I have two Collection objects, I want to associate each object of these two in a readable way (HashMap, Object created on purpose, you choose).
I was thinking of two loops one nested into the other, but maybe it's a well known problem and has a commonly understandable solution...
What if the number of Collection objects raises above two?
EDIT after Joseph Daigle comment: The items of the Collection objects are all of the same type, they are rooms of hotels found to be bookable under certain conditions.
Collection<Room> roomsFromA = getRoomsFromA();
Collection<Room> roomsFromB = getRoomsFromB();
for(Room roomA : roomsFromA){
for(Room roomB : roomsFromB){
//add roomA and roomB to something, this is not important for what I need
//the important part is how you handle the part before
//especially if Collection objects number grows beyond two
}
}
EDIT 2: I'll try to explain better, sorry for the question being unclear.
Follows an example:
A user requests for a double and a single room.
The hotel has 3 double and 4 single rooms available.
I need to associate every "double room" to every "single room", this is because each Room has its own peculiarity say internet, a more pleasant view, and so on. So i need to give the user all the combinations to let him choose.
This is the simple case, in which only two Collection of Room objects are involved, how do you manage the problem when say both hotel and user can offer / request more Room types?
What you are trying to do here is to get all possible permutations of choosing X from a set of Y. This is a well known problem in discrete mathematics and I think it is just called Combinatorial Mathematics.
To solve your problem you need to create a super collection containing all your Room types. If this is an array or a List you can then use this example to calculate all possible ways of choosing X from the set of Y. The example will give you the indices from the list/array.
Do the collections line up exactly?
HashMap map = new HashMap();
for (int i=0; i<c1.Size(); i++) {
map.put(c1[i], c2[i]);
}
Well, since I don't know if you will need to search for both of them having only one, the HashMap won't work.
I would create a class that receives a Pair.. sort of:
private static class Pair<K, T> {
private K one;
private T two;
public Pair(K one, T two) {
this.one = one;
this.two = two;
}
/**
* #return the one
*/
public K getOne() {
return one;
}
/**
* #return the two
*/
public T getTwo() {
return two;
}
}
And create a List with them.
Your example implies that the return value from "roomsFromB" is a subcollection of the return value of "roomsFromA", so it'd be more natural to model it that way:
class Room {
public Collection<Room> getRoomsFromB { ...
}
which would then let you do :
//Collection rooms
for (Room a: rooms)
{
for(Room b a.getRoomsFromB){ ...
This is assuming that they're modeled hierarchically, of course. If they're not then this is inappropriate, but then the question you're asking, it seems to me, is really how to model the relationship between them, and you haven't yet made that explicit.
You might reconsider whether you need exactly this logic. You're introducing an O(n^2) operation, which can quickly get out of hand. (Technically O(mn), but I'm guessing m and n are roughly the same order.)
Is there another solution to your problem? Perhaps you could create a 'set' which includes all of A and all of B, and then each object in A and B could point to this set, instead?
I assume that:
Each element in collection 1 will
match a single element in
collection 2
The collections have the same
size
The collections can be ordered and
the order matches each element in
both collections
Order both collections (in the same
order) by the property that
identifies each object.
Iterate through both collections with a single loop, build a relation object and add it into a new collection.
See if this helps you:
public static class Room {
private int number;
private String name;
public Room(int number, String name) {
super();
this.number = number;
this.name = name;
}
public int getNumber() {
return number;
}
public String getName() {
return name;
}
}
public static class RoomRelation {
private Room a;
private Room b;
public RoomRelation(Room a, Room b) {
super();
this.a = a;
this.b = b;
}
public Room getA() {
return a;
}
public Room getB() {
return b;
}
#Override
public String toString() {
return a.getName() + "(" + a.getNumber() + ") " + b.getName() + "(" + b.getNumber() + ")";
}
}
public static void main(String[] args) {
List<Room> roomsFromA = new ArrayList<Room>();
List<Room> roomsFromB = new ArrayList<Room>();
roomsFromA.add(new Room(1,"Room A"));
roomsFromA.add(new Room(2,"Room A"));
roomsFromB.add(new Room(1,"Room B"));
roomsFromB.add(new Room(2,"Room B"));
Comparator<Room> c = new Comparator<Room>() {
#Override
public int compare(Room o1, Room o2) {
return o1.getNumber() - o2.getNumber();
} };
Collections.sort(roomsFromA, c);
Collections.sort(roomsFromB, c);
List<RoomRelation> relations = new ArrayList<RoomRelation>();
for (int i = 0; i < roomsFromA.size(); i++) {
relations.add(new RoomRelation(roomsFromA.get(i), roomsFromB.get(i)));
}
for (RoomRelation roomRelation : relations) {
System.out.println(roomRelation);
}
}
Your question is quite unclear. As I understand you want to list all combinations of rooms, minus duplicates. Here us some code to build up a 2d array of all the room combinations. For more kinds of room, put in another nested loop.
Collection<Room> roomsFromA = getRoomsFromA();
Collection<Room> roomsFromB = getRoomsFromB();
Room[][] combinations = new Room[roomsFromA .size()][roomsFromB .size()];
int a = 0;
int b = 0;
for(Room roomA : roomsFromA){
for(Room roomB : roomsFromB){
combinations [a][b] = [roomA][roomB]; //Build up array
b++;
}
a++;
}
return combinations;
It is a common problem. It's called a Cartesian product. If you have two collections like in your case, I would not hesitate to have two nested loops. Otherwise, see this question.
Related
This is mainly a question intended for me to learn about various performant ways of filtering and assigning objects to Lists.
Assume
public class A implements Comparable<A> {
private String id;
public String getId() {
return id;
}
public void setId(String id) {
this.id = id;
}
#Override
public int compareTo(A o) {
return o.getId().compareTo(this.getId());
}
}
public class B implements Comparable<B>{
private String id;
private List<A> aList = new ArrayList<>();
public String getId() {
return id;
}
public void setId(String id) {
this.id = id;
}
public void addA(A a)
{
aList.add(a);
}
#Override
public int compareTo(B o) {
return o.getId().compareTo(this.getId());
}
}
public class Main {
public static void main(String[] args) {
SortedSet<A> aSet = new TreeSet<>();
SortedSet<B> bSet = new TreeSet<>();
for(int i=0;i<100000;i++)
{
UUID uuid = UUID.randomUUID();
String uuidAsString = uuid.toString();
A a1 = new A();
a1.setId(uuidAsString);
aSet.add(a1);
A a2 = new A();
a2.setId(uuidAsString);
aSet.add(a2);
B b = new B();
b.setId(uuidAsString);
bSet.add(b);
}
//this is where the performance part comes in
//scenario: For each B I want to find A whose Id matches B's Id, and assign it to B
//assume B can have 1-5 instances of A (even though for this example I only initialized 2)
bSet.parallelStream().forEach(b -> {
aSet.parallelStream().filter(a -> {
return b.getId().equals(a.getId());
}).forEach(a -> {
b.addA(a);
});
});
}
}
The solution I came up with was to combine parallelstreams and filters to find the matching IDs between the two types of objects and then to loops through the filtered results to add the instances of A to B.
I used TreeSets here because I thought the ordered IDs might help speed things up, same reason I used parallelStreams.
This is mostly abstracted out from a scenario from a project I am doing at the office which I cant post here. The classes in the actual project have a lot more variables, and in the worst case - have sublists of lists (I resolved that using flatMaps in streams).
However my inexperienced gut tells me there is a more performant way to solve this problem.
I am primarily looking for practical ways to speed this up.
Some ways I thought of speeding this up:
Switch the lists and sets to Eclipse Collections
Assuming the starting point of these classes are CSV files -> Maybe write an apache spark application that will map these(I assumed that Spark could have some internal clever way of doing this faster than Streams).
I dunno......write them all to sql tables....map them via foreign keys and then query them again?
Speed is the name of the game, solutions using vanilla java, different librarys (like Eclipse Collections), or entire engines like Spark are acceptable
Assume the minimum list size is atleast 50,000
Bonus complexity: You can add another class 'C', with multiple instances of 'B' in it. My inexperienced self can only think of writing another similar streaming operation as A->B and run it after the first stream is done. Is there a way to combine both A->B and B->C operations together so that they happen at once. That will definitely speed things up.
Sorry about my inexperienced self and sorry again if this is a duplicate too
In your code, you use b.addA(a); where b is an instance of B while B doesn't have a method addA(A). Is B supposed to keep a list of A's?
However, the answer to your question is hashing. You are looking for a multimap, to be specific. As a quick fix you can use a TreeMap that stores a List of A's by their id:
public static void main(String[] args) {
TreeMap<String, ArrayList<A>> aSet = new TreeMap<>();
ArrayList<B> bSet = new ArrayList<>();
for (int i = 0; i < 100000; i++) {
UUID uuid = UUID.randomUUID();
String uuidAsString = uuid.toString();
A a1 = new A();
a1.setId(uuidAsString);
ArrayList<A> list = aSet.get(a1.getId());
if (list == null) {
list = new ArrayList<>();
aSet.put(a1.getId(), list);
}
list.add(a1);
A a2 = new A();
a2.setId(uuidAsString);
list = aSet.get(a2.getId());
if (list == null) {
list = new ArrayList<>();
aSet.put(a2.getId(), list);
}
list.add(a2);
B b = new B();
b.setId(uuidAsString);
bSet.add(b);
}
for (B b : bSet) {
System.out.println(aSet.get(b.getId()));
}
}
Please note that this isn't a good implementation and instead you should write your own multimap or use the one in guava
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I'm looking for some tip on how to do a percentage thing for my game I want all flowers in a range of 1-98 and white/black flowers 99-100 to make it more rarerity thanks for the help :)
public enum FlowerSuit {
WHITE_FLOWERS("white", ":white:", "470419377456414720", 1),
YELLOW_FLOWERS("yellow", ":yellow:", "470419561267855360", 1 ),
RED_FLOWERS("red", ":red:", "470419583250202644", 1),
RAINBOW_FLOWERS("rainbow", ":rainbow:", "470419602841665536", 1),
PASTEL_FLOWERS("pastel", ":pastel:", "470419629450199040", 1),
ORANGE_FLOWERS("orange", ":orange:", "470419647900942366", 1),
BLUE_FLOWERS("blue", ":blue:", "470419688753594368", 1),
BLACK_FLOWERS("black", ":black:", "470419706751352842", 1);
private final String displayName;
private final String emoticon;
private int value;
private final String id;
FlowerSuit(String displayName, String emoticon, String id, int value ) {
this.displayName = displayName;
this.emoticon = emoticon;
this.value = value;
this.id = id;
}
public String getDisplayName() {
return displayName;
}
public String getEmoticon() {
return emoticon;
}
public String getId() {
return id;
}
public int getValue() {
// TODO Auto-generated method stub
return value;
}
}
This is how I'd do it, but it can probably be improved, for starters by using Java 8 streams etc.
public enum FlowerSuit {
WHITE_FLOWERS("white", ":white:", "470419377456414720", 1,3),
YELLOW_FLOWERS("yellow", ":yellow:", "470419561267855360", 1,2),
RED_FLOWERS("red", ":red:", "470419583250202644", 1,2),
RAINBOW_FLOWERS("rainbow", ":rainbow:", "470419602841665536", 1,2),
PASTEL_FLOWERS("pastel", ":pastel:", "470419629450199040", 1,2),
ORANGE_FLOWERS("orange", ":orange:", "470419647900942366", 1,2),
BLUE_FLOWERS("blue", ":blue:", "470419688753594368", 1,2),
BLACK_FLOWERS("black", ":black:", "470419706751352842", 1,1);
private static Random random = new Random();
private final String displayName;
private final String emoticon;
private int value;
private final String id;
private final int freq;
private FlowerSuit(String displayName, String emoticon, String id, int value, int freq ) {
this.displayName = displayName;
this.emoticon = emoticon;
this.value = value;
this.id = id;
this.freq = freq;
}
public String getDisplayName() {return displayName;}
public String getEmoticon() {return emoticon;}
public String getId() {return id;}
public int getValue() {return value;}
/**
* Choose a flower
* white has a 3 in 16 (about a 5:1) chance of being picked
* Black has a 1 in 16 chance, everything else 2/16
* #return
*/
public static FlowerSuit pick() {
//first sum all the chances (currently it's 16)
int sum = 0;
for (FlowerSuit f:FlowerSuit.values()) sum+= f.freq;
//now choose a random number
int r = FlowerSuit.random.nextInt(sum) + 1;
//now find out which flower to pick
sum = 0;
for (FlowerSuit f:FlowerSuit.values()) {
sum += f.freq;
if (r<=sum) return f;
}
//code will never get here
return FlowerSuit.WHITE_FLOWERS;
}
public static void main(final String[] args) throws Exception {
//Test it
Map<FlowerSuit,Integer>count = new HashMap<FlowerSuit,Integer>();
for (int a=0;a<1000000;a++) {
FlowerSuit f = FlowerSuit.pick();
Integer i = (count.get(f)!=null)?count.get(f):new Integer(0);
i = new Integer(i+1);
count.put(f,i);
}
int sum = 0;
for (Map.Entry<FlowerSuit,Integer>e:count.entrySet()) sum+=e.getValue();
float f = Float.valueOf(sum);
for (Map.Entry<FlowerSuit,Integer>e:count.entrySet()) {
System.out.println(e.getKey() + " was chosen " + ((e.getValue() / f) * 100f) + "% of the time");
}
}
}
gives
BLUE_FLOWERS was chosen 12.4986% of the time
PASTEL_FLOWERS was chosen 12.4707% of the time
WHITE_FLOWERS was chosen 18.7365% of the time
BLACK_FLOWERS was chosen 6.2632003% of the time
ORANGE_FLOWERS was chosen 12.4986% of the time
RED_FLOWERS was chosen 12.5241995% of the time
YELLOW_FLOWERS was chosen 12.501401% of the time
RAINBOW_FLOWERS was chosen 12.5068% of the time
You can use a TreeMap to map all of the integers from 0 to 99 to a particular FlowerSuit. Take advantage of the floorEntry method to choose a FlowerSuit for each number. It might look something like this.
public class FlowerChooser {
private static final NavigableMap<Integer, FlowerSuit> FLOWER_SUITS;
private static final Random RANDOMS = new Random();
public FlowerChooser() {
FLOWER_SUITS = new TreeMap<>();
FLOWER_SUITS.put(0, FlowerSuit.RED_FLOWERS);
FLOWER_SUITS.put(14, FlowerSuit.ORANGE_FLOWERS);
FLOWER_SUITS.put(28, FlowerSuit.YELLOW_FLOWERS);
FLOWER_SUITS.put(42, FlowerSuit.GREEN_FLOWERS);
FLOWER_SUITS.put(56, FlowerSuit.BLUE_FLOWERS);
FLOWER_SUITS.put(70, FlowerSuit.INDIGO_FLOWERS);
FLOWER_SUITS.put(84, FlowerSuit.VIOLET_FLOWERS);
FLOWER_SUITS.put(98, FlowerSuit.WHITE_FLOWERS);
FLOWER_SUITS.put(99, FlowerSuit.BLACK_FLOWERS);
}
public FlowerSuit randomFlowerSuit() {
int index = RANDOMS.nextInt(100);
return FLOWER_SUITS.floorEntry(index).getValue();
}
}
Create just one object of this class, then whenever you want a FlowerSuit, call the randomFlowerSuit method.
The randomFlowerSuit method picks a random number from 0 to 99, then finds an appropriate entry in the map. The floorEntry method chooses an entry whose key is less than or equal to the chosen number. This means that numbers from 0 to 13 get mapped to red, 14 to 27 get mapped to orange, and so on. The only number that gets mapped to white is 98, and the only number that gets mapped to black is 99.
No matter what solution you implement, you want to include a frequency measure in your enum. As an example, you can do something like this:
public enum FlowerSuit {
WHITE_FLOWERS("white", ":white:", "470419377456414720", 1, 1),
YELLOW_FLOWERS("yellow", ":yellow:", "470419561267855360", 1, 20),
// More declarations here
// Add this variable
private final int frequency;
// Do just as you did before in the constructor, but with the frequency
FlowerSuit(String displayName, String emoticon, String id, int value, int frequency){
this.frequency = frequency;
// More assignments here
}
public int getFrequency(){
return frequency;
}
// More getters here
}
This addition is critical, and no matter what method you use to weight flower selection, you will want this addition to your FlowerSuit enum.
Now, we can explore a few different ways to perform this selection.
Note 1: I use ThreadLocalRandom for random numbers in a range, which is from java.util.concurrent.ThreadLocalRandom.
Note 2: For each of these, make a single instance of FlowerPicker, and use the pickFlower() method to pick the next flower. This avoid running costly setup code over and over.
Method 1: Bag of Flowers
This method is probably the easiest to implement. It entails creating a list of enums where each is represented frequency times, and then selecting a random entry from this list. It is similar to throwing a bunch of flowers in a bag, shaking it, and then reaching your hand in and grabbing the first flower you touch. Here's the implementation:
public class FlowerPicker(){
private ArrayList<FlowerSuit> bag;
public FlowerPicker(){
// Get all possible FlowerSuits
FlowerSuit[] options = FlowerSuit.values();
// You can use an array here or an array list with a defined length if you know the total of the frequencies
bag = new ArrayList<FlowerSuit>();
// Add each flower from options frequency times
for (FlowerSuit flower : options)
for (int i=0; i<flower.getFrequency(); i++)
bag.add(flower);
}
public FlowerBag pickFlower(){
// Now, select a random flower from this list
int randomIndex = ThreadLocalRandom.current().nextInt(0, bag.size());
return bag.get(randomIndex);
}
}
This method has the advantage of being simple enough to understand very easily. However, it can be inefficient if your frequencies are extremely specific (like if you want a rainbow flower to be returned 499,999,999 times out of 1,000,000,000). Let's move on to the next method.
Note 1: You could make this better by reducing the fractions representing the frequency of being chosen, but I'll leave this to you.
Note 2: You could also make this slightly better by storing identification numbers, not FlowerSuit objects in the bag list.
Method 2: Navigable Map
This method is a little bit more difficult. It uses a [NavigableMap][1], which is an implementation of [TreeMap][2]. This method is fairly similar to the Bag of Flowers method, but it is a little bit more efficient. Put simply, it uses the TreeMap to give each FlowerSuit a range of numbers that can be selected to return that FlowerSuit. Here's a full example:
public class FlowerPicker(){
private NavigableMap<Double, FlowerSuit> map;
public FlowerPicker(){
// Get all possible FlowerSuits
FlowerSuit[] options = FlowerSuit.values();
map = new TreeMap<Double, FlowerSuit>();
int runningTotal = 0;
// Add each flower with the proper range
for (FlowerSuit flower : options){
runningTotal += flower.getFrequency();
map.put(runningTotal, flower);
}
}
public FlowerBag pickFlower(){
// Now, select a random number and get the flower with this number in its range
int randomRange = ThreadLocalRandom.current().nextInt(0, bag.size());
return map.higherEntry(randomRange).getValue();
}
}
This is a solid method, and it scales well for very specific frequencies. If you have a bunch of different types of flowers, it will be slightly worse, but this method is still a good option at large scales. There's one more option though.
Method 3: Enumerated Distribution
This method is really nice because you barely have to do anything. However, it uses [EnumeratedDistribution][3] from Apache Commons. Enumerated Distribution requires a list of pairs of objects and weights. Anyway, lets jump into it:
public class FlowerPicker(){
private EnumeratedDistribution distribution;
public FlowerPicker(){
// Get all possible FlowerSuits
FlowerSuit[] options = FlowerSuit.values();
List<Pair<FlowerSuit, Double>> weights = new List<Pair<FlowerSuit, Double>>();
// Add each flower and weight to the list
for (FlowerSuit flower : options){
weights.add(new Pair(flower, flower.getFrequency()));
// Construct the actual distribution
distribution = new EnumeratedDistribution(weights);
}
public FlowerBag pickFlower(){
// Now, sample the distribution
return distribution.sample();
}
}
This is my favorite method, simply because so much of it is done for you. Many problems like this have been solved, so why not use solutions that always exist? However, there is some value to writing the solution yourself.
In conclusion, each of these methods are perfectly fine to use at your scale, but I would recommend the second or third method.
I am looking to implement a sort feature for my address book application.
I want to sort an ArrayList<Contact> contactArray. Contact is a class which contains four fields: name, home number, mobile number and address. I want to sort on name.
How can I write a custom sort function to do this?
Here's a tutorial about ordering objects:
The Java Tutorials - Collections - Object Ordering
Although I will give some examples, I would recommend to read it anyway.
There are various way to sort an ArrayList. If you want to define a natural (default) ordering, then you need to let Contact implement Comparable. Assuming that you want to sort by default on name, then do (nullchecks omitted for simplicity):
public class Contact implements Comparable<Contact> {
private String name;
private String phone;
private Address address;
#Override
public int compareTo(Contact other) {
return name.compareTo(other.name);
}
// Add/generate getters/setters and other boilerplate.
}
so that you can just do
List<Contact> contacts = new ArrayList<Contact>();
// Fill it.
Collections.sort(contacts);
If you want to define an external controllable ordering (which overrides the natural ordering), then you need to create a Comparator:
List<Contact> contacts = new ArrayList<Contact>();
// Fill it.
// Now sort by address instead of name (default).
Collections.sort(contacts, new Comparator<Contact>() {
public int compare(Contact one, Contact other) {
return one.getAddress().compareTo(other.getAddress());
}
});
You can even define the Comparators in the Contact itself so that you can reuse them instead of recreating them everytime:
public class Contact {
private String name;
private String phone;
private Address address;
// ...
public static Comparator<Contact> COMPARE_BY_PHONE = new Comparator<Contact>() {
public int compare(Contact one, Contact other) {
return one.phone.compareTo(other.phone);
}
};
public static Comparator<Contact> COMPARE_BY_ADDRESS = new Comparator<Contact>() {
public int compare(Contact one, Contact other) {
return one.address.compareTo(other.address);
}
};
}
which can be used as follows:
List<Contact> contacts = new ArrayList<Contact>();
// Fill it.
// Sort by address.
Collections.sort(contacts, Contact.COMPARE_BY_ADDRESS);
// Sort later by phone.
Collections.sort(contacts, Contact.COMPARE_BY_PHONE);
And to cream the top off, you could consider to use a generic javabean comparator:
public class BeanComparator implements Comparator<Object> {
private String getter;
public BeanComparator(String field) {
this.getter = "get" + field.substring(0, 1).toUpperCase() + field.substring(1);
}
public int compare(Object o1, Object o2) {
try {
if (o1 != null && o2 != null) {
o1 = o1.getClass().getMethod(getter, new Class[0]).invoke(o1, new Object[0]);
o2 = o2.getClass().getMethod(getter, new Class[0]).invoke(o2, new Object[0]);
}
} catch (Exception e) {
// If this exception occurs, then it is usually a fault of the developer.
throw new RuntimeException("Cannot compare " + o1 + " with " + o2 + " on " + getter, e);
}
return (o1 == null) ? -1 : ((o2 == null) ? 1 : ((Comparable<Object>) o1).compareTo(o2));
}
}
which you can use as follows:
// Sort on "phone" field of the Contact bean.
Collections.sort(contacts, new BeanComparator("phone"));
(as you see in the code, possibly null fields are already covered to avoid NPE's during sort)
In addition to what was already posted by BalusC it may be worth pointing that since Java 8 we can shorten our code and write it like:
Collection.sort(yourList, Comparator.comparing(YourClass::getSomeComparableField));
or since List now have sort method also like
yourList.sort(Comparator.comparing(YourClass::getSomeComparableField));
Explanation:
Since Java 8, functional interfaces (interfaces with only one abstract method - they can have more default or static methods) can be easily implemented using:
lambdas arguments -> body
or method references source::method.
Since Comparator<T> has only one abstract method int compare(T o1, T o2) it is functional interface.
So instead of (example from #BalusC answer)
Collections.sort(contacts, new Comparator<Contact>() {
public int compare(Contact one, Contact other) {
return one.getAddress().compareTo(other.getAddress());
}
});
we can reduce this code to:
Collections.sort(contacts, (Contact one, Contact other) -> {
return one.getAddress().compareTo(other.getAddress());
});
We can simplify this (or any) lambda by skipping
argument types (Java will infer them based on method signature)
or {return ... }
So instead of
(Contact one, Contact other) -> {
return one.getAddress().compareTo(other.getAddress();
}
we can write
(one, other) -> one.getAddress().compareTo(other.getAddress())
Also now Comparator has static methods like comparing(FunctionToComparableValue) or comparing(FunctionToValue, ValueComparator) which we could use to easily create Comparators which should compare some specific values from objects.
In other words we can rewrite above code as
Collections.sort(contacts, Comparator.comparing(Contact::getAddress));
//assuming that Address implements Comparable (provides default order).
This page tells you all you need to know about sorting collections, such as ArrayList.
Basically you need to
make your Contact class implement the Comparable interface by
creating a method public int compareTo(Contact anotherContact) within it.
Once you do this, you can just call Collections.sort(myContactList);,
where myContactList is ArrayList<Contact> (or any other collection of Contact).
There's another way as well, involving creating a Comparator class, and you can read about that from the linked page as well.
Example:
public class Contact implements Comparable<Contact> {
....
//return -1 for less than, 0 for equals, and 1 for more than
public compareTo(Contact anotherContact) {
int result = 0;
result = getName().compareTo(anotherContact.getName());
if (result != 0)
{
return result;
}
result = getNunmber().compareTo(anotherContact.getNumber());
if (result != 0)
{
return result;
}
...
}
}
BalusC and bguiz have already given very complete answers on how to use Java's built-in Comparators.
I just want to add that google-collections has an Ordering class which is more "powerful" than the standard Comparators.
It might be worth checking out. You can do cool things such as compounding Orderings, reversing them, ordering depending on a function's result for your objects...
Here is a blog post that mentions some of its benefits.
You need make your Contact classes implement Comparable, and then implement the compareTo(Contact) method. That way, the Collections.sort will be able to sort them for you. Per the page I linked to, compareTo 'returns a negative integer, zero, or a positive integer as this object is less than, equal to, or greater than the specified object.'
For example, if you wanted to sort by name (A to Z), your class would look like this:
public class Contact implements Comparable<Contact> {
private String name;
// all the other attributes and methods
public compareTo(Contact other) {
return this.name.compareTo(other.name);
}
}
By using lambdaj you can sort a collection of your contacts (for example by their name) as it follows
sort(contacts, on(Contact.class).getName());
or by their address:
sort(contacts, on(Contacts.class).getAddress());
and so on. More in general, it offers a DSL to access and manipulate your collections in many ways, like filtering or grouping your contacts based on some conditions, aggregate some of their property values, etc.
Ok, I know this was answered a long time ago... but, here's some new info:
Say the Contact class in question already has a defined natural ordering via implementing Comparable, but you want to override that ordering, say by name. Here's the modern way to do it:
List<Contact> contacts = ...;
contacts.sort(Comparator.comparing(Contact::getName).reversed().thenComparing(Comparator.naturalOrder());
This way it will sort by name first (in reverse order), and then for name collisions it will fall back to the 'natural' ordering implemented by the Contact class itself.
The Collections.sort is a good sort implementation. If you don't have The comparable implemented for Contact, you will need to pass in a Comparator implementation
Of note:
The sorting algorithm is a modified mergesort (in which the merge is omitted if the highest element in the low sublist is less than the lowest element in the high sublist). This algorithm offers guaranteed n log(n) performance. The specified list must be modifiable, but need not be resizable. This implementation dumps the specified list into an array, sorts the array, and iterates over the list resetting each element from the corresponding position in the array. This avoids the n2 log(n) performance that would result from attempting to sort a linked list in place.
The merge sort is probably better than most search algorithm you can do.
I did it by the following way.
number and name are two arraylist. I have to sort name .If any change happen to name arralist order then the number arraylist also change its order.
public void sortval(){
String tempname="",tempnum="";
if (name.size()>1) // check if the number of orders is larger than 1
{
for (int x=0; x<name.size(); x++) // bubble sort outer loop
{
for (int i=0; i < name.size()-x-1; i++) {
if (name.get(i).compareTo(name.get(i+1)) > 0)
{
tempname = name.get(i);
tempnum=number.get(i);
name.set(i,name.get(i+1) );
name.set(i+1, tempname);
number.set(i,number.get(i+1) );
number.set(i+1, tempnum);
}
}
}
}
}
use this method:
private ArrayList<myClass> sortList(ArrayList<myClass> list) {
if (list != null && list.size() > 1) {
Collections.sort(list, new Comparator<myClass>() {
public int compare(myClass o1, myClass o2) {
if (o1.getsortnumber() == o2.getsortnumber()) return 0;
return o1.getsortnumber() < o2.getsortnumber() ? 1 : -1;
}
});
}
return list;
}
`
and use: mySortedlist = sortList(myList);
No need to implement comparator in your class.
If you want inverse order swap 1 and -1
With java 8 feature
List<Contact> contact = contactArray.stream().sorted((c1, c2) -> ((c1.getName().compareTo(c2.getName())))).collect(Collectors.toList());
You shoud use the Arrays.sort function. The containing classes should implement Comparable.
i am trying to sort an array of strings which are terms of a polynomial. every position is 1 term of the polynomial as a string, and signed approapriately, however i want to sort them in order by the power.
eg
+3x^5
+5
-8x
-4x^2
how i have approached this is by creating a second array storing just the power, and i want to sort them both based off this array. ie
for (int i=0; i<sortArray.length; i++) {
if (sortArray[i].indexOf("^")!= -1)
sortArrayDegree[i] = Integer.parseInt((sortArray[i].
substring(sortArray[i].indexOf("^") + 1, sortArray[i].length())));
else if (sortArray[i].indexOf("x")!= -1)
sortArrayDegree[i]=1;
else
sortArrayDegree[i]=0;
}
however i am not sure how to link the two, so any changes to the second happen to the first
currently that means the second array looks like this
5
0
1
2
i thought i could make a new array and store this as the second column(clash of data types), but that still leaves the sorting problem
I'm not sure that the way you want achieve this is the wisest way, but this is how you could do it:
Create a class of both the power and the number of the polynomial member. Make that class Comparable, then put it in one array and the sort method will use the comparable method you have overridden from the Comparable interface.
public class PolynomialMember implements Comparable<PolynomialMember> {
public int power; // public for brevity, but should be private with getters and setters
public String number; // public for brevity, but should be private with getters and setters
public PolynomialMember(String number, int power) {
this.number = number;
this.power = power;
}
#Override
public int compareTo(PolynomialMember o) {
return Integer.compare(this.power, o.power);
}
// optional: override for pretty printing
#Override
public String toString() {
if(!number.equals("0")) {
if(number.charAt(0) == '-') {
return number + "x^" + power;
} else {
return "+" + number + "x^" + power;
}
} else {
return "";
}
}
}
this way you don't need two arrays, and you certainly shouldn't "link" two arrays.
You can use this class like this:
public static void main(String[] args) {
List<PolynomialMember> polynom = new ArrayList<PolynomialMember>();
polynom.add(new PolynomialMember("-5", 3));
polynom.add(new PolynomialMember("7", 1));
polynom.add(new PolynomialMember("4", 0));
polynom.add(new PolynomialMember("8", 2));
for(PolynomialMember pm : polynom) {
System.out.print(pm + " ");
// prints: -5x^3 +7x^1 +4x^0 +8x^2
}
System.out.println();
Collections.sort(polynom); //this is where the magic happens.
for(PolynomialMember pm : polynom) {
System.out.print(pm + " ");
// prints: +4x^0 +7x^1 +8x^2 -5x^3
}
}
If I understand correctly, which I'm really not sure, you want to bind the data of 2 arrays containing value types\immutables. The easiest way i know to bind data from 2 arrays is to create a class containing both of them as private members and exposing public methods to control them. in these methods you could implement the logic that defines the relationship between them.
First of all sorry if my English bad, its not my first language..
I'm working on and android app project, that needed to sort ArrayList of an object..so I made this method to deal with that...
Lets say that I have an object of Restaurant that will contain this data:
private String name;
private float distance ;
And I sort it using the value of the variable distance from lowest to highest:
public void sort(RArrayList<RestaurantData> datas) {
RestaurantData tmp = new RestaurantData();
int swapped;
boolean b = true;
while (b) {
swapped = 0;
for (int i = 0; i < datas.size()-1; i++) {
if (datas.get(i).getDistance() > datas.get(i+1).getDistance()) {
tmp = datas.get(i);
datas.set(i, datas.get(i+1));
datas.set(i+1, tmp);
swapped = 1;
System.err.println("Swapped happening");
}
}
if (swapped == 0) {
System.err.println("Swapped end");
break;
}
}
But when i try the program..the result of an ArrayList is still random, is there any problem with my logic to sort the ArrayList of an object..
Please Help...Thankyou..
Why not use the Collections.sort method?
Here's how you could do it in your project:
public void sort(RArrayList<RestaurantData> datas) {
Collections.sort(datas, new Comparator<RestaurantData>() {
#Override
public int compare(RestaurantData lhs, RestaurantData rhs) {
return lhs.getDistance() - rhs.getDistance();
}
});
}
The above solution is a bit "destructive" in the sense that it changes the order of the elements in the original array - datas. If that's fine for you go ahead and use it. Personally I prefer things less destructive and if you have the memory to spare (meaning your array is small) you could consider this solution which copies the array before sorting. It also assumes your RArrayList is an implementation of ArrayList or backed up by it:
public List<RestaurantData> sort(RArrayList<RestaurantData> datas) {
// Create a list with enough capacity for all elements
List<RestaurantData> newList = new RArrayList<RestaurantData>(datas.size());
Collections.copy(newList, datas);
Collections.sort(newList, new Comparator<RestaurantData>() {
#Override
public int compare(RestaurantData lhs, RestaurantData rhs) {
return lhs.getDistance() - rhs.getDistance();
}
});
return newList;
}
Another thing to consider is also to create a single instance of the Comparator used in the method, since this implementation will create one instance per call. Not sure if it's worth it though, because it will also be destroyed quite soon since the scope is local.
Here's the documentation for the Collections api
One last thing, the comparator simply needs to return a value less than 0 if the elements are in the right order, bigger than 0 if they're in the wrong order or 0 if they're the same. Therefore it seems to be that it's enough to simply subtract the distances of each restaurant. However, if this isn't the case, please implement the comparator suiting your needs.