How to sort List of objects by some property
JavaSortingSort ObjectJava Problem Overview
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.
Java Solutions
Solution 1 - Java
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;
}
}
Solution 2 - 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 */ });
Solution 3 - Java
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.
Solution 4 - Java
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);
Solution 5 - Java
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))
Solution 6 - Java
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.
Solution 7 - Java
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);
Solution 8 - Java
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));
Solution 9 - Java
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();
}});
Solution 10 - Java
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;
}
}
Solution 11 - Java
You can use Collections.sort and pass your own Comparator<ActiveAlarm>
Solution 12 - Java
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;
}
}
});
}
Solution 13 - Java
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]
Solution 14 - Java
You can call Collections.sort() and pass in a Comparator which you need to write to compare different properties of the object.
Solution 15 - Java
As mentioned you can sort by:
- Making your object implement
Comparable - Or pass a
ComparatortoCollections.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.
Solution 16 - Java
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