How to sort List of objects by some property

问题:

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.

回答1:

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;
  }
}

回答2:

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 */ });

回答3:

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.

回答4:

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);

回答5:

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.

回答6:

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))

回答7:

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);

回答8:

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();
            }});

回答9:

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;
            }
        }
    });
}

回答10:

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

回答11:

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

回答12:

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.