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This is a static method of the Lotus formula @Unique, where you pass it a vector and it returns you a string without duplicate values.
Code: Usage:
String returnList = Unique(vector value);
static String Unique(Vector vValue) {
try
{
Vector hList = vValue;
String gListH = hList.toString();
String gListH1 = gListH.replace('[', ' ');
String gList = gListH1.replace(']', ' ');
String gListClean = gList.trim();
Vector v1 = new Vector();
Vector v2 = new Vector();
Vector temp=new Vector();
String test = gListClean;
StringTokenizer st = new StringTokenizer(test,",");
while(st.hasMoreTokens())
{
String tmp = st.nextToken();
v1.add(tmp);
v2.add(tmp);
}
String item2="";
for(int k=0; k < v1.size(); k++)
{
String item=(String)v1.get(k);
if(!temp.contains(item))
temp.addElement(item);
}
v2 = temp;
Vector hReturnList = v2;
String gReturnListH = hReturnList.toString();
String gReturnListH1 = gReturnListH.replace('[', ' ');
String gReturnListH2 = gReturnListH1.replace(']', ' ');
String gReturnListClean = gReturnListH2.trim();
return gReturnListClean;
}
catch(Exception e)
{
return "";
}
}
MEMBER FEEDBACK TO THIS TIP
I have a feeling that @Unique can be done with some help from Java by using Set.
A Set is a collection that cannot contain duplicate elements.
See: http://java.sun.com/docs/books/tutorial/collections/index.html
The code fragment to get unique entries is only one line of code:
Set set = new TreeSet(vector);
I include the whole program to show the Set usage:
(the Unique method is included as well)
import java.util.Arrays;
import java.util.Set;
import java.util.StringTokenizer;
import java.util.TreeSet;
import java.util.Vector;
public class Main {
static String Unique(Vector vValue) {
try {
Vector hList = vValue;
String gListH = hList.toString();
String gListH1 = gListH.replace('[', ' ');
String gList = gListH1.replace(']', ' ');
String gListClean = gList.trim();
Vector v1 = new Vector();
Vector v2 = new Vector();
Vector temp = new Vector();
String test = gListClean;
StringTokenizer st = new StringTokenizer(test, ",");
while (st.hasMoreTokens()) {
String tmp = st.nextToken();
v1.add(tmp);
v2.add(tmp);
}
String item2 = "";
for (int k = 0; k < v1.size(); k++) {
String item = (String) v1.get(k);
if (!temp.contains(item))
temp.addElement(item);
}
v2 = temp;
Vector hReturnList = v2;
String gReturnListH = hReturnList.toString();
String gReturnListH1 = gReturnListH.replace('[', ' ');
String gReturnListH2 =
gReturnListH1.replace(']', ' ');
String gReturnListClean = gReturnListH2.trim();
return gReturnListClean;
} catch (Exception e) {
return "";
}
}
/**
* Return some data for the test.
* @return
*/
private static Vector getNewVector() {
Integer inputArray[] = new Integer[5];
for (int index = 0; index < inputArray.length; index++) {
inputArray[index] = new Integer(index + 1);
}
Vector vector = new Vector();
vector.addAll(Arrays.asList(inputArray));
vector.addAll(Arrays.asList(inputArray));
vector.addAll(Arrays.asList(inputArray));
return vector;
}
public static void main(String[] args) {
Vector vector = new Vector(getNewVector());
System.out.println("The original vector: " + vector);
System.out.println("The result of Unique Method: \n"
+ "(I think see a bug there.. the '1'
appears twice \n"
+ Unique(vector));
// Using the set interface
Set set = new TreeSet(vector);
System.out.println("set content: " + set);
vector.removeAllElements();
vector.addAll(set);
System.out.println("vector after removeAllElements()"
+ "and addAll(set) : " + vector);
}
}
This is the result:
The original vector: [1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5] The result
of Unique Method:
(I think see a bug there.. the '1' appears twice 1, 2, 3, 4, 5, 1
setcontent: [1, 2, 3, 4, 5] vector after removeAllElements()and addAll(set) :
[1, 2, 3, 4, 5]
—Eitan R.
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