Sort List In Groovy

If we have a simple list like

List list = [4, 2, 9, 6, 7, 8, 2]

We can easily sort this list using

assert list.sort() == [2, 2, 4, 6, 7, 8, 9]

or by using closure like

assert list.sort { it } == [2, 2, 4, 6, 7, 8, 9]

Both gives us a list sorted in ascending order.

sort modifies the original list i.e., original list is also sorted. If you do not want to modify the original list then just pass false as argument like

list.sort(false)
list.sort(false) { it }

Using closure we can also sort object by its property:

class User {
    String name
    Integer age

    @Override
    String toString() {
        return name
    }
}

List userList = [new User(name: 'Manish', age: 25), 
        new User(name: 'Kumar', age: 26), 
        new User(name: 'JFT', age: 24)]

assert userList.sort { it.name } == ['JFT', 'Kumar', 'Manish']
assert userList.sort { it.age } == ['JFT', 'Manish', 'Kumar']
assert userList.sort { it.name.size() } == ['JFT', 'Kumar', 'Manish']

We can also sort list in descending order but if we compare numeric fields only

assert list.sort { -it } == [9, 8, 7, 6, 4, 2, 2]
assert userList.sort { -it.age } == ['Kumar', 'Manish', 'JFT']
assert userList.sort { -it.name.size() } == ['Manish', 'Kumar', 'JFT']

Hope this helps 🙂 .