排序Java对象列表

Question

i have a Java Object List

List<MyObject> list = new ArrayList<>();
    list.add(new MyObject("1", "1"));
    list.add(new MyObject("2", "2"));
    list.add(new MyObject("3", "3"));
    list.add(new MyObject("4", "12"));
    list.add(new MyObject("5", "1d2"));

My Objects Are

Public MyObject {
String id;
String myNumber;
...
}

now i want to sort my list by myNumber by only numbers so i want a result:

MyObject("5", "1d3");
MyObject("4", "12");
MyObject("3", "3");
MyObject("2", "2");
MyObject("1", "1");

How can i do this?

with list.sort(Comparator.comparing(MyObject::getMyNumber)); gives me

MyObject("3", "3");
MyObject("2", "2");
MyObject("5", "1d3");
MyObject("4", "12");
MyObject("1", "1");

Answer 1

I would suggest you write your own compare method inside your object while implementing "Comparable"

something like

public class CompareObject implements Comparable{

String id;
String myNumber;

public CompareObject(String id, String myNumber) {
    this.id = id;
    this.myNumber = myNumber;
}

public int compareTo(CompareObject c) {
    String myNumberOnlyNumbers = "";
    String otherNumberOnlyNumbers = "";
    for(int i = 0; i < myNumber.length(); i++) {
        try {
            myNumberOnlyNumbers += Integer.parseInt(myNumber.charAt(i)+"") + "";
        } catch (NumberFormatException e) {

        }
    }

    for(int i = 0; i < c.myNumber.length(); i++) {
        try {
            otherNumberOnlyNumbers += Integer.parseInt(c.myNumber.charAt(i)+"") + "";
        } catch (NumberFormatException e) {

        }
    }

    return Integer.compare(Integer.parseInt(myNumberOnlyNumbers), Integer.parseInt(otherNumberOnlyNumbers));
}

}

in your case "CompareObject" would be "MyObject"

after that you can do

    List<CompareObject> list = new ArrayList<>();
    list.add(new CompareObject("1", "1"));
    list.add(new CompareObject("2", "2"));
    list.add(new CompareObject("3", "3"));
    list.add(new CompareObject("4", "12"));
    list.add(new CompareObject("5", "1d3"));

    Collections.sort(list, Collections.reverseOrder()); //Collections.reverseOrder() meaning in descending order here

and if i do

    for(int i = 0; i < list.size(); i++) {
        System.out.println(list.get(i).id + " - " + list.get(i).myNumber);
    }

to print the list i get

5 - 1d3
4 - 12
3 - 3
2 - 2
1 - 1

Answer 2

AFAIK there is no java in built solution, but among java developers this is famous as AlphaNumeric sorting and you can find some sample code here.

https://github.com/my-flow/alphanum/blob/master/src/AlphanumComparator.java

Answer 3

Collections.sort(list, new Comparator<MyObject>() {
        @Override
        public int compare(MyObject o1, MyObject o2) {
            String one = o1.myNumber;
            String two = o2.myNumber;
            if (isNumeric(one) && isNumeric(two)) {
                return Integer.valueOf(two).compareTo(Integer.valueOf(one));
            } else {
                return two.compareTo(one); //add new fuction here to compare strings having alphabets
            }
        }
    });

for(int i=0; i<list.size(); i++){
        System.out.println(list.get(i));
}

isNumeric method:

    public static boolean isNumeric(String str) {
    try {
        Double.parseDouble(str);
        return true;
    } catch(NumberFormatException e){
        return false;
    }
}

Answer 4

list.sort(Comparator.comparingLong(myobj ->
                        Long.parseLong(myobj.getMyNumber().replaceAll("\\D", "")
                                                          .replaceFirst("^$", "0"))
                    .reversed());

This assumes you want:

• to discard all non-digit characters
• sort numerically descending

(That is what I read from your example: 13, 12, 3, 2, 1.)

The easiest is to provide a long/int function that derives a long from a MyObject.

(The replaceFirst ensures that as value without any digit will yield 0.)

Answer 5

List<MyObject> list = new ArrayList<>();
    list.add(new MyObject("1", "1"));
    list.add(new MyObject("2", "2"));
    list.add(new MyObject("3", "3"));
    list.add(new MyObject("4", "12"));
    list.add(new MyObject("5", "1d2"));
    Collections.sort(list, new Comparator<MyObject>() {
        @Override
        public int compare(MyObject o1, MyObject o2) {
            if(Integer.parseInt(o1.getId())<=Integer.parseInt(o2.getId()))
            {
                return 1;
            }else
                return -1;
        }
    });
    for(int i=0;i<list.size();i++)
    {
        System.out.println(list.get(i));
    }

Answer 6

I think you problem behind your question shall be 'how to extract numbers from a string', which is well answered in this case. Extract digits from a string in Java

myNumber= myNumber.replaceAll("\\D+","");

And you can create you own comparator to handle the result easily.