How can I sort a vector array??? 3

[DCCoolBreeze]

I have a vector of class objects. Within the class object, I have several properties one of which is an integer. Is there a way I can sort the vector array relative to that specific integer property?

Example: Class Person
String fname
String lname
int age

Vector v = new Vector

v.add(...)
v.add(...)
v.add(...)

now sort v by age

Thanks for the help

 

[toolkit]

An example:

import java.util.*;

public class SortedVector {

  static class Person implements Comparable {
    String name;
    String surname;
    int age;

    Person(String name, String surname, int age) {
      this.name = name;
      this.surname = surname;
      this.age = age;
    }
    public String toString() {
      return
        ("Name: "+name+
        "\nSurname: "+surname+
        "\nAge: "+age);
    }
    public int compareTo(Object o) {
      int age1 = ((Person) this).age;
      int age2 = ((Person) o).age;
      return (age1 - age2);
    }
  }

  public static void main(String[] args) {
    Vector v = new Vector();
    v.add( new Person( "John", "Smith", 20 ) );
    v.add( new Person( "Jane", "Doe",   30 ) );
    v.add( new Person( "Fred", "Black", 10 ) );
    Collections.sort(v);
    Enumeration e = v.elements();
    while(e.hasMoreElements()) {
      System.out.println(e.nextElement().toString());
    }
  }
}

Cheers, Neil

 

[DCCoolBreeze]

Thanks! Works great. Now I have one more question. What if I want to sometimes sort on one field in the class and then another field in the class. Say perhaps, sometimes age...sometimes lastname and sometimes both

 

[toolkit]

Collections.sort is overloaded with a method that requires the user to supply an implementation of Comparator:

Collections.sort(Collection coll, Comparator comp);

The Comparator interface is defined as:

package java.util;
public interface Comparator {
  public abstract int compare(Object o1, Object o2);
  public abstract boolean equals(Object obj);
}

Therefore you can use code like:

import java.util.*;

public class SortedVector {

  static class Person {
    String name;
    String surname;
    int age;

    Person(String name, String surname, int age) {
      this.name = name;
      this.surname = surname;
      this.age = age;
    }
    public String toString() {
      return
        ("Name: "+name+
        "\nSurname: "+surname+
        "\nAge: "+age);
    }
  }

  public static void main(String[] args) {
    Vector v = new Vector();
    v.add( new Person( "John", "Smith", 20 ) );
    v.add( new Person( "Jane", "Smith", 30 ) );
    v.add( new Person( "Fred", "Black", 10 ) );
    Collections.sort(v, new Comparator() {
        // sort by surname, then age
        public int compare(Object o1, Object o2) {
          Person p1 = (Person) o1;
          Person p2 = (Person) o2;
          int res = p1.surname.compareTo(p2.surname);
          if(res != 0)
            return res;
          else
            return p1.age - p2.age;
        }
        public boolean equals(Object obj) {
          return this == obj;
        }
    });
    Enumeration e = v.elements();
    while(e.hasMoreElements()) {
      System.out.println(e.nextElement().toString());
    }
  }
}

Cheers, Neil

 

[DCCoolBreeze]

outstanding! Thanks!

 

[DCCoolBreeze]

OK. Just one more question (do I sound like Columbo?). Is there a way to extract all objects that are the same in some way? For example, let's say that I want to move all objects whose age values are above 40 to another vector. I know that I can do this in about 6 lines but I noticed a method like copyAll that moves all objects in a Vector to another Vector.

or

Can I sort a vector using tags? for example, Let's say that I want to sort the Collection using the following tags in the order of the listing of the tags:

Germany, Alaska, North Pole.

Notice that they are not in alphabetical order.

I guess if I understood the logic behind sort I might be able to manipulate it to do the above

 

[palbano]

>> Can I sort a vector using tags?

In your implementation of Comparable.compareTo() you can compare whatever you like. Also you can have more than one <Comparable> implementation so you can sort the Vector in several different ways.


>> Is there a way to extract all objects that are the
>> same in some way?

Not how I would do this but for a school project it should be fine. You can use Hashtables for that by putting a Vector in each location and filling up the Vector.

Hashtable locations = new Hashtable();
locations.put(&quot;Germany&quot;, new Vector());
((Vector)locations.get(&quot;Germany&quot).add( new MyObject(...));
locations.put(&quot;North Pole&quot;, new Vector());
((Vector)locations.get(&quot;North Pole&quot).add( new MyObject(...));

Now if you are in an OOAD course you might want to consider deriving from Hashtable but that is another subject.

Good luck
-pete

 

[DCCoolBreeze]

thanks again!

 

[DCCoolBreeze]

I am looking at Hashtables now. The problem with Hashtables is they reference a unique key...say, I can have more than one John Smith who is 40. Now I can generate my own key but then would we not be back to Vectors? Actually, let me ask this? What would be the best practise methodology for a situation like this?

 

[toolkit]

One approach to finding people within a certain age range.

  public static void main(String[] args) {
    Vector v = new Vector();
    v.add( new Person( "a", "z", 3 ) );
    v.add( new Person( "b", "y", 5 ) );
    v.add( new Person( "c", "x", 23 ) );
    v.add( new Person( "d", "w", 17 ) );
    v.add( new Person( "e", "v", 45 ) );
    v.add( new Person( "f", "u", 99 ) );
    v.add( new Person( "g", "t", 40 ) );
    v.add( new Person( "h", "s", 77 ) );
    v.add( new Person( "i", "r", 32 ) );
    v.add( new Person( "j", "q", 12 ) );
    v.add( new Person( "k", "p", 43 ) );
    v.add( new Person( "l", "o", 22 ) );
    v.add( new Person( "m", "n", 35 ) );

    Comparator byAge = new Comparator() {
        public int compare(Object o1, Object o2) {
          Person p1 = (Person) o1;
          Person p2 = (Person) o2;
          return p1.age - p2.age;
        }
    };

    Collections.sort(v, byAge);
    int min = Collections.binarySearch(v, new Person("","",40), byAge);
    int max = Collections.binarySearch(v, new Person("","",80), byAge);
    System.out.println("min: "+min+", max: "+max);
    min = (min < 0) ? -(min+1) : min;
    max = (max < 0) ? -(max+1) : max;
    System.out.println(&quot;min: &quot;+min+&quot;, max: &quot;+max);
    Iterator it = v.subList(min, max).iterator();
    while(it.hasNext())
      System.out.println(it.next().toString());
  }

See

http://java.sun.com/j2se/1.4/docs/api/java/util/Collections.html

for information on the binarySearch method.

A similar thing can be done for a specific surname. See next post

 

[toolkit]

And find all surnames beginning with &quot;J&quot;.

  public static void main(String[] args) {
    Vector v = new Vector();
    v.add( new Person( "a", "Smith", 10 ) );
    v.add( new Person( "b", "Smith", 10 ) );
    v.add( new Person( "c", "Smith", 10 ) );
    v.add( new Person( "d", "Jones", 10 ) );
    v.add( new Person( "e", "Jones", 10 ) );
    v.add( new Person( "f", "Jones", 10 ) );
    v.add( new Person( "g", "Jones", 10 ) );
    v.add( new Person( "h", "Jones", 10 ) );
    v.add( new Person( "i", "Jones", 10 ) );
    v.add( new Person( "j", "Flintstone", 10 ) );
    v.add( new Person( "k", "Flintstone", 10 ) );
    v.add( new Person( "l", "Flintstone", 10 ) );
    v.add( new Person( "m", "Flintstone", 10 ) );

    Comparator bySurname = new Comparator() {
        public int compare(Object o1, Object o2) {
          Person p1 = (Person) o1;
          Person p2 = (Person) o2;
          return p1.surname.compareTo(p2.surname);
        }
    };

    Collections.sort(v, bySurname);
    int min = Collections.binarySearch(v, new Person("","J",0), bySurname);
    int max = Collections.binarySearch(v, new Person("","K",0), bySurname);
    System.out.println("min: "+min+", max: "+max);
    min = (min < 0) ? -(min+1) : min;
    max = (max < 0) ? -(max+1) : max;
    System.out.println(&quot;min: &quot;+min+&quot;, max: &quot;+max);
    Iterator it = v.subList(min, max).iterator();
    while(it.hasNext())
      System.out.println(it.next().toString());
  }

Cheers, Neil

 

[palbano]

When I need to work with complex data using different methods to organize and extract data I look at using XML. The combination of XPath query results and XSLT transformations can provide a simple and powerful data mechanism.

&quot;But, that's just my opinion... I could be wrong&quot;.
-pete