Sort String Array by alphabet and then by number

Question

I have an String Array i have fileterd from a database with output like this in Java

[ABCD XYZ M1210, 
ABCD XYZ M149, 
ABCD XYZ M5130, 
ABCD XYZ N1420, 
ABCD XYZ T11299, 
ABCD XYZ S11044]

Im looking to sort my array as follows:

[ABCD XYZ M149, 
ABCD XYZ M1210, 
ABCD XYZ M5130, 
ABCD XYZ N1420,  
ABCD XYZ S11044,
ABCD XYZ T11299]

and it is the last element i specifically want

 ==> String theStringIWant = myStringArray.get(myStringArray.size() - 1);

What i need to dois first sort the letter after "XYZ" alphabetically and then sort the numerically after that so for example ABCD XYZ M1210 < ABCD XYZ M5130 as 5130 is greater than 1210.

Any help here would be much appreciated

*Any referencing to suitable libraries in java etc

Cheers

I would make my own `Comparator` to process the string in two parts. — PurkkaKoodari, Nov 14 '13 at 17:06
@Pietu1998 Could you point to any similar comparator example for reference? Thanks — user1694873, Nov 14 '13 at 17:08
possible duplicate of [Sort ArrayList of custom Objects by property](http://stackoverflow.com/questions/2784514/sort-arraylist-of-custom-objects-by-property) — , Nov 14 '13 at 17:10

score 4 · Answer 1 · answered Nov 14 '13 at 17:11

Arrays.sort(array, new Comparator<String>() {
    @Override
    public int compare(String s1, String s2) {
        String stringPart1 = extractStringPart(s1); 
        String stringPart2 = extractStringPart(s2); 
        int result = stringPart1.compare(stringPart2);
        if (result == 0) {
            int intPart1 = extractIntPart(s1);
            int intPart2 = extractIntPart(s2);
            result = Integer.compare(intPart1, intPart2);
        }
        return result;
    }
});

The two extract methods are left as an exercise. Read the javadoc for String and/or Matcher to find out how you could do.

Mengjun · Answer 2 · 2013-11-14T17:42:09.517

You can use Collections.sort to sort the contents.

In the mean while, write a comparator for comparison, like,

1 Write a Comparator

public class MyComparator implements Comparator {

@Override
public int compare(Object o1, Object o2) {

    String s1 = (String)o1;
    String s2 = (String)o2;
    //String part has 10 characters, It is fixed.
    String strPart1 = s1.substring(0,10); 
    int intPart1 = Integer.valueOf(s1.substring(10));

    String strPart2 = s2.substring(0,10);
    int intPart2 = Integer.valueOf(s2.substring(10));

    int strCompareResult = strPart1.compareTo(strPart2);
    if(0 == strCompareResult )
    {
        return intPart1 - intPart2;
    }
    else
    {
        return strCompareResult;
    }
}

}

2 Use Collections.sort and Comparator to complete the sorting

    List<String> results = Arrays.asList("ABCD XYZ M1210", "ABCD XYZ M149",
            "ABCD XYZ M5130", "ABCD XYZ N1420", "ABCD XYZ T11299",
            "ABCD XYZ S11044");

    System.out.println("Before sorting... ...");
    System.out.println(results);

    System.out.println("After sorting... ... ");
    Collections.sort(results, new MyComparator());
    System.out.println(results);

Output in Console:

Before sorting... ...

[ABCD XYZ M1210, ABCD XYZ M149, ABCD XYZ M5130, ABCD XYZ N1420, ABCD XYZ T11299, ABCD XYZ S11044]

After sorting... ...

[ABCD XYZ M149, ABCD XYZ M1210, ABCD XYZ M5130, ABCD XYZ N1420, ABCD XYZ S11044, ABCD XYZ T11299]

@Pietu1998 - Thanks for the correction. Changes are made accordingly. :) — Mengjun, Nov 14 '13 at 17:42

score 1 · Accepted Answer · answered Nov 14 '13 at 17:21

and it is the last element i specifically want

The following program first extract the last integer part of your string element. And then return by comparing them.

List<String> results = Arrays.asList("ABCD XYZ M1210", "ABCD XYZ M149",
        "ABCD XYZ M5130", "ABCD XYZ N1420", "ABCD XYZ T11299",
        "ABCD XYZ S11044");

        Collections.sort(results, new Comparator<String>(){

            @Override
            public int compare(String o1, String o2) {
                Integer x1 = Integer.parseInt(o1.substring(o1.lastIndexOf(" ")+2, o1.length()));
                Integer x2 = Integer.parseInt(o2.substring(o2.lastIndexOf(" ")+2, o2.length()));

                return x1.compareTo(x2);

            }
        });

        System.out.println(results);

Output:

[ABCD XYZ M149, 
ABCD XYZ M1210, 
ABCD XYZ N1420, 
ABCD XYZ M5130, 
ABCD XYZ S11044, 
ABCD XYZ T11299]

@user1694873 notice the output carefully! it doesn't work. `ABCD XYZ N1420,` comes before `ABCD XYZ M5130` which shouldn't. — Rahul Sharma, Aug 25 '16 at 10:19

score 1 · Answer 4 · answered Apr 17 '23 at 06:41

A question can also be revers of it like sort number followed by alphabate i know this is revers of it but posting so anyone can get idea

expected output:B01 A02 A03 A04 A10 A11 C13 A20 A112


var items = ["A01", "A112","A02", "A03", "B01","C13","A04","A10", "A11", "A20"];
// console.log(   items.sort());
let obj={}
items.map((ele)=>{
   let y=ele.slice(1)
   // console.log(y);
   obj[parseInt(ele.slice(1))]=ele

})
console.log(obj);
for(let yy in obj)
{
console.log(obj[yy]);
}```

Sort String Array by alphabet and then by number

4 Answers4