Sorting An ArrayList of Objects by Object Parameters - Integers

Question

I have This method below which works for sorting out Student objects in an arraylist using String parameters, for example a Student Object would have a String name,int age and String course parameter

//This Sorts with String parameters
public void sortArrayListByName()
{
    Collections.sort(sList,new Comparator<Student>() {
         public int compare(Student s1, Student s2) {
                 return s1.getName().compareTo(s2.getName());
         }
     });
}

Now I want to implement this code with integer parameters

//This should sort the Arraylist of Objects
public void sortArrayListByAge()
{
    Collections.sort(sList,new Comparator<Student>() {
         public int compare(Student s1, Student s2) {
                 return s1.getAge().compareToIgnoreCase(s2.getAge()); //This Part gives me errors
         }
     });
}

But i get an error message saying Cannot invoke compareToIgnoreCase(int) on the primitive type int

How can i fix this code so that it will be able to sort the list with integer values?

If age in the Student class is of type int then just use a simple equality comparison operator. — Ryan J
– Ryan J, Commented Aug 14, 2014 at 17:11

Luiggi Mendoza · Accepted Answer · 2014-08-14 17:13:13Z

3

You're comparing int values, which are primitive types and int does not have any methods. This is why s1.getAge().compareToIgnoreCase(s2.getAge()) fails.

If you happen to use Java 7 or prior, use Integer#compare:

public int compare(Student s1, Student s2) {
    return Integer.compare(s1.getAge(), s2.getAge()); 
}

Otherwise, compare the int variables manually (code adapted from Integer#compare source, there's no need to reinvent the wheel):

public int compare(Student s1, Student s2) {
    return (s1.getAge() < s2.getAge()) ? -1 : ((s1.getAge() == s2.getAge()) ? 0 : 1); 
}

answered Aug 14, 2014 at 17:13

Luiggi Mendoza

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3 Comments

Kebab Programmer Over a year ago

haha yea, this works, i don't know why i didn't think of that, thanks for your input

Ryan J Over a year ago

Take care to check for null on either of your objects or this solution could break.

Luiggi Mendoza Over a year ago

@RyanJ if there's a NPE I would not worry on this method but why I added null elements in the list.

Ryan J · Accepted Answer · 2014-08-14 17:14:46Z

2

Try this:

//This should sort the Arraylist of Objects
public void sortArrayListByAge()
{
    Collections.sort(sList,new Comparator<Student>() {
         public int compare(Student s1, Student s2) {
                 int result = 0;   // assume equal
                 if ( (s1 == null && s2 != null )|| (s1.getAge() > s2.getAge()) ) {
                     result = 1;
                 }
                 if ( (s1 != null && s2 == null) || s1.getAge() < s2.getAge() ) {
                     result = -1;
                 }
                 return result;
         }
     });
}

answered Aug 14, 2014 at 17:14

Ryan J

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Comments

Branislav Lazic · Accepted Answer · 2014-08-14 17:26:03Z

2

You already got your answer, but one of the options is to also use sorted method from Java 8 Stream API:

sList.stream().sorted((s1, s2) -> Integer.compare(s1.getAge(), s2.getAge()))
              .forEach(s -> System.out.println(s.getName() + ", " + s.getAge()));

answered Aug 14, 2014 at 17:26

Branislav Lazic

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Comments

Eran · Accepted Answer · 2014-08-14 17:13:56Z

0

You can use normal comparison operators :

public void sortArrayListByAge()
{
    Collections.sort(sList,new Comparator<Student>() {
         public int compare(Student s1, Student s2) {
                 if (s1 != null && s2 != null) {
                    if (s1.getAge() < s2.getAge())
                        return -1;
                    else {
                        if (s1.getAge() > s2.getAge())
                            return 1;
                        else
                            return 0;
                    }
                 } else if (s1 != null) {
                    return 1; // s2 is null
                 } else if (s2 != null) {
                    return -1; // s1 is null
                 } else {
                    return 0; // both s1 and s2 are null
                 }
         }
     });

}

answered Aug 14, 2014 at 17:13

Eran

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2 Comments

Jashaszun Over a year ago

Wouldn't it be the other way? (I think you mixed up negative and positive returns.)

Eran Over a year ago

@Jashaszun

a negative integer, zero, or a positive integer as the first argument is less than, equal to, or greater than the second.

- I think I got it right.

elias · Accepted Answer · 2014-08-14 17:20:40Z

0

The compareTo and compareToIgnoreCase methods are inverted.

To sort integers, like age, use:

return s1.getAge().compareTo(s2.getAge());

And to sort Strings, like names, and ignoring the case, use:

return s1.getName().compareToIgnoreCase(s2.getName());

answered Aug 14, 2014 at 17:20

elias

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