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How to order cities name by their letters, not plate code in the list? [duplicate]

I have simple class
public class ActiveAlarm {
public long timeStarted;
public long timeEnded;
private String name = "";
private String description = "";
private String event;
private boolean live = false;
}
and List<ActiveAlarm> con. How to sort in ascending order by timeStarted, then by timeEnded? Can anybody help? I know in C++ with generic algorithm and overload operator <, but I am new to Java.

Using Comparator
For Example:
class Score {
private String name;
private List<Integer> scores;
// +accessor methods
}
Collections.sort(scores, new Comparator<Score>() {
public int compare(Score o1, Score o2) {
// compare two instance of `Score` and return `int` as result.
return o2.getScores().get(0).compareTo(o1.getScores().get(0));
}
});
With Java 8 onwards, you can simply use lambda expression to represent Comparator instance.
Collections.sort(scores, (s1, s2) -> { /* compute and return int */ });

Either make ActiveAlarm implement Comparable<ActiveAlarm> or implement Comparator<ActiveAlarm> in a separate class. Then call:
Collections.sort(list);
or
Collections.sort(list, comparator);
In general, it's a good idea to implement Comparable<T> if there's a single "natural" sort order... otherwise (if you happen to want to sort in a particular order, but might equally easily want a different one) it's better to implement Comparator<T>. This particular situation could go either way, to be honest... but I'd probably stick with the more flexible Comparator<T> option.
EDIT: Sample implementation:
public class AlarmByTimesComparer implements Comparator<ActiveAlarm> {
#Override
public int compare(ActiveAlarm x, ActiveAlarm y) {
// TODO: Handle null x or y values
int startComparison = compare(x.timeStarted, y.timeStarted);
return startComparison != 0 ? startComparison
: compare(x.timeEnded, y.timeEnded);
}
// I don't know why this isn't in Long...
private static int compare(long a, long b) {
return a < b ? -1
: a > b ? 1
: 0;
}
}

JAVA 8 and Above Answer (Using Lambda Expressions)
In Java 8, Lambda expressions were introduced to make this even easier! Instead of creating a Comparator() object with all of it's scaffolding, you can simplify it as follows: (Using your object as an example)
Collections.sort(list, (ActiveAlarm a1, ActiveAlarm a2) -> a1.timeStarted-a2.timeStarted);
or even shorter:
Collections.sort(list, Comparator.comparingInt(ActiveAlarm ::getterMethod));
That one statement is equivalent to the following:
Collections.sort(list, new Comparator<ActiveAlarm>() {
#Override
public int compare(ActiveAlarm a1, ActiveAlarm a2) {
return a1.timeStarted - a2.timeStarted;
}
});
Think of Lambda expressions as only requiring you to put in the relevant parts of the code: the method signature and what gets returned.
Another part of your question was how to compare against multiple fields. To do that with Lambda expressions, you can use the .thenComparing() function to effectively combine two comparisons into one:
Collections.sort(list, (ActiveAlarm a1, ActiveAlarm a2) -> a1.timeStarted-a2.timeStarted
.thenComparing ((ActiveAlarm a1, ActiveAlarm a2) -> a1.timeEnded-a2.timeEnded)
);
The above code will sort the list first by timeStarted, and then by timeEnded (for those records that have the same timeStarted).
One last note: It is easy to compare 'long' or 'int' primitives, you can just subtract one from the other. If you are comparing objects ('Long' or 'String'), I suggest you use their built-in comparison. Example:
Collections.sort(list, (ActiveAlarm a1, ActiveAlarm a2) -> a1.name.compareTo(a2.name) );
EDIT: Thanks to Lukas Eder for pointing me to .thenComparing() function.

We can sort the list in one of two ways:
1. Using Comparator : When required to use the sort logic in multiple places
If you want to use the sorting logic in a single place, then you can write an anonymous inner class as follows, or else extract the comparator and use it in multiple places
Collections.sort(arrayList, new Comparator<ActiveAlarm>() {
public int compare(ActiveAlarm o1, ActiveAlarm o2) {
//Sorts by 'TimeStarted' property
return o1.getTimeStarted()<o2.getTimeStarted()?-1:o1.getTimeStarted()>o2.getTimeStarted()?1:doSecodaryOrderSort(o1,o2);
}
//If 'TimeStarted' property is equal sorts by 'TimeEnded' property
public int doSecodaryOrderSort(ActiveAlarm o1,ActiveAlarm o2) {
return o1.getTimeEnded()<o2.getTimeEnded()?-1:o1.getTimeEnded()>o2.getTimeEnded()?1:0;
}
});
We can have null check for the properties, if we could have used 'Long' instead of 'long'.
2. Using Comparable(natural ordering): If sort algorithm always stick to one property:
write a class that implements 'Comparable' and override 'compareTo' method as defined below
class ActiveAlarm implements Comparable<ActiveAlarm>{
public long timeStarted;
public long timeEnded;
private String name = "";
private String description = "";
private String event;
private boolean live = false;
public ActiveAlarm(long timeStarted,long timeEnded) {
this.timeStarted=timeStarted;
this.timeEnded=timeEnded;
}
public long getTimeStarted() {
return timeStarted;
}
public long getTimeEnded() {
return timeEnded;
}
public int compareTo(ActiveAlarm o) {
return timeStarted<o.getTimeStarted()?-1:timeStarted>o.getTimeStarted()?1:doSecodaryOrderSort(o);
}
public int doSecodaryOrderSort(ActiveAlarm o) {
return timeEnded<o.getTimeEnded()?-1:timeEnded>o.getTimeEnded()?1:0;
}
}
call sort method to sort based on natural ordering
Collections.sort(list);

In java8+ this can be written in single line as follows:
collectionObjec.sort(comparator_lamda) or comparator.comparing(CollectionType::getterOfProperty)
code:
ListOfActiveAlarmObj.sort((a,b->a.getTimeStarted().compareTo(b.getTimeStarted())))
or
ListOfActiveAlarmObj.sort(Comparator.comparing(ActiveAlarm::getTimeStarted))

public class ActiveAlarm implements Comparable<ActiveAlarm> {
public long timeStarted;
public long timeEnded;
private String name = "";
private String description = "";
private String event;
private boolean live = false;
public int compareTo(ActiveAlarm a) {
if ( this.timeStarted > a.timeStarted )
return 1;
else if ( this.timeStarted < a.timeStarted )
return -1;
else {
if ( this.timeEnded > a.timeEnded )
return 1;
else
return -1;
}
}
That should give you a rough idea. Once that's done, you can call Collections.sort() on the list.

Since Java8 this can be done even cleaner using a combination of Comparator and Lambda expressions
For Example:
class Student{
private String name;
private List<Score> scores;
// +accessor methods
}
class Score {
private int grade;
// +accessor methods
}
Collections.sort(student.getScores(), Comparator.comparing(Score::getGrade);

Java-8 solution using Stream API:
A. When timeStarted and timeEnded are public (as mentioned in the requirement) and therefore do not (need to) have public getter methods:
List<ActiveAlarm> sorted =
list.stream()
.sorted(Comparator.comparingLong((ActiveAlarm alarm) -> alarm.timeStarted)
.thenComparingLong((ActiveAlarm alarm) -> alarm.timeEnded))
.collect(Collectors.toList());
B. When timeStarted and timeEnded have public getter methods:
List<ActiveAlarm> sorted =
list.stream()
.sorted(Comparator.comparingLong(ActiveAlarm::getTimeStarted)
.thenComparingLong(ActiveAlarm::getTimeEnded))
.collect(Collectors.toList());
If you want to sort the original list itself:
A. When timeStarted and timeEnded are public (as mentioned in the requirement) and therefore do not (need to) have public getter methods:
list.sort(Comparator.comparingLong((ActiveAlarm alarm) -> alarm.timeStarted)
.thenComparingLong((ActiveAlarm alarm) -> alarm.timeEnded));
B. When timeStarted and timeEnded have public getter methods:
list.sort(Comparator.comparingLong(ActiveAlarm::getTimeStarted)
.thenComparingLong(ActiveAlarm::getTimeEnded));

Guava's ComparisonChain:
Collections.sort(list, new Comparator<ActiveAlarm>(){
#Override
public int compare(ActiveAlarm a1, ActiveAlarm a2) {
return ComparisonChain.start()
.compare(a1.timestarted, a2.timestarted)
//...
.compare(a1.timeEnded, a1.timeEnded).result();
}});

We can use the Comparator.comparing() method to sort a list based on an object's property.
class SortTest{
public static void main(String[] args) {
ArrayList<ActiveAlarm> activeAlarms = new ArrayList<>(){{
add(new ActiveAlarm("Alarm 1", 5, 10));
add(new ActiveAlarm("Alarm 2", 2, 12));
add(new ActiveAlarm("Alarm 3", 0, 8));
}};
/* I sort the arraylist here using the getter methods */
activeAlarms.sort(Comparator.comparing(ActiveAlarm::getTimeStarted)
.thenComparing(ActiveAlarm::getTimeEnded));
System.out.println(activeAlarms);
}
}
Note that before doing it, you'll have to define at least the getter methods of the properties you want to base your sort on.
public class ActiveAlarm {
public long timeStarted;
public long timeEnded;
private String name = "";
private String description = "";
private String event;
private boolean live = false;
public ActiveAlarm(String name, long timeStarted, long timeEnded) {
this.name = name;
this.timeStarted = timeStarted;
this.timeEnded = timeEnded;
}
public long getTimeStarted() {
return timeStarted;
}
public long getTimeEnded() {
return timeEnded;
}
#Override
public String toString() {
return name;
}
}
Output:
[Alarm 3, Alarm 2, Alarm 1]

Employee POJO Class
package in.ac.adit.oop.sort;
public class Employee {
private int id;
private String name;
private String department;
public int getId() {
return id;
}
public Employee() {
super();
}
public Employee(int id, String name, String department) {
super();
this.id = id;
this.name = name;
this.department = department;
}
#Override
public String toString() {
return "Employee [id=" + id + ", name=" + name + ", department=" + department + "]";
}
public void setId(int id) {
this.id = id;
}
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
public String getDepartment() {
return department;
}
public void setDepartment(String department) {
this.department = department;
}
}
Employee Class To Manage Employee
package in.ac.adit.oop.sort;
import java.util.ArrayList;
import java.util.List;
public class Example {
public static void main(String[] args) {
/*
* Create 10 Employee Object
*/
Employee emp1 = new Employee(1, "Nayan", "IT");
Employee emp2 = new Employee(2, "Siddarth", "CP");
Employee emp3 = new Employee(3, "Samarth", "AE");
Employee emp4 = new Employee(4, "Bhavesh", "CV");
Employee emp5 = new Employee(5, "Sam", "FT");
Employee emp6 = new Employee(6, "Keyur", "IT");
Employee emp7 = new Employee(7, "Bala", "ME");
Employee emp8 = new Employee(8, "Mitul", "ME");
Employee emp9 = new Employee(9, "Kamlesh", "EE");
Employee emp10 = new Employee(10, "Piyush", "EE");
/*
* List of Employee Object
*/
List<Employee> employeeList = new ArrayList<Employee>();
employeeList.add(emp1);
employeeList.add(emp2);
employeeList.add(emp3);
employeeList.add(emp4);
employeeList.add(emp5);
employeeList.add(emp6);
employeeList.add(emp7);
employeeList.add(emp8);
employeeList.add(emp9);
employeeList.add(emp10);
CustomObjectSort customObjectSort = new CustomObjectSort();
List<Employee> sortByDepartment = customObjectSort.sortByDepartment(employeeList);
/*
* Sorted By Department
*/
for (Employee employee : sortByDepartment) {
System.out.println(employee);
}
/*
* Sorted By Name
*/
List<Employee> sortByName = customObjectSort.sortByName(employeeList);
for (Employee employee : sortByName) {
System.out.println(employee);
}
/*
* Sorted By Id
*/
List<Employee> sortById = customObjectSort.sortById(employeeList);
for (Employee employee : sortById) {
System.out.println(employee);
}
}
}
Custom Sorting
package in.ac.adit.oop.sort;
import java.util.Collections;
import java.util.Comparator;
import java.util.List;
public class CustomObjectSort {
public List<Employee> sortByName(List<Employee> employeeList) {
Collections.sort(employeeList, new Comparator<Employee>() {
#Override
public int compare(Employee employee1, Employee employee2) {
return employee1.getName().compareTo(employee2.getName());
}
});
return employeeList;
}
public List<Employee> sortByDepartment(List<Employee> employeeList) {
Collections.sort(employeeList, new Comparator<Employee>() {
#Override
public int compare(Employee employee1, Employee employee2) {
return employee1.getDepartment().compareTo(employee2.getDepartment());
}
});
return employeeList;
}
public List<Employee> sortById(List<Employee> employeeList) {
Collections.sort(employeeList, new Comparator<Employee>() {
#Override
public int compare(Employee employee1, Employee employee2) {
return employee1.getId() - employee2.getId();
}
});
return employeeList;
}
}

You can use Collections.sort and pass your own Comparator<ActiveAlarm>

In java you need to use the static Collections.sort method. Here is an example for a list of CompanyRole objects, sorted first by begin and then by end. You can easily adapt for your own object.
private static void order(List<TextComponent> roles) {
Collections.sort(roles, new Comparator() {
#Override
public int compare(Object o1, Object o2) {
int x1 = ((CompanyRole) o1).getBegin();
int x2 = ((CompanyRole) o2).getBegin();
if (x1 != x2) {
return x1 - x2;
} else {
int y1 = ((CompanyRole) o1).getEnd();
int y2 = ((CompanyRole) o2).getEnd();
return y2 - y1;
}
}
});
}

You can call Collections.sort() and pass in a Comparator which you need to write to compare different properties of the object.

As mentioned you can sort by:
Making your object implement Comparable
Or pass a Comparator to Collections.sort
If you do both, the Comparable will be ignored and Comparator will be used. This helps that the value objects has their own logical Comparable which is most reasonable sort for your value object, while each individual use case has its own implementation.

public class ActiveAlarm {
public long timeStarted;
public long timeEnded;
private String name = "";
private String description = "";
private String event;
private boolean live = false;
}
List<ActiveAlarm> con = new ArrayList<ActiveAlarm>();
Collections.sort(con , (a1, a2) -> a1.timeStarted.compareTo(a2.timeStarted));
Collections.sort(con , (a1, a2) -> a1.timeEnded.compareTo(a2.timeEnded));

Here's what did the trick for me.
Was much shorter and easier than everything else I found:
Collections.sort(listName, Comparator.comparing(Object::getProperty).reversed());
The ".reversed()" part at the end was a requirement for my specific project but I'm sharing it too, as it took a while to find it

The best and the easiest way to sort any list of objects in Java (Java 8 and above).
Lets sort a basket of fruits based on the property "fruitName"
Fruit POJO:
class Fruit
{
int price;
String fruitName;
public Fruit(int price, String fruitName) {
super();
this.price = price;
this.fruitName = fruitName;
}
public int getPrice() {
return price;
}
public void setPrice(int price) {
this.price = price;
}
public String getFruitName() {
return fruitName;
}
public void setFruitName(String fruitName) {
this.fruitName = fruitName;
}
#Override
public String toString() {
return "Fruits [price=" + price + ", fruitName=" + fruitName + "]";
}
}
Now lets add fruits into a list and then sort it
List<Fruit> basketOfFruits = new ArrayList<>();
basketOfFruits.add(new Fruit(123, "oranges"));
basketOfFruits.add(new Fruit(45, "nectarine"));
basketOfFruits.add(new Fruit(369, "blueberries"));
basketOfFruits.add(new Fruit(248, "apple"));
basketOfFruits.add(new Fruit(968, "peaches"));
basketOfFruits.add(new Fruit(436, "grapes"));
basketOfFruits.add(new Fruit(596, "figs"));
//sorting by the property fruitName
Collections.sort(basketOfFruits, (f1, f2)->{return f1.getFruitName().compareTo(f2.getFruitName());});
You can now print the list (i.e basketOfFruits) and the fruits in the list would be sorted in ASCENDING order (lexicographically).
The output would look like this:
[Fruits [price=248, fruitName=apple], Fruits [price=369, fruitName=blueberries], Fruits [price=596, fruitName=figs], Fruits [price=436, fruitName=grapes], Fruits [price=45, fruitName=nectarine], Fruits [price=123, fruitName=oranges], Fruits [price=968, fruitName=peaches]]
Instead of Collections.sort(), Java streams can also be used (Java 8 and above). The following is the code using Java streams
List<Fruit> sortedFruits = basketOfFruits.stream().sorted( (f1, f2)->{return f1.getFruitName().compareTo(f2.getFruitName());}).collect(Collectors.toList());
here the list is sorted in the same manner as Collections.sort(), but the sorted items would be stored/collected in another list "sortedFruits". So, if we want to print the sorted items of the list, we need to print "sortedFruits" instead of "basketOfFruits" in this case

Filter an ArrayList of Objects using Java Streams to return an ArrayList of type String

I can't find an exact solution for this on SO. I have a Crowd class which consists of a Crowd object which is an arraylist of type People. People is a class with properties String name, Double bankBalance, Integer numberOfCarsOwned.
In my crowd class I have the following method whereby I seek to filter by names beginning with the letter P and return these an arraylist of type String:
public ArrayList<String> filterByLetterP(){
ArrayList<String> filteredNames = this.crowd.stream()
.filter(name -> name.getName().contains("P"));
return filteredNames;
}
My error is:
required type ArrayList<String> provided Stream<People>
Note: My solution must make use of streams. How can I correct my solution to get it to work?
Reference info below.
People class definition:
public class People {
private String name;
private Double bankBalance;
private Integer numberOfCarsOwned;
public People(String name, Double bankBalance, Integer numberOfCarsOwned) {
this.name = name;
this.bankBalance = bankBalance;
this.numberOfCarsOwned = numberOfCarsOwned;
}
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
public Double getBankBalance() {
return bankBalance;
}
public void setBankBalance(Double bankBalance) {
this.bankBalance = bankBalance;
}
public Integer getNumberOfCarsOwned() {
return numberOfCarsOwned;
}
public void setNumberOfCarsOwned(Integer numberOfCarsOwned) {
this.numberOfCarsOwned = numberOfCarsOwned;
}
}
Crowdclass definition:
public class Crowd {
private ArrayList<People> crowd;
public Crowd() {
this.crowd = new ArrayList<>();
}
public ArrayList<People> getCrowd() {
return crowd;
}
public void setCrowd(ArrayList<People> crowd) {
this.crowd = crowd;
}
public void addPeopleToCrowd(People people){
this.crowd.add(people);
}
public ArrayList<String> filterByLetterP(){
ArrayList<String> filteredNames = this.crowd.stream()
.filter(name -> name.getName().contains("P"));
return filteredNames;
}
}

A couple of things needs to be addressed:
You should use String#startsWith instead of String#contains.
You have to map the Stream to People#name.
You have missed collecting the Stream.
Do it as shown below:
public List<String> filterByLetterP() {
List<String> filteredNames = this.crowd.stream()
.map(p -> p.getName())
.filter(s -> s.startsWith("P"))
.collect(Collectors.toList());
return filteredNames;
}

You should change your filter because it doesn't check names with beginning letter 'P':
public List<String> filterByLetterP(){
return this.crowd.stream()
.map(People::getName)
.filter(name -> name.charAt(0) == 'P')
.collect(Collectors.toList());
}

java stream: group by an field and select other as grouping key

I'd like group by a field a select
Guess this class:
public class Person {
private Name name;
private Address address;
...
}
public class Name {
private String name;
private String fatherName;
private String motherName;
}
I'd like to group by Person.getName().getMotherName() and select Person.getName() as grouping object key.
Collection<Person> persons = new ...
Map<Name, List<Person>> group = persons.stream()
.collect(groupingby(p -> p.getName().getMotherName());
As you can see, I'd like to get a Map<Name, List<Person>>, instead of a Map<String, List<Person>>
Any ideas?

As you can't override equals() I can think of this way:
Create a wrapper around Name which overrides equals() and hashCode() for Name.getMotherName():
public class NameWrapper {
private final Name name;
public NameWrapper(Name name) {
this.name = name;
}
#Override
public int hashcode() {
return name.getMotherName().hashCode();
}
#Override
public boolean equals(Object obj) {
if(obj == this) return true;
if(!(obj instanceof NameWrapper)) return false;
NameWrapper other = (NameWrapper) obj;
return name.getMotherName().equals(other.name.getMotherName());
}
}
Then you can map the Person#getName() to a NameWrapper and group by that:
Map<NameWrapper, List<Person>> group = persons.stream()
.collect(Collectors.groupingBy(p -> new NameWrapper(p.getName())));

You can use :
Map<Name, List<Person>> group = persons.stream()
.collect(Collectors.groupingBy(p -> p.getName()));
And you have to Overide the hashCode and equald in your Name class.
#Override
public boolean equals(Object o) {
// your equals code
}
#Override
public int hashCode() {
// your hash code
}

Java List - Sorting by certain column e.g FirstName [duplicate]

I have simple class
public class ActiveAlarm {
public long timeStarted;
public long timeEnded;
private String name = "";
private String description = "";
private String event;
private boolean live = false;
}
and List<ActiveAlarm> con. How to sort in ascending order by timeStarted, then by timeEnded? Can anybody help? I know in C++ with generic algorithm and overload operator <, but I am new to Java.

Using Comparator
For Example:
class Score {
private String name;
private List<Integer> scores;
// +accessor methods
}
Collections.sort(scores, new Comparator<Score>() {
public int compare(Score o1, Score o2) {
// compare two instance of `Score` and return `int` as result.
return o2.getScores().get(0).compareTo(o1.getScores().get(0));
}
});
With Java 8 onwards, you can simply use lambda expression to represent Comparator instance.
Collections.sort(scores, (s1, s2) -> { /* compute and return int */ });

Either make ActiveAlarm implement Comparable<ActiveAlarm> or implement Comparator<ActiveAlarm> in a separate class. Then call:
Collections.sort(list);
or
Collections.sort(list, comparator);
In general, it's a good idea to implement Comparable<T> if there's a single "natural" sort order... otherwise (if you happen to want to sort in a particular order, but might equally easily want a different one) it's better to implement Comparator<T>. This particular situation could go either way, to be honest... but I'd probably stick with the more flexible Comparator<T> option.
EDIT: Sample implementation:
public class AlarmByTimesComparer implements Comparator<ActiveAlarm> {
#Override
public int compare(ActiveAlarm x, ActiveAlarm y) {
// TODO: Handle null x or y values
int startComparison = compare(x.timeStarted, y.timeStarted);
return startComparison != 0 ? startComparison
: compare(x.timeEnded, y.timeEnded);
}
// I don't know why this isn't in Long...
private static int compare(long a, long b) {
return a < b ? -1
: a > b ? 1
: 0;
}
}

JAVA 8 and Above Answer (Using Lambda Expressions)
In Java 8, Lambda expressions were introduced to make this even easier! Instead of creating a Comparator() object with all of it's scaffolding, you can simplify it as follows: (Using your object as an example)
Collections.sort(list, (ActiveAlarm a1, ActiveAlarm a2) -> a1.timeStarted-a2.timeStarted);
or even shorter:
Collections.sort(list, Comparator.comparingInt(ActiveAlarm ::getterMethod));
That one statement is equivalent to the following:
Collections.sort(list, new Comparator<ActiveAlarm>() {
#Override
public int compare(ActiveAlarm a1, ActiveAlarm a2) {
return a1.timeStarted - a2.timeStarted;
}
});
Think of Lambda expressions as only requiring you to put in the relevant parts of the code: the method signature and what gets returned.
Another part of your question was how to compare against multiple fields. To do that with Lambda expressions, you can use the .thenComparing() function to effectively combine two comparisons into one:
Collections.sort(list, (ActiveAlarm a1, ActiveAlarm a2) -> a1.timeStarted-a2.timeStarted
.thenComparing ((ActiveAlarm a1, ActiveAlarm a2) -> a1.timeEnded-a2.timeEnded)
);
The above code will sort the list first by timeStarted, and then by timeEnded (for those records that have the same timeStarted).
One last note: It is easy to compare 'long' or 'int' primitives, you can just subtract one from the other. If you are comparing objects ('Long' or 'String'), I suggest you use their built-in comparison. Example:
Collections.sort(list, (ActiveAlarm a1, ActiveAlarm a2) -> a1.name.compareTo(a2.name) );
EDIT: Thanks to Lukas Eder for pointing me to .thenComparing() function.

We can sort the list in one of two ways:
1. Using Comparator : When required to use the sort logic in multiple places
If you want to use the sorting logic in a single place, then you can write an anonymous inner class as follows, or else extract the comparator and use it in multiple places
Collections.sort(arrayList, new Comparator<ActiveAlarm>() {
public int compare(ActiveAlarm o1, ActiveAlarm o2) {
//Sorts by 'TimeStarted' property
return o1.getTimeStarted()<o2.getTimeStarted()?-1:o1.getTimeStarted()>o2.getTimeStarted()?1:doSecodaryOrderSort(o1,o2);
}
//If 'TimeStarted' property is equal sorts by 'TimeEnded' property
public int doSecodaryOrderSort(ActiveAlarm o1,ActiveAlarm o2) {
return o1.getTimeEnded()<o2.getTimeEnded()?-1:o1.getTimeEnded()>o2.getTimeEnded()?1:0;
}
});
We can have null check for the properties, if we could have used 'Long' instead of 'long'.
2. Using Comparable(natural ordering): If sort algorithm always stick to one property:
write a class that implements 'Comparable' and override 'compareTo' method as defined below
class ActiveAlarm implements Comparable<ActiveAlarm>{
public long timeStarted;
public long timeEnded;
private String name = "";
private String description = "";
private String event;
private boolean live = false;
public ActiveAlarm(long timeStarted,long timeEnded) {
this.timeStarted=timeStarted;
this.timeEnded=timeEnded;
}
public long getTimeStarted() {
return timeStarted;
}
public long getTimeEnded() {
return timeEnded;
}
public int compareTo(ActiveAlarm o) {
return timeStarted<o.getTimeStarted()?-1:timeStarted>o.getTimeStarted()?1:doSecodaryOrderSort(o);
}
public int doSecodaryOrderSort(ActiveAlarm o) {
return timeEnded<o.getTimeEnded()?-1:timeEnded>o.getTimeEnded()?1:0;
}
}
call sort method to sort based on natural ordering
Collections.sort(list);

In java8+ this can be written in single line as follows:
collectionObjec.sort(comparator_lamda) or comparator.comparing(CollectionType::getterOfProperty)
code:
ListOfActiveAlarmObj.sort((a,b->a.getTimeStarted().compareTo(b.getTimeStarted())))
or
ListOfActiveAlarmObj.sort(Comparator.comparing(ActiveAlarm::getTimeStarted))

public class ActiveAlarm implements Comparable<ActiveAlarm> {
public long timeStarted;
public long timeEnded;
private String name = "";
private String description = "";
private String event;
private boolean live = false;
public int compareTo(ActiveAlarm a) {
if ( this.timeStarted > a.timeStarted )
return 1;
else if ( this.timeStarted < a.timeStarted )
return -1;
else {
if ( this.timeEnded > a.timeEnded )
return 1;
else
return -1;
}
}
That should give you a rough idea. Once that's done, you can call Collections.sort() on the list.

Since Java8 this can be done even cleaner using a combination of Comparator and Lambda expressions
For Example:
class Student{
private String name;
private List<Score> scores;
// +accessor methods
}
class Score {
private int grade;
// +accessor methods
}
Collections.sort(student.getScores(), Comparator.comparing(Score::getGrade);

Java-8 solution using Stream API:
A. When timeStarted and timeEnded are public (as mentioned in the requirement) and therefore do not (need to) have public getter methods:
List<ActiveAlarm> sorted =
list.stream()
.sorted(Comparator.comparingLong((ActiveAlarm alarm) -> alarm.timeStarted)
.thenComparingLong((ActiveAlarm alarm) -> alarm.timeEnded))
.collect(Collectors.toList());
B. When timeStarted and timeEnded have public getter methods:
List<ActiveAlarm> sorted =
list.stream()
.sorted(Comparator.comparingLong(ActiveAlarm::getTimeStarted)
.thenComparingLong(ActiveAlarm::getTimeEnded))
.collect(Collectors.toList());
If you want to sort the original list itself:
A. When timeStarted and timeEnded are public (as mentioned in the requirement) and therefore do not (need to) have public getter methods:
list.sort(Comparator.comparingLong((ActiveAlarm alarm) -> alarm.timeStarted)
.thenComparingLong((ActiveAlarm alarm) -> alarm.timeEnded));
B. When timeStarted and timeEnded have public getter methods:
list.sort(Comparator.comparingLong(ActiveAlarm::getTimeStarted)
.thenComparingLong(ActiveAlarm::getTimeEnded));

Guava's ComparisonChain:
Collections.sort(list, new Comparator<ActiveAlarm>(){
#Override
public int compare(ActiveAlarm a1, ActiveAlarm a2) {
return ComparisonChain.start()
.compare(a1.timestarted, a2.timestarted)
//...
.compare(a1.timeEnded, a1.timeEnded).result();
}});

We can use the Comparator.comparing() method to sort a list based on an object's property.
class SortTest{
public static void main(String[] args) {
ArrayList<ActiveAlarm> activeAlarms = new ArrayList<>(){{
add(new ActiveAlarm("Alarm 1", 5, 10));
add(new ActiveAlarm("Alarm 2", 2, 12));
add(new ActiveAlarm("Alarm 3", 0, 8));
}};
/* I sort the arraylist here using the getter methods */
activeAlarms.sort(Comparator.comparing(ActiveAlarm::getTimeStarted)
.thenComparing(ActiveAlarm::getTimeEnded));
System.out.println(activeAlarms);
}
}
Note that before doing it, you'll have to define at least the getter methods of the properties you want to base your sort on.
public class ActiveAlarm {
public long timeStarted;
public long timeEnded;
private String name = "";
private String description = "";
private String event;
private boolean live = false;
public ActiveAlarm(String name, long timeStarted, long timeEnded) {
this.name = name;
this.timeStarted = timeStarted;
this.timeEnded = timeEnded;
}
public long getTimeStarted() {
return timeStarted;
}
public long getTimeEnded() {
return timeEnded;
}
#Override
public String toString() {
return name;
}
}
Output:
[Alarm 3, Alarm 2, Alarm 1]

Employee POJO Class
package in.ac.adit.oop.sort;
public class Employee {
private int id;
private String name;
private String department;
public int getId() {
return id;
}
public Employee() {
super();
}
public Employee(int id, String name, String department) {
super();
this.id = id;
this.name = name;
this.department = department;
}
#Override
public String toString() {
return "Employee [id=" + id + ", name=" + name + ", department=" + department + "]";
}
public void setId(int id) {
this.id = id;
}
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
public String getDepartment() {
return department;
}
public void setDepartment(String department) {
this.department = department;
}
}
Employee Class To Manage Employee
package in.ac.adit.oop.sort;
import java.util.ArrayList;
import java.util.List;
public class Example {
public static void main(String[] args) {
/*
* Create 10 Employee Object
*/
Employee emp1 = new Employee(1, "Nayan", "IT");
Employee emp2 = new Employee(2, "Siddarth", "CP");
Employee emp3 = new Employee(3, "Samarth", "AE");
Employee emp4 = new Employee(4, "Bhavesh", "CV");
Employee emp5 = new Employee(5, "Sam", "FT");
Employee emp6 = new Employee(6, "Keyur", "IT");
Employee emp7 = new Employee(7, "Bala", "ME");
Employee emp8 = new Employee(8, "Mitul", "ME");
Employee emp9 = new Employee(9, "Kamlesh", "EE");
Employee emp10 = new Employee(10, "Piyush", "EE");
/*
* List of Employee Object
*/
List<Employee> employeeList = new ArrayList<Employee>();
employeeList.add(emp1);
employeeList.add(emp2);
employeeList.add(emp3);
employeeList.add(emp4);
employeeList.add(emp5);
employeeList.add(emp6);
employeeList.add(emp7);
employeeList.add(emp8);
employeeList.add(emp9);
employeeList.add(emp10);
CustomObjectSort customObjectSort = new CustomObjectSort();
List<Employee> sortByDepartment = customObjectSort.sortByDepartment(employeeList);
/*
* Sorted By Department
*/
for (Employee employee : sortByDepartment) {
System.out.println(employee);
}
/*
* Sorted By Name
*/
List<Employee> sortByName = customObjectSort.sortByName(employeeList);
for (Employee employee : sortByName) {
System.out.println(employee);
}
/*
* Sorted By Id
*/
List<Employee> sortById = customObjectSort.sortById(employeeList);
for (Employee employee : sortById) {
System.out.println(employee);
}
}
}
Custom Sorting
package in.ac.adit.oop.sort;
import java.util.Collections;
import java.util.Comparator;
import java.util.List;
public class CustomObjectSort {
public List<Employee> sortByName(List<Employee> employeeList) {
Collections.sort(employeeList, new Comparator<Employee>() {
#Override
public int compare(Employee employee1, Employee employee2) {
return employee1.getName().compareTo(employee2.getName());
}
});
return employeeList;
}
public List<Employee> sortByDepartment(List<Employee> employeeList) {
Collections.sort(employeeList, new Comparator<Employee>() {
#Override
public int compare(Employee employee1, Employee employee2) {
return employee1.getDepartment().compareTo(employee2.getDepartment());
}
});
return employeeList;
}
public List<Employee> sortById(List<Employee> employeeList) {
Collections.sort(employeeList, new Comparator<Employee>() {
#Override
public int compare(Employee employee1, Employee employee2) {
return employee1.getId() - employee2.getId();
}
});
return employeeList;
}
}

You can use Collections.sort and pass your own Comparator<ActiveAlarm>

In java you need to use the static Collections.sort method. Here is an example for a list of CompanyRole objects, sorted first by begin and then by end. You can easily adapt for your own object.
private static void order(List<TextComponent> roles) {
Collections.sort(roles, new Comparator() {
#Override
public int compare(Object o1, Object o2) {
int x1 = ((CompanyRole) o1).getBegin();
int x2 = ((CompanyRole) o2).getBegin();
if (x1 != x2) {
return x1 - x2;
} else {
int y1 = ((CompanyRole) o1).getEnd();
int y2 = ((CompanyRole) o2).getEnd();
return y2 - y1;
}
}
});
}

You can call Collections.sort() and pass in a Comparator which you need to write to compare different properties of the object.

As mentioned you can sort by:
Making your object implement Comparable
Or pass a Comparator to Collections.sort
If you do both, the Comparable will be ignored and Comparator will be used. This helps that the value objects has their own logical Comparable which is most reasonable sort for your value object, while each individual use case has its own implementation.

public class ActiveAlarm {
public long timeStarted;
public long timeEnded;
private String name = "";
private String description = "";
private String event;
private boolean live = false;
}
List<ActiveAlarm> con = new ArrayList<ActiveAlarm>();
Collections.sort(con , (a1, a2) -> a1.timeStarted.compareTo(a2.timeStarted));
Collections.sort(con , (a1, a2) -> a1.timeEnded.compareTo(a2.timeEnded));

Here's what did the trick for me.
Was much shorter and easier than everything else I found:
Collections.sort(listName, Comparator.comparing(Object::getProperty).reversed());
The ".reversed()" part at the end was a requirement for my specific project but I'm sharing it too, as it took a while to find it

The best and the easiest way to sort any list of objects in Java (Java 8 and above).
Lets sort a basket of fruits based on the property "fruitName"
Fruit POJO:
class Fruit
{
int price;
String fruitName;
public Fruit(int price, String fruitName) {
super();
this.price = price;
this.fruitName = fruitName;
}
public int getPrice() {
return price;
}
public void setPrice(int price) {
this.price = price;
}
public String getFruitName() {
return fruitName;
}
public void setFruitName(String fruitName) {
this.fruitName = fruitName;
}
#Override
public String toString() {
return "Fruits [price=" + price + ", fruitName=" + fruitName + "]";
}
}
Now lets add fruits into a list and then sort it
List<Fruit> basketOfFruits = new ArrayList<>();
basketOfFruits.add(new Fruit(123, "oranges"));
basketOfFruits.add(new Fruit(45, "nectarine"));
basketOfFruits.add(new Fruit(369, "blueberries"));
basketOfFruits.add(new Fruit(248, "apple"));
basketOfFruits.add(new Fruit(968, "peaches"));
basketOfFruits.add(new Fruit(436, "grapes"));
basketOfFruits.add(new Fruit(596, "figs"));
//sorting by the property fruitName
Collections.sort(basketOfFruits, (f1, f2)->{return f1.getFruitName().compareTo(f2.getFruitName());});
You can now print the list (i.e basketOfFruits) and the fruits in the list would be sorted in ASCENDING order (lexicographically).
The output would look like this:
[Fruits [price=248, fruitName=apple], Fruits [price=369, fruitName=blueberries], Fruits [price=596, fruitName=figs], Fruits [price=436, fruitName=grapes], Fruits [price=45, fruitName=nectarine], Fruits [price=123, fruitName=oranges], Fruits [price=968, fruitName=peaches]]
Instead of Collections.sort(), Java streams can also be used (Java 8 and above). The following is the code using Java streams
List<Fruit> sortedFruits = basketOfFruits.stream().sorted( (f1, f2)->{return f1.getFruitName().compareTo(f2.getFruitName());}).collect(Collectors.toList());
here the list is sorted in the same manner as Collections.sort(), but the sorted items would be stored/collected in another list "sortedFruits". So, if we want to print the sorted items of the list, we need to print "sortedFruits" instead of "basketOfFruits" in this case

java: Comparator and Treeset to remove duplicates

i have a java class like this
public class A {
private String field1;
private String field2;
// getters, setters but no equals and hashcode
}
and a list of objects of this class, i want to remove from this list all the duplicates elements that has the same field1 or the same field2, so i have 2 Comparators
public class Comparator1 implements Comparator<A> {
public int compare(A o1, A o2) {
return o1.getField1().compareToIgnoreCase( o2.getField1() );
}
}
public class Comparator2 implements Comparator<A> {
public int compare(A o1, A o2) {
return o1.getField2().compareToIgnoreCase(o2.getField2());
}
}
so to do the task i use treeset like
TreeSet<A> ts1 = new TreeSet<A>(new Comparator1())
ts1.addAll(list)
TreeSet<A> ts2 = new TreeSet<A>(new Comparator2())
ts2.addAll(ts1)
list.clear()
list.addAll(ts2)
but how can i do the same using just one comparator and one treeset ?
Thanks for the help
Update:
Thanks all for the answers, but after reading them i don't know if this is the right approach to the real problem.
In my real case field1 is like a phone number and field2 is like a name.
So i don't want to call the same phone number more than one time (this is the first treeset to removes duplicates) and i don't want to call more than one time the same name (the second treeset to removes duplicates)
You can modify the class but i'd like to know if this approach is ok to resolve the real problem.
If this approach is correct, from your question, i see that without modifying the class is not possible to use just one comparator
Thanks

You can't use one comparator to sort by two criteria at the same time, so there is no real way to go better than two TreeSets in your case. Of course, you can wrap them in one data structure.
(Alternatively you could use two HashMaps, each having one of the strings as key - this will be faster on average, but is more complicated to program.)

You can't, and it's not clear to me that what you're trying to do is well-defined.
Are you aware that your current approach depends both on the order in which elements are added and on whether you check field1 or field2 first for duplicates? Imagine you had these objects of class A:
A ab = new A("a", "b");
A cb = new A("c", "b");
A cd = new A("c", "d");
Checking field1 first gives the result [ab] or [ab, cd], depending on the order added.
Checking field2 first gives the result [cb] or [ab, cd], depending on the order added.
This is pretty strange behavior. Is this what you intended? I don't think it is possible to reproduce this with a single TreeSet and Comparator in the general case.

public static <A extends Comparable<?>> TreeSet<A> getTreeSet(Collection<A> list){
TreeSet<A> result = new TreeSet<A>();
HashSet<A> unique = new HashSet<A>();
unique.addAll(list);
result.addAll(unique);
return result;
}
Generic function that adds items to hashset to make them unique, and then drop them to TreeSet to sort. You can use it with: TreeSet<A> ts1 = getTreeSet(list);.
This approach works well for a fixed list.
#BalusC No, this assumes
public class A implements Comparable<A> {
private String field1;
private String field2;
#Override
public int compareTo(A o) {
// No null checks, because it's illegal anyways.
int tmp = 0;
if ((tmp = field1.compareToIgnoreCase(o.field1)) != 0)
return tmp;
if ((tmp = field2.compareToIgnoreCase(o.field2)) != 0)
return tmp;
return tmp;
}
// getters, setters but no equals and hashcode
}

If your intention is to do two levels of sorting(first: PhoneNumber and second:Name), then you can use the following code, where the duplicate check will be done against both the fields(field1 and field2). As we are already using compareTo for both the fields, it is not required to use equals and hashcode. But it is always good practice to use hashcode and equals.
public class A implements Comparable<A> {
private String field1;
private String field2;
public A(String number, String name) {
this.field1 = number;
this.field2 = name;
}
// First level sorting will be done by field1.
// If field1 is equal then second level sorting will be done on field2
#Override
public int compareTo(A o) {
int compareTo = field1.compareTo(o.getNumber());
if(compareTo==0){
return field2.compareTo(o.getName());
}
return compareTo;
}
public String getNumber() {
return field1;
}
public String getName() {
return field2;
}
}

public class RemoveDuplicate {
public static void main(String[] args) {
final ArrayList<Student> students = new ArrayList<Student>();
Set<Student> set = new TreeSet<Student>();
Student[] starr = new Student[6];
starr[0] = new Student("Student1", "1005");
starr[1] = new Student("Student2", "1004");
starr[2] = new Student("Student3", "1003");
starr[3] = new Student("Student6", "1002");
starr[4] = new Student("Student5", "1001");
starr[5] = new Student("Student6", "1000");
Arrays.sort(starr, Student.StudentIdComparator);
for (Student s : starr) {
students.add(s);
}
System.out.println(students);
set.addAll(students);
System.out.println("\n***** After removing duplicates *******\n");
final ArrayList<Student> newList = new ArrayList<Student>(set);
/** Printing original list **/
System.out.println(newList);
}}

import java.util.ArrayList;
import java.util.Arrays;
import java.util.Set;
import java.util.TreeSet;
import java.util.Comparator;
import java.util.List;
public class RemoveDuplicate {
public static void main(String[] args) {
Set<Student> set = new TreeSet<Student>();
List<Student> students = Arrays.asList(new Student("Student1", "1005"), new Student("Student2", "1004"),
new Student("Student3", "1003"), new Student("Student6", "1002"), new Student("Student5", "1001"),
new Student("Student6", "1000"));
// Sorting Using Lambda
students.sort(new Comparator<Student>() {
#Override
public int compare(Student s1, Student s2) {
return s1.getId().compareTo(s2.getId());
}
});
System.out.println(students);
set.addAll(students);
System.out.println("\n***** After removing duplicates *******\n");
final ArrayList<Student> newList = new ArrayList<Student>(set);
/** Printing original list **/
System.out.println(newList);
}
}
class Student implements Comparable<Student> {
private String name;
private String id;
public Student(String name, String id) {
this.name = name;
this.id = id;
}
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
public String getId() {
return id;
}
public void setId(String id) {
this.id = id;
}
#Override
public String toString() {
return "\n" + "Name=" + name + " Id=" + id;
}
#Override
public int compareTo(Student o1) {
if (o1.getName().equalsIgnoreCase(this.name)) {
return 0;
}
return 1;
}
// public static Comparator<Student> StudentIdComparator = (Student
// s1,Student s2) -> s1.getId().compareTo(s2.getId());
}