How to sort an alphanumeric string

Question

I have a problem with sorting strings that include integers. If I use the code below, I sort: 1 some, 2 some, 20 some, 21 some, 3 some, some

However, I want it to be sorted as: 1 part, 2 part, 3 part, 20 part, 21 part, some

How can i do this?

Thank!

Collections.sort(selectedNodes,
    new Comparator<DefaultMutableTreeNode>() {
    @Override
    public int compare(DefaultMutableTreeNode o1,
        DefaultMutableTreeNode o2) {
        return o1.getUserObject().toString()
            .compareTo(o2.getUserObject().toString());
    }
    });

Answer 1

Here is an example of how to do this (not particularly optimized):

final Pattern p = Pattern.compile("^\\d+");
String[] examples = { 
   "1some", "2some", "20some", "21some", "3some", "some", "1abc", "abc"
};
Comparator<String> c = new Comparator<String>() {
    @Override
    public int compare(String object1, String object2) {
        Matcher m = p.matcher(object1);
        Integer number1 = null;
        if (!m.find()) {
            return object1.compareTo(object2);
        }
        else {
            Integer number2 = null;
            number1 = Integer.parseInt(m.group());
            m = p.matcher(object2);
            if (!m.find()) {
                return object1.compareTo(object2);
            }
            else {
                number2 = Integer.parseInt(m.group());
                int comparison = number1.compareTo(number2);
                if (comparison != 0) {
                    return comparison;
                }
                else {
                    return object1.compareTo(object2);
                }
            }
        }
    }
};
List<String> examplesList = new ArrayList<String>(Arrays.asList(examples));
Collections.sort(examplesList, c);
System.out.println(examplesList);

Output

[1abc, 1some, 2some, 3some, 20some, 21some, abc, some]

Explanation

• The example uses a constant Patternto determine if the number is in the starting position String.
• If not in the first String, he compares it as with the second.
• If he is present in the first, he checks the second.
• , String ,
• , Integer String s, , ,
• , String ( MihaiC , )

Answer 2

, this

Answer 3

- , .

public class AlphaNumericalComparator implements Comparator<String> {
    @Override
    public int compare(String o1, String o2) {
        List<Object> parts1 = partsOf(o1);
        List<Object> parts2 = partsOf(o2);
        while (!parts1.isEmpty() && !parts2.isEmpty()) {
            Object part1 = parts1.remove(0);
            Object part2 = parts2.remove(0);
            int cmp = 0;
            if (part1 instanceof Integer && part2 instanceof Integer) {
                cmp = Integer.compare((Integer)part1, (Integer)part2);
            } else if (part1 instanceof String && part2 instanceof String) {
                cmp = ((String) part1).compareTo((String) part2);
            } else {
                cmp = part1 instanceof String ? 1 : -1; // XXXa > XXX1
            }
            if (cmp != 0) {
                return cmp;
            }
        }
        if (parts1.isEmpty() && parts2.isEmpty()) {
            return 0;
        }
        return parts1.isEmpty() ? -1 : 1;
    }

    private List<Object> partsOf(String s) {
        List<Object> parts = new LinkedList<>();
        int pos0 = 0;
        int pos = 0;
        boolean wasDigit = false;
        while (true) {
            if (pos >= s.length()
                    || Character.isDigit(s.charAt(pos)) != wasDigit) {
                if (pos > pos0) {
                    String part = s.substring(pos0, pos);
                    parts.add(wasDigit? Integer.valueOf(part) : part);
                    pos0 = pos;
                }
                if (pos >= s.length()) {
                    break;
                }
                wasDigit = !wasDigit;
            }
            ++pos;
        }
        return parts;
    }
};

, Java 8 Comparator.

Answer 4

Comparator, . String.compareTo() . 0 20some s 3some, 0 , .
, : . , String ( ?). , Alphanum, .
, , . int Integer.parse(). int , String s, . , , , , .

Answer 5

String compareTo() , , .

• .
• , .
• , String compareTo()

.

Answer 6

​​ , :

Collections.sort(selectedNodes, new Comparator<DefaultMutableTreeNode>() {
    @Override
    public int compare(DefaultMutableTreeNode o1,
        DefaultMutableTreeNode o2) {
        return Integer.parseInt(o1.getUserObject().toString().replaceAll("\\D", "")) -
            Integer.parseInt(o2.getUserObject().toString().replaceAll("\\D", ""));
    }
});

Answer 7

This is a working Java solution. If you have any suggestions on the code, please let me know in Gist .

import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Comparator;
import java.util.List;

public class FB {

    public static int comparator(String s1, String s2) {

        String[] pt1 = s1.split("((?<=[a-z])(?=[0-9]))|((?<=[0-9])(?=[a-z]))"); 
        String[] pt2 = s2.split("((?<=[a-z])(?=[0-9]))|((?<=[0-9])(?=[a-z]))"); 
//pt1 and pt2 arrays will have the string split in alphabets and numbers

        int i=0;
        if(Arrays.equals(pt1, pt2))
            return 0;
        else{
            for(i=0;i<Math.min(pt1.length, pt2.length);i++)
                if(!pt1[i].equals(pt2[i])) {
                    if(!isNumber(pt1[i],pt2[i])) {
                        if(pt1[i].compareTo(pt2[i])>0)
                            return 1;
                        else
                            return -1;
                    }
                    else {
                        int nu1 = Integer.parseInt(pt1[i]);
                        int nu2 = Integer.parseInt(pt2[i]);
                        if(nu1>nu2)
                            return 1;
                        else
                            return -1;
                    }
                }
        }

        if(pt1.length>i)
            return 1;
        else
            return -1;
    }

    private static Boolean isNumber(String n1, String n2) {
        // TODO Auto-generated method stub
        try {
            int nu1 = Integer.parseInt(n1);
            int nu2 = Integer.parseInt(n2);
            return true;
        }
        catch(Exception x) {
            return false;
        }

    }

    public static void main(String[] args) {
        // TODO Auto-generated method stub

        String[] examples = {"1some", "2some", "20some", "21some", "3some", "some", "1abc", "abc"};
        List<String> values = new ArrayList<String>(Arrays.asList(examples));

        System.out.println(values);
        Comparator<String> com = (o1,o2) -> {return comparator(o1,o2);}; //lambda expression

        Collections.sort(values,com);
        System.out.println(values);
    }
}

Output:

[1some, 2some, 20some, 21some, 3some, some, 1abc, abc]
[1abc, 1some, 2some, 3some, 20some, 21some, abc, some]

Answer 8

If you know that the pattern is always NUMALPHA or ALPHANUM and alpha are always the same:

if(str1.length() != str2.length()){
   return str1.length() - str2.length();
}

return str1.compareTo(str2);

Answer 9

If you have an alphanumeric array of strings, you can sort it directly with

Arrays.sort(Array_name)

and then print:

for(String a : Array_name)
    System.out.print(a);

Answer 10

    String [] str = new String[]{"1some", "2some", "20some", "21some", "3some", "some"};
    List<String> list = Arrays.asList(str);

    Collections.sort(list, String.CASE_INSENSITIVE_ORDER);
    System.out.println(list);