根据另一个列表的顺序对列表进行排序 [英] Sorting list based on another list's order

问题描述

我需要对Person个对象的列表进行排序(List<Person>，其中每个Person对象具有一些属性，例如id(唯一)，name，age…等).

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

排序顺序基于另一个列表.该列表包含一组Person id(已排序的A List<String>).

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

使用Kotlin或Java以与id列表相同的顺序订购List<Person>的最佳方法是什么.

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

示例:

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

订购的ID列表:

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

排序的Person列表应为:

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

如果Person列表包含id列表中未提及的任何id，则这些值应位于已排序列表的末尾.

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.

这是我目前使用Java的方式.我希望有一个比这更好的方法:

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;
}

推荐答案

一种有效的解决方案是首先创建从ids中的ID(您所需的ID顺序)到该列表中的索引的映射:

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 }

，然后在此映射中按id的顺序对people列表进行排序:

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

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

注意:如果某人具有ids中不存在的ID，则他们将被放置在列表的第一位.要将它们放置在最后，可以使用 nullsLast 比较器:

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] })

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