Java sort array of objects by integer field and if those are identical then sort by another integer value

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英文:

Java sort array of objects by integer field and if those are identical then sort by another integer value

问题

我在处理一个涉及对对象数组进行排序的 Java 问题。

我已经解决了按特定字段对对象数组进行排序的方法,可以在以下代码中看到:

public void printPrioritized() {
    System.out.println("Prioritized todo:");
    System.out.println("-----------------");
    List<Task> sortedList = new ArrayList<Task>(taskList);

    Collections.sort(sortedList, new Comparator<Task>() {
        public int compare(Task o1, Task o2) {
            int priorityComparison = Integer.valueOf(o1.getPriority()).compareTo(o2.getPriority());
            if (priorityComparison == 0) {
                // If priorities are the same, sort by another value (e.g., time)
                return Integer.valueOf(o1.getTime()).compareTo(o2.getTime());
            }
            return priorityComparison;
        }
    });

    sortedList.forEach((e) -> {
        System.out.println(e);
    });
}

我的问题是,如果两个对象的某些字段相同,那么我应该按另一个值进行排序。这意味着我需要按 1 到 4 的值进行排序(通过 getPriority() 方法),但如果两个对象例如都是 2,则我必须按另一个值排序,比如时间。希望有人能够帮助。

英文:

I have a java problem in relation too sorting an arrayList of objects.

I have already figured out to sort an arrayList of object by specific fields, which can be seen in the following code

public void printPrioritized() {
    System.out.println(&quot;Prioritized todo:&quot;);
    System.out.println(&quot;-----------------&quot;);
    List&lt;Task&gt; sortedList = new ArrayList&lt;Task&gt;(taskList);

    Collections.sort(sortedList, new Comparator&lt;Task&gt;() {
        public int compare(Task o1, Task o2) {
            return Integer.valueOf(o1.getPriority()).compareTo(o2.getPriority());
        }
    });

    sortedList.forEach((e) -&gt; {
        System.out.println(e);
    });

My problem is that if to object fields are the same then i am supposed to sort by another value. This means that i have to sort by an value of 1 to 4 (getPriority() method), but if two objects for instance both are 2 then i have to sort by another value which for instance could be time. Hope someone can help.

答案1

得分: 3

假设你的 Task 类看起来类似于:

class Task {
    int priority;
    int anotherValue;
    // 获取器、设置器...
}

你可以创建自定义比较器并在排序时链式使用它们,示例如下:

List<Task> myList = new ArrayList<>();

Comparator<Task> byPriority = (t1, t2) -> Integer.compare(t1.getPriority(), t2.getPriority());
Comparator<Task> byAnotherValue = (t1, t2) -> Integer.compare(t1.getAnotherValue(), t2.getAnotherValue());

myList.sort(byPriority.thenComparing(byAnotherValue));

或者你可以将这些排序组合起来,如下所示;

List<Task> myList = new ArrayList<>();

Comparator<Task> sortedComparator = (t1, t2) -> {
    if (t1.getPriority() != t2.getPriority()) {
        return Integer.compare(t1.getPriority(), t2.getPriority());
    } else if (t1.getAnotherValue() != t2.getAnotherValue()) {
        return Integer.compare(t1.getAnotherValue(), t2.getAnotherValue());
    }
};

myList.sort(sortedComparator);
英文:

Assuming your Task class looks something like:

class Task {
    int priority;
    int anotherValue;
    // getters, setters ...
}

you can create custom compartors and chain them while sorting, example:

List&lt;Task&gt; myList = new ArrayList&lt;&gt;();

Comparator&lt;Task&gt; byPriority = (t1,t2) -&gt; Integer.compare(t1.getPriority(), t2.getPriority());
Comparator&lt;Task&gt; byAnotherValue = (t1,t2) -&gt; Integer.compare(t1.getAnotherValue(), t2.getAnotherValue());

myList.sort(byPriority.thenComparing(byAnotherValue));

OR
you can combine those sortings ->

List&lt;Task&gt; myList = new ArrayList&lt;&gt;();

Comparator&lt;Task&gt; sortedComparator = (t1,t2) -&gt; {
    if (t1.getPriority() != t2.getPriority()) {
        return Integer.compare(t1.getPriority(), t2.getPriority()); 
    }
    else if (t1.getAnotherValue() != t2.getAnotherValue()) {
        return Integer.compare(t1.getAnotherValue(), t2.getAnotherValue());
    }
};

myList.sort(sortedComparator);

答案2

得分: 2

尝试自定义compare方法。
例如:

if(o1.getPriority() != o2.getPriority())
   return Integer.valueOf(o1.getPriority()).compareTo(o2.getPriority());

if(o1.getTime() != o2.getTime())
   return Integer.valueOf(o1.getTime()).compareTo(o2.getTime());

return 0; //它们在所有字段上都相等
英文:

Try to customize the compare method.
e.g.

if(o1.getPriority() != o2.getPriority()) 
   return Integer.valueOf(o1.getPriority()).compareTo(o2.getPriority());

if(o1.getTime() != o2.getTime())
   return Integer.valueOf(o1.getTime()).compareTo(o2.getTime());

return 0; //they are equal with all fields

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  • 本文由 发表于 2020年10月20日 02:01:23
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