Sorting a list based on another list
I need to sort a list of objects Person
( List<Person>
where each object Person
has several attributes, such as the id
(unique) name
, age
... etc).
The sort order is based on a different list. This list contains the set Person
id
(A List<String>
, which is already sorted).
What is the best way to order List<Person>
in the same order as a list id
using Kotlin or Java.
Example:
List Person {
("ID1","PERSON1",22,..), ("ID-2","PERSON2",20,..) ), ("ID-3","PERSON3",19,..),…..
}
List of ordered identifiers:
List of ID {("ID2"), ("ID1"),("ID3")….}
The sorted Person
list should be:
List PERSON {
("ID-2","PERSON 2",20,..) ), ("ID1","PERSON 2",22,..), ("ID-3","PERSON 2",19,..),…..
}
If the list Person
contains anything id
that is not in the list id
, then these values must be at the end of the sorted list.
Edited: This is my current way in Java. I hope 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;
}
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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 search will be easier if you have a card (id-> person) and not listOfPersons.
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An efficient solution is to first create a mapping from ID in ids
(your desired order of IDs) to an index in that list:
val orderById = ids.withIndex().associate { it.value to it.index }
And then sort your list people
in order of them id
in this mapping:
val sortedPeople = people.sortedBy { orderById[it.id] }
Note. If the person has an identifier that is not in ids
, it will be placed first in the list. To put them one last time, you can use a nullsLast
comparator:
val sortedPeople = people.sortedWith(compareBy(nullsLast<String>) { orderById[it.id] })
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Try the below code.
Map<String, Person> personMap=new HashMap<>(); // create a map with key as ID and value as Person object
List<String> orderList=new ArrayList<>(); // create a list or array with ID order
List<Person> outputList=new ArrayList<>(); //list to hold sorted values
//logic
// to sort Person based on ID list order
for (String order : orderList) {
if(personMap.containsKey(order)){
outputList.add(personMap.get(order));
personMap.remove(order);
}
}
// logic to add the Person object whose id is not present in ID order list
for (Entry<String, Person> entry : personMap.entrySet())
{
int lastIndex=outputList.size();
outputList.add(lastIndex, entry.getValue());
lastIndex++;
}
outputList
Will now have the values you expect ...
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The following code converts the list Person to Map
, where the key will be the identifier and the value will be the Person object itself. This Map
will help you find quickly. Then go through the list of IDs to get the value from Map
and appended to another List
.
fun main(args: Array<String>) {
// List of ID
val listId = listOf(2, 1, 3)
val list = listOf(Person(id = 1, name = "A"),
Person(id = 2, name = "B"),
Person(id = 3, name = "C"))
val map: Map<Int, Person> = list.associateBy ({it.id}, {it})
val sortedList = mutableListOf<Person>()
listId.forEach({
sortedList.add(map[it]!!)
})
sortedList.forEach({
println(it)
})
}
data class Person(val id: Int, val name: String)
Output
Person(id=2, name=B)
Person(id=1, name=A)
Person(id=3, name=C)
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