英文:
How to combine two object lists by id and select non-null values in kotlin
问题
我想要合并两个对象列表。每个对象都有一个特定的id。如果同一个对象存在于两个列表中，我想要将它们合并。而且如果一个值为空，我想要获取非空的值。例如：val list1 = listOf(Object(id = 1, hello, 100),Object(id = 2, null, 40))val list2 = listOf(Object(id = 1, null, 100),Object(id = 2, test, 40),Object(id = 3, hi, 13))我想要实现的结果如下：val result = listOf(Object(id = 1, hello, 100),Object(id = 2, test, 40),Object(id = 3, hi, 13))**注意：如果一个值不为空，我知道它是相同的。**
英文:
I want to combine two object lists. Each object has a specific id. If the same object exists in two lists, I want to merge them. And if a value is null I want to get the non-null one. For example:
val list1 = listOf(Object(id = 1, hello, 100),Object(id = 2, null, 40))val list2 = listOf(Object(id = 1, null, 100),Object(id = 2, test, 40),Object(id = 3, hi, 13))
The result I want to achieve is as follows:
val result = listOf(Object(id = 1, hello, 100),Object(id = 2, test, 40),Object(id = 3, hi, 13))
NOTE: if a value is not empty I know it is not different.
答案1
得分: 3
fun main() {val list1 = listOf(Obj(id = 1, "你好", 100),Obj(id = 2, null, 40))val list2 = listOf(Obj(id = 1, null, 100),Obj(id = 2, "测试", 40),Obj(id = 3, "嗨", 13))val result = (list1.asSequence() + list2).groupingBy { it.id }.reduce { _, accumulator: Obj, element: Obj ->accumulator.copy(second = accumulator.second ?: element.second)}.values.toList()println(result)}data class Obj(val id: Int, val second: String?, val third: Int)
英文:
fun main() {val list1 = listOf(Obj(id = 1, "hello", 100),Obj(id = 2, null, 40))val list2 = listOf(Obj(id = 1, null, 100),Obj(id = 2, "test", 40),Obj(id = 3, "hi", 13))val result = (list1.asSequence() + list2).groupingBy { it.id }.reduce { _, accumulator: Obj, element: Obj ->accumulator.copy(second = accumulator.second ?: element.second)}.values.toList()println(result)}data class Obj(val id: Int, val second: String?, val third: Int)
答案2
得分: 0
有两种方法可以解决这个问题，一种是O(n^2)的方法，其中你要遍历两个列表并检查每个对象。另一种是O(n)的方法，其中你会浪费一些额外的空间，通过在一个字典中创建映射，然后在一个列表上进行循环，同时在另一个列表上进行检查。
英文:
There can be 2 approaches to the problem, O(n^2) approach where you loop through both the lists and check for each object. Or O(n) approach where you waste some extra space by creating a mapping in one dictionary for your list and looping on one while checking in other.
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