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("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) {return Integer.valueOf(o1.getPriority()).compareTo(o2.getPriority());}});sortedList.forEach((e) -> {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<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));
OR
you can combine those sortings ->
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);
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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