In the contacts class (implements actionListener) I have a JList that contains objects (elements from a file). There are sorting buttons, each button has an actionListener
by firstname by last name by city The sorting is done by the elements of objects.
the user should give the first and last name and city for a contact. How to do the sorting ??
And to put those into the list I used this code:
Map<String, contacts> ma = readFromFile();
for (Map.Entry<String, contacts> entry : ma.entrySet()) {
contacts c = (contacts) entry.getValue();
d.addElement(c.getLastName() + "" + c.getFirstName() + "" + c.getCity());
}
How to do the sorting ?? plz help
Instead of JList<String>, create a JList<contacts>
Implement class ContactsRenderer extends JLabel implements ListCellRenderer.
See java doc for more details: http://docs.oracle.com/javase/7/docs/api/javax/swing/JList.html
Implement your custom list data model:
public class ContactsModel extends AbstractListModel<contacts>() {
private final List<contacts> backingList = ArrayList<contacts>;
public ContactsModel(Map<String, contacts> ma) {
//populate backing list here from your map
}
public int getSize() {
return backingList.size();
}
public contacts getElementAt(int index) {
return backingList.get(index);
}
};
Add a sort method to your model:
public void sort(Comparator<contacts> comparator) {
Collections.sort(backingList, comparator);
fireContentsChanged(this, 0, backingList.size());
}
Implement comparators for every sort use case:
public class LastNameAscendingComparator implements Comparator<contacts> {
#Override
public int compare(contacts c1, contacts c2){
return c1.getLastName().compareTo(c2.getLastName());
}
}
Finally call your model.sort() with a corresponding comparator
In java you can sort a list like so:
Collections.sort(yourListOfContacts, new Comparator<Contacts>(){
#Override
public int compare(Contacts obj1, Contacts obj2){
obj1.getFirstName().compareTo(obj2.getFirstName());
}
});
Notice that the return type of the compare function is an integer. This integer tells the sorting algorithm which one of the two should come first. Note that a -1 means the first element is "less" than the second, a 0 means that they are the same, and a 1 means that the first element is "greater" than the second. So, to reverse the order, you would use: obj2.getFirstName().compareTo(obj1.getFirstName()); or, you could just multiply by -1.
The other fields you want to sort by will follow the same sort of pattern, this is just an example.
In Java8 you can sort a list like this:
Collections.sort(contactList, (contact1, contact2)
-> contact1.getName().compareTo(contact2.getName()));
Related
Working with a list of Object where one of the item has an empty String. Trying to write method which would return a sorted list. By sorting means the first item value of the list should always be an empty String.
Since I don't want to manipulate the unsorted list, I am creating a new list to sort.
So far my code is:
private List<LoggerConfig> sort(List<LoggerConfig> unSortedList) {
List<LoggerConfig> sortedList = new ArrayList<LoggerConfig>(unSortedList);
//What to do here
return sortedList;
}
Looked at lot of SO posts but very confused.
You can trust the String.compareTo to match the order you seek. Here is a Comparator :
new Comparator<LoggerConfig>() {
#Override
public int compare(LoggerConfig o1, LoggerConfig o2) {
return (o1.getName().compareTo(o2.getName()));
}
};
or directly implementing Comparable in the specific class (here Dummy)
class Dummy implements Comparable<Dummy>{
String name;
public int compareTo(Dummy o) {
return this.name.compareTo(o.name);
}
}
The why :
The String.compareTo check the first characters of both to find a difference (until the smallest length of both), if they match, the lengths are use to make the difference, the longest will be after the shortest (shortest.compareTo(longuest) will return an negative value (the length difference)).
In this case, "".compareTo("abc"), there is no character in the empty String, so the first check is skipped and the length is use to compare the Strings, so an empty String will always be seen as first compare to any "non-empty" String
An example with the previous Dummy class (just need to add the Constructor Dummy(String):
public class Main {
public static void main(String[] args) {
List<Dummy> dummies = new LinkedList<Dummy>();
dummies.add(new Dummy("abc.com.core"));
dummies.add(new Dummy(""));
dummies.add(new Dummy("abc.com.core.def"));
System.out.println("BEFORE : " + dummies);
Collections.sort(dummies);
System.out.println("AFTER : " + dummies);
}
}
Output :
BEFORE : [abc.com.core, , abc.com.core.def]
AFTER : [, abc.com.core, abc.com.core.def]
You can place this condition in your comparator so that elements with an empty value are considered "less" than other elements, so that it shows up at the beginning of the sorted list. Try something like this:
Collections.sort(sortedList, new Comparator<LoggerConfig>() {
#Override
public int compare(LoggerConfig o1, LoggerConfig o2) {
if(o1.getName().isEmpty(){
return -1;
}
if(o2.getName().isEmpty(){
return 1;
}
return (o1.getName().compareTo(o2.getName()));
}
});
I didn't test this, but this should make the idea clear. If the empty element shows up at the end of the list, swap the -1 and the 1.
If your List is huge and sorting takes a lot of time, it might be a better idea to remove the empty element before sorting, then sort, then place the element at the beginning.
The Comparator solution seems feasible to me; what you're missing is implementing the compare method so that it does what you want.
Collections.sort(sortedList, new Comparator<LoggerConfig>() {
#Override
public int compare(LoggerConfig o1, LoggerConfig o2) {
if(o1.getName().equals("")){
return -1;
} else if(o2.getName().equals("")) {
return 1;
} else {
return (o1.getName().compareTo(o2.getName()));
}
}
});
As per Java docs, the Comparator has a compare method that returns an int which is
less than 0 if the first argument is less than the second
0 if the arguments are equal
greater than 0 if the first argument is greater than the second
So the Comparator you need should return the comparison of the two strings if they're both different from "", and -1 (or 1) if the first (or second) String is empty.
This is what i have so far, i'm trying to sort a bunch of List<String>'s based on the value of an index.
LinkedHashSet<List<String>> sorted = new LinkedHashSet<List<String>>();
How do i sort the LinkedHashSet in order from Highest to Lowest index 2 value of the List's?
Example input:
List<String> data1 = Database.getData(uuid);
double price = Double.valueOf(data1.get(2))
data1.add("testval");
data1.add("testval");
data1.add("100.00");
sorted.add(data1);
and on another seperate List:
List<String> data2 = Database.getData(uuid);
double price = Double.valueOf(data2.get(2))
data2.add("anotherval");
data2.add("anotherval");
data2.add("50.00");
sorted.add(data2);
Output of the sorted LinkedHashSet in descending order.
testval testval 100.00
anotherval anotherval 50.00
Sorry if this is confusing, im not sure where to go about sorting like this.
Create a new class to represent you complex objects. There is no need to store multiple values in a list when you can do it in objects.
public class ComplexObject {
private String description1;
private String description2;
private Double value;
public ComplexObject(String description1, String description2, Double value) {
this.description1 = description1;
this.description2 = description2;
this.value = value;
}
public void setDescription1(String description1) {
this.description1 = description1;
}
public String getDescription1() {
return description1;
}
public void setDescription2(String description2) {
this.description2 = description2;
}
public String getDescription2() {
return description2;
}
public void setValue(Double value) {
this.value = value;
}
public Double getValue() {
return value;
}
}
Then add elements to the list and sort it using a new, custom, comparator:
public static void main(String[] args) {
List<ComplexObject> complexObjectList = new ArrayList<ComplexObject>();
//add elements to the list
complexObjectList.add(new ComplexObject("testval","testval",100.00d));
complexObjectList.add(new ComplexObject("anotherval","anotherval",50.00d));
//sort the list in descending order based on the value attribute of complexObject
Collections.sort(complexObjectList, new Comparator<ComplexObject>() {
public int compare(ComplexObject obj1, ComplexObject obj2) {
return obj2.getValue().compareTo(obj1.getValue()); //compares 2 Double values, -1 if less , 0 if equal, 1 if greater
}
});
//print objects from sorted list
for(ComplexObject co : complexObjectList){
System.out.println(co.getDescription1()+" "+co.getDescription2()+" "+co.getValue());
}
}
Output:
testval testval 100.0
anotherval anotherval 50.0
Firstly, you shouldn't use a LinkedHashSet but a TreeSet. LinkedHashSet will retain the insertion order without sorting.
Secondly, you need to initialize your TreeSet with a Comparator that compares based on whichever value of your List is required, that is, if you know the index of the String that will represent a double value in advance. Otherwise I would recommend using custom objects instead of List.
If you decide to use custom objects, you don't necessarily need to initialize your TreeSet with a Comparator as second argument.
Instead, you could have your custom objects implement Comparable, and implement a one-time comparation logic there.
It all depends on whether you only need to sort in a particular order.
Finally, custom objects will require you to override equals and hashCode.
First, and extracted from Oracle's Java reference:
This linked list defines the iteration ordering, which is the order in which elements were inserted into the set
So you can't sort your data just inserting it into the LinkedHashSet.
You may be confusing that set implementation with SortedSet. SortedSet allows you to pass a comparator which will determine the elements order in the data structure.
On the other hand, I don't know whether you chose you List<String> arbitrarily but it seems to me a wiser option to aggregate your the 3 strings as a class attributes. The point is that, if your elements are always going to be 3 elements, being the last one a double value: Why do you need a dynamic structure as a List?
EDIT
Here you have a possible better implementation of what you want:
public class Element
{
public Element(String a, String b, double val) {
this.a = a;
this.b = b;
this.val = val;
}
#Override
public String toString() {
return a + "\t" + b + "\t" + val;
}
public String a;
public String b;
public double val;
}
And you can use this class to store your elements. An example of use:
SortedSet<Element> sorted = new TreeSet<>(new Comparator<Element>() {
#Override
public int compare(Element o1, Element o2) {
return (new Double(o1.val)).compareTo(o2.val);
}
});
sorted.add(new Element("testval", "testval", 100.0));
sorted.add(new Element("anotherval", "anotherval", 50.0));
for(Element el: sorted)
{
System.out.println(el);
}
Note that the comparator is given as an instance of an anonympous inner class implementing Java's Comparator interface.
i have a class Car representing name and IDs of cars:
public class Car {
String name;
int ID;
}
and another class representing races in which i need to sort the cars by their order in race:
public class Race {
private Set<Car> cars = new TreeSet<>();
private Map<Integer, Integer> races = new TreeMap<>();//key represents the order in race, value represents the ID of a car, so i need to sort cars by the keys in races
...
public Collection getSortedCars() { ??? }
}
-any ideas how to get sorted cars? thanks much
EDIT: Im sorry, i used very bad example with values, so heres it with identifiers, i hope you get what i need..
I would not do this with a SortedSet, even though a custom Comparator could be used. The reason is because the races could be modified and thus invalidate any structure inside the TreeSet making the behavior "unpredictable".
Instead, I would make getSortedCars first get a sequence (e.g. a List) from the Set, and then sort and return such a sequence.
The actual sorting is "trivial" with Collections.sort and a custom Comparator as this is really a "sort by" operation, for instance:
class CompareCarsByWins implements Comparator<Car> {
Map<Car,Integer> wins;
public CompareCarsByWins(Map<Car,Integer> wins) {
this.wins = wins;
}
public int compareTo (Car a, Car b) {
// Actual code should handle "not found" cars as appropriate
int winsA = wins.get(a);
int winsB = wins.get(b);
if (winsA == winsB) {
// Tie, uhm, let's .. choose by name
return a.getName().compareTo(b.getName());
} else {
// Sort most wins first
return winsB - winsA;
}
}
// ..
}
// Usage:
List<Car> results = new ArrayList<Car>(cars);
Collections.sort(results, new CompareCarsByWins(races));
I need a Collection that sorts the element, but does not removes the duplicates.
I have gone for a TreeSet, since TreeSet actually adds the values to a backed TreeMap:
public boolean add(E e) {
return m.put(e, PRESENT)==null;
}
And the TreeMap removes the duplicates using the Comparators compare logic
I have written a Comparator that returns 1 instead of 0 in case of equal elements. Hence in the case of equal elements the TreeSet with this Comparator will not overwrite the duplicate and will just sort it.
I have tested it for simple String objects, but I need a Set of Custom objects.
public static void main(String[] args)
{
List<String> strList = Arrays.asList( new String[]{"d","b","c","z","s","b","d","a"} );
Set<String> strSet = new TreeSet<String>(new StringComparator());
strSet.addAll(strList);
System.out.println(strSet);
}
class StringComparator implements Comparator<String>
{
#Override
public int compare(String s1, String s2)
{
if(s1.compareTo(s2) == 0){
return 1;
}
else{
return s1.compareTo(s2);
}
}
}
Is this approach fine or is there a better way to achieve this?
EDIT
Actually I am having a ArrayList of the following class:
class Fund
{
String fundCode;
BigDecimal fundValue;
.....
public boolean equals(Object obj) {
// uses fundCode for equality
}
}
I need all the fundCode with highest fundValue
You can use a PriorityQueue.
PriorityQueue<Integer> pQueue = new PriorityQueue<Integer>();
PriorityQueue(): Creates a PriorityQueue with the default initial capacity (11) that orders its elements according to their natural ordering.
This is a link to doc: https://docs.oracle.com/javase/8/docs/api/java/util/PriorityQueue.html
I need all the fundCode with highest fundValue
If that's the only reason why you want to sort I would recommend not to sort at all. Sorting comes mostly with a complexity of O(n log(n)). Finding the maximum has only a complexity of O(n) and is implemented in a simple iteration over your list:
List<Fund> maxFunds = new ArrayList<Fund>();
int max = 0;
for (Fund fund : funds) {
if (fund.getFundValue() > max) {
maxFunds.clear();
max = fund.getFundValue();
}
if (fund.getFundValue() == max) {
maxFunds.add(fund);
}
}
You can avoid that code by using a third level library like Guava. See: How to get max() element from List in Guava
you can sort a List using Collections.sort.
given your Fund:
List<Fund> sortMe = new ArrayList(...);
Collections.sort(sortMe, new Comparator<Fund>() {
#Override
public int compare(Fund left, Fund right) {
return left.fundValue.compareTo(right.fundValue);
}
});
// sortMe is now sorted
In case of TreeSet either Comparator or Comparable is used to compare and store objects . Equals are not called and that is why it does not recognize the duplicate one
Instead of the TreeSet we can use List and implement the Comparable interface.
public class Fund implements Comparable<Fund> {
String fundCode;
int fundValue;
public Fund(String fundCode, int fundValue) {
super();
this.fundCode = fundCode;
this.fundValue = fundValue;
}
public String getFundCode() {
return fundCode;
}
public void setFundCode(String fundCode) {
this.fundCode = fundCode;
}
public int getFundValue() {
return fundValue;
}
public void setFundValue(int fundValue) {
this.fundValue = fundValue;
}
public int compareTo(Fund compareFund) {
int compare = ((Fund) compareFund).getFundValue();
return compare - this.fundValue;
}
public static void main(String args[]){
List<Fund> funds = new ArrayList<Fund>();
Fund fund1 = new Fund("a",100);
Fund fund2 = new Fund("b",20);
Fund fund3 = new Fund("c",70);
Fund fund4 = new Fund("a",100);
funds.add(fund1);
funds.add(fund2);
funds.add(fund3);
funds.add(fund4);
Collections.sort(funds);
for(Fund fund : funds){
System.out.println("Fund code: " + fund.getFundCode() + " Fund value : " + fund.getFundValue());
}
}
}
Add the elements to the arraylist and then sort the elements using utility Collections.sort,. then implement comparable and write your own compareTo method according to your key.
wont remove duplicates as well, can be sorted also:
List<Integer> list = new ArrayList<>();
Collections.sort(list,new Comparator<Integer>()
{
#Override
public int compare(Objectleft, Object right) {
**your logic**
return '';
}
}
)
;
I found a way to get TreeSet to store duplicate keys.
The problem originated when I wrote some code in python using SortedContainers. I have a spatial index of objects where I want to find all objects between a start/end longitude.
The longitudes could be duplicates but I still need the ability to efficiently add/remove specific objects from the index. Unfortunately I could not find the Java equivalent of the Python SortedKeyList that separates the sort key from the type being stored.
To illustrate this consider that we have a large list of retail purchases and we want to get all purchases where the cost is in a specific range.
// We are using TreeSet as a SortedList
TreeSet _index = new TreeSet<PriceBase>()
// populate the index with the purchases.
// Note that 2 of these have the same cost
_index.add(new Purchase("candy", 1.03));
Purchase _bananas = new Purchase("bananas", 1.45);
_index.add(new Purchase(_bananas);
_index.add(new Purchase("celery", 1.45));
_index.add(new Purchase("chicken", 4.99));
// Range scan. This iterator should return "candy", "bananas", "celery"
NavigableSet<PriceBase> _iterator = _index.subset(
new PriceKey(0.99), new PriceKey(3.99));
// we can also remove specific items from the list and
// it finds the specific object even through the sort
// key is the same
_index.remove(_bananas);
There are 3 classes created for the list
PriceBase: Base class that returns the sort key (the price).
Purchase: subclass that contains transaction data.
PriceKey: subclass used for the range search.
When I initially implemented this with TreeSet it worked except in the case where the prices are the same. The trick is to define the compareTo() so that it is polymorphic:
If we are comparing Purchase to PriceKey then only compare the price.
If we are comparing Purchase to Purchase then compare the price and the name if the prices are the same.
For example here are the compareTo() functions for the PriceBase and Purchase classes.
// in PriceBase
#Override
public int compareTo(PriceBase _other) {
return Double.compare(this.getPrice(), _other.getPrice());
}
// in Purchase
#Override
public int compareTo(PriceBase _other) {
// compare by price
int _compare = super.compareTo(_other);
if(_compare != 0) {
// prices are not equal
return _compare;
}
if(_other instanceof Purchase == false) {
throw new RuntimeException("Right compare must be a Purchase");
}
// compare by item name
Purchase _otherPurchase = (Purchase)_other;
return this.getName().compareTo(_otherChild.getName());
}
This trick allows the TreeSet to sort the purchases by price but still do a real comparison when one needs to be uniquely identified.
In summary I needed an object index to support a range scan where the key is a continuous value like double and add/remove is efficient.
I understand there are many other ways to solve this problem but I wanted to avoid writing my own tree class. My solution seems like a hack and I am surprised that I can't find anything else. if you know of a better way then please comment.
I know that TreeSet stores objects in a sorted manner. But is there a way that i can customize the order?
For example if i have a treeSet:
TreeSet t1 = new TreeSet();
t1.add("c");
t1.add("d");
t1.add("a");
And now if i iterate over it>
Iterator it1 =t1.iterator();
while(it1.hasNext()){
Object o1 = it1.next();
System.out.println(o1);
}
i will always get the order as: a>c>d, however i want it to return the order the same as i added the elements in it i.e c>d>a?
Use LinkedHashSet for it,
TreeSet sorts the element, and for string it sorts based on natural order( this is how its comparator is implemented), If you want to manage the insertion order then you need to user LinkedHashSet
and if you don't need the uniquness (feature of set) then go for List
Since you mention about being bound to use TreeSet, something like this comes to my mind:
Set<String> result = new TreeSet<String>(new Comparator<String>(){
#Override
public int compare(String arg0, String arg1) {
return returnCode(arg0).compareTo(returnCode(arg1));
}
});
where:
private Integer returnCode(String p){
int code = 0;
String id = p.toLowerCase();
if ("a".equalsIgnoreCase(id)) code = 3;
else if ("b".equalsIgnoreCase(id)) code = 2;
else if ("c".equalsIgnoreCase(id)) code = 1;
//etc
return new Integer(code);
}
So basically you are implementing your own comparator which is nothing but assigning certain integer values to the inserted String (which i assume you know already).
NOTE: this solution will not work, incase you do not catch the option in your returnCode() method. I assume you already know the data that is being fed to the TreeSet.
It is much more easy:
TreeSet treeSetObj = new TreeSet(Collections.reverseOrder());
I have got a good answer. It's my first one.
import java.util.*;
class M
{
public static void main(String args[])
{
TreeSet<String> t=new TreeSet<>(new MyComparator());
t.add("c");
t.add("d");
t.add("a");
System.out.println(t);//[c,d,a] not [a,c,d] ....your answer
}
}
class MyComparator implements Comparator
{
public int compare(Object o1,Object o2)
{
String i1=(String)o1;
String i2=(String)o2;
return +1; //Store elements as they come
}
}
What basically +1 does store the elements in the order in which they come.
Always remember
if I write return +1 it means to store the elements in the order in which they come.
if I write return -1 it means to store the elements in reverse order in which they come.