Sorting list based on another list's order [duplicate]

Question

I need to sort a list of Person objects(List<Person>, where each Person object has few attributes like id(unique), name, age… etc).

The sorting order is based on another list. That list contains a set of Person id's (A List<String> which is already sorted).

What is the best way to order the List<Person> in the same order as the list of id's using Kotlin or Java.

Example:

List Person {
(“ID1”,”PERSON1”,22,..), (“ID-2”,”PERSON2”,20,..) ), (“ID-3”,”PERSON3”,19,..),…..
}

Ordered Id List :

List of ID {(“ID2”), (“ID1”),(”ID3”)….}

Sorted Person list should be:

List PERSON {
 (“ID-2”,”PERSON 2”,20,..) ), (“ID1”,”PERSON 2”,22,..),  (“ID-3”,”PERSON 2”,19,..),…..
}

If the Person list contains any id's which are not mentioned in the id list then those values should be at the end of the sorted list.

---

Edited: This is my current way in Java. I am hoping for a better way than this:

public static List<Person> getSortList(List <Person> unsortedList, List<String> orderList){

    if(unsortedList!=null && !unsortedList.isEmpty() && orderList!=null && !orderList.isEmpty()){
        List sortedList = new ArrayList<OpenHABWidget>();
        for(String id : orderList){
            Person found= getPersonIfFound(unsortedList, id); // search for the item on the list by ID
            if(found!=null)sortedList.add(found);       // if found add to sorted list
            unsortedList.remove(found);        // remove added item
        }
        sortedList.addAll(unsortedList);        // append the reaming items on the unsorted list to new sorted list
        return sortedList;
    }
    else{
        return unsortedList;
    }

}

public static Person getPersonIfFound(List <Person> list, String key){
    for(Person person : list){
        if(person.getId().equals(key)){
            return person;
        }
    }
    return null;
}

Answer 1

An efficient solution is to first create the mapping from the ID in the ids (your desired IDs order) to the index in that list:

val orderById = ids.withIndex().associate { it.value to it.index }

And then sort your list of people by the order of their id in this mapping:

val sortedPeople = people.sortedBy { orderById[it.id] }

Note: if a person has an ID that is not present in the ids, they will be placed first in the list. To place them last, you can use a nullsLast comparator:

val sortedPeople = people.sortedWith(compareBy(nullsLast<String>) { orderById[it.id] })

Answer 2

I would do something like (in pseudocode, since i don't know what your code looks like)

listOfPersons = [{2,Bob},{3,Claire},{1,Alice}]
orderList = [1,3,2]
sortedList = []
for(id in orderList)
    person = listOfPersons.lookup(id)
    sortedList.add(person)  

And the lookup would be easier if you would have a map (id-> person) rather than a listOfPersons.