problem solved new one occurs

Now my question, it´s the TreeMap the RIGHT tool to do this, or should i try something else ?


If i put the names like suggested, i got something like this:

14 Ana, Mike, ..
15 John, Travis ...


Give for each name one variable, c1 (Ana), c11 (Mike), c2 (John), c21 (Travis) where at begin they have the same values c1 = c11 = 14, c2 = c21 = 15.
Sort with TreeMap, no problem.

Now, i increase each of them randomly, exeample:

c1 = c1 + (int)Math.round( Math.random() * 7 ); 

now when two of my variables have the same value, my initial problem appears again

means after first round random, c1 = 18, c11 = 19, c2 = 19, c21 = 22

now i see:

18 Ana
19 John <------------------------------- Mike is missing here
22 Travis


Why i need this? In my quiz ( here the old version without the TreeMap modifications QUIZ ) i give out a updated list with the "dummy" users + the real user.

For each dummy user i give a variable that increase randomly, so the one user who play have the impresion that 10-15 other guys play at the same time with him .

[Rob Spoor]

In a Map, each key can only have one value. For key 19 you want two values, but that's simply not possible with maps. Unless you use a Map<X,List<Y>>, but it will require some more work.

[Lucian Botezatul]

@ Rob

if i get it right, when i have a Map i can use two different keys with the same value?

In the Map If i have the keys = my variable c1, c11, c2, c21, ... c6, c61 and the names "John, Mike ... ".

when the value of my two keys is the same c1 = c11 = 19 ... in TreeMap i dont have key 1 value John and key 2 value Mike, key 2 override my key 1?

i allready got the hint:

# if (s == null) {  
#     s = "Anna";  // new entry  
# }  
# else {  
#     s = s + ", Anna";  // use whatever was there and add Anna to it  
# }  
# NewSorted_list.put(c1, s);  

and it works on the first time, at new sorting again if two keys the same show me one name....

Your idea something like that? I tried so, but the same, only one name per integer

TreeMap<Integer,ArrayList<String>>sorted_List = new TreeMap<Integer,ArrayList<String>> (); 
    
    
       ArrayList<String> names; 
       names = new ArrayList<String>();
       names .add("Anna");    
       names .add("Anna 2");              <------------------------- then Anna and Anna 2 stay allways together
       sorted_List.put(c1 , names );      
       
       names = new ArrayList<String>();      
       names .add("Mike");       
       sorted_List.put(c11 , names );
       
       names = new ArrayList<String>();       
       names .add("John");
       sorted_List.put(c2 , names );

       names = new ArrayList<String>();       
       names .add("Travis");
       sorted_List.put(c22 , names );

[Jesper de Jong]

Lucian Botezatul wrote:if i get it right, when i have a Map i can use two different keys with the same value?

Yes, only the keys have to be unique, for the values it doesn't matter.

But you want to store multiple values under the same key; the key is the number, and the values are the names. You can do that with a Map that looks like this:

Map<Integer, List<String>> data = new TreeMap<Integer, List<String>>();

But my suggestion was to use a Map that doesn't store a List of Strings as the values, but just a single string - and that one string contains the names separated by commas, for example "Anna, Mike".

With your second piece of code: You almost got it. You already noticed that "Anna" and "Anna 2" are both under the same key. Now, what would you have to do to store "Anna" and "Mike" under the same key?

[Lucian Botezatul]

Thanks Jesper!

Now i got the key concept of the TreeMap.
I was thinking all the time that the key is the variable herself and the TreeMap sort after the value of the variable , so i followed that having two keys different keys (c1, c11) with the same value the names comes automaticaly on the same raw .

[Lucian Botezatul]

First, Thank you for all the help!

Now it works well, the Names are sorted in the right oder.

Have 2 more Questions, when i try to give the result in a window, it dont looks so nice:

1 QUESTION :

 JOptionPane myStatisticWindow = new JOptionPane();
       JTextArea MyList = new JTextArea("Number of the correct Answers \n\n" + Sortierte_liste);   
       MyList.setEditable(false);
       MyList.setOpaque(false);
       MyList.setFont(new Font("Arial", Font.PLAIN, 20));       
       myStatisticWindow.showMessageDialog(null, MyList, "LIST WITH CORRECT ANSWERS", JOptionPane.DEFAULT_OPTION);

how can i remoe the Strings so i get something like this:

1 NameX
2 NameY
3 NameZ

2 QUESTION :

does treeMap provide a descending sorting:

3 NameZ
2 NameY
1 NameX

Do i need that my class implements the NavigableMap<K,V> interface ? ... after 6 hours of coding, searching ... looks that my brain got a buffer overflow

[Lucian Botezatul]

Reverse Order was easier as i tought

SortedMap<Integer,ArrayList<String>>Sortierte_liste = new TreeMap<Integer,ArrayList<String>>(Collections.reverseOrder());

instead of

SortedMap<Integer,ArrayList<String>>Sortierte_liste = new TreeMap<Integer,ArrayList<String>>();

sometime the solution is simpler as we think

[Lucian Botezatul]

Resolved Problem, Code here:

http://www.coderanch.com/t/526309/java/java/Problems-printing-Treemap-nd-run