Removing Duplicates.

[dashwall]

Hi I'm fairly new to Java and have hit a small problem.

I have a 2D array all ready set up. Now I am searching through that array to populate a JOptionPane for a user to select from.

Ok I can populate the list no problems but what I would like to do is after the String Array has been populated I would like to remove all duplicates first before the list is shown to the user.

Is it possible and how could I achieve this.

Thanks

Code is as follows.

 String[] code = new String[oceanLink.length];

      for (int row = 0; row < oceanLink.length; row++)
      {
         code[row] = oceanLink[row][1];
      }

      Object routeCode = JOptionPane.showInputDialog(null, "Please choose a option", "Route Codes", JOptionPane.QUESTION_MESSAGE, null, code, code[1]);
      System.out.println(routeCode);
      String port = routeCode.toString();
      return port;

[adz]

You could merely use a Set.

[dashwall]

Ok I have never used "Set" done a small reading here and there on it but I am unsure where to use it.

Would I use a .add?

a little confussed at the moment.

[AndreB]

yes, an .add(object) is sufficient.

see java.sun.com/javase/6/docs/api/java/util/HashSet.html#add(E) for example

[adz]

Do something like this:

  String[] myArray = {"whatever","whatever","whateverx","whateverx"};
  Set<String> noDups = new HashSet<String>(Arrays.asList(myArray));
  myArray = new String[noDups.size()];
  System.arraycopy(noDups.toArray(), 0, myArray, 0, noDups.size());

Note: Don't know if this is correct since I made it up or if its even the best way to do it. Considering you're just passing the array back to the method in JOptionPane you can simply use the toArray method actually instead of the array copying crap I posted.

Meh I'm tired so I've just re-read what I posted and decided its well, wrong. You could do this:

      Set codeSet = new LinkedHashSet<String>();
      for (int row = 0, n = oceanLink.length; row < n; row++)
      {
         codeSet.add(oceanLink[row][1]);
      }

      Object routeCode = JOptionPane.showInputDialog(null, "Please choose a option", "Route Codes", JOptionPane.QUESTION_MESSAGE, null, codeSet.toArray(), codeSet.get(1));
      System.out.println(routeCode);
      String port = routeCode.toString();
      return port;

But er yeah, I'm tired, so apologies if this isn't right too :) Should at least give you the idea, but look up Sets. I suspect if I read this tomorrow I'll wonder why I posted it :/

[AndreB]

The error in the first example is that although the set contains 2 elements you use arrayCopy method for the two elements. And of course they are equal.

You propably should use the toArray method to achieve desired results

String[] myArray = {"whatever","whatever","whateverx","whateverx"};
Set<String> noDups = new HashSet<String>(Arrays.asList(myArray)); 
myArray = noDups.toArray(new String[noDups.size()]);

[mac]

May i interest you in this solution?

private static String[] uniqueArguments(String[] arguments) {
		ArrayList<String> b = new ArrayList<String>();
		ArrayList<String> c = new ArrayList<String>();

		for (int i = 0; i < arguments.length; i++) {
			b.add(arguments[i].trim());
		}

		int busted = 0;

		for (int ix = 0; ix < b.size(); ix++) {
			busted = 0;
			for (int j = 0; j < b.size(); j++) {
				if (b.get(ix).equals(b.get(j))) {
					busted++;
				}
			}
			if ((busted == 1)) {
				c.add(b.get(ix));
			} else if (busted > 1) {
				b.set(ix, null);
			}
		}

		// TODO: I don't understand this statement,
		// why do we need the "(new String[0]) part???

		logger.info("Done processing -v list. Looks good ..!");
		return c.toArray(new String[0]);
	}
}

[AndreB]

Well this algorithm is slower (quadratic runtime!) than using sets and requires more memory.

To answer your other question ist the "new String[0]" is used to achive additional type cast (see javadoc of toArray method): the method will return an array of String and not an array of Object.

[AndreB]

Oh, and the other thing: if you really need that preprocessing and want to use ArrayList then this should do the same as above

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

for (String s : arguments) {
	String myProcessedString = s.trim();
	if (!c.contains(myProcessedString)) {
		c.add(myProcessedString);
	}
}
		
return c.toArray(new String[c.size()]);

However i'm a fan of HashSets ;-)

[dashwall]

Thank you all so very much for the help i used the second method that adz posted thank you again.

Dash