[英]Get items from List A where Id is common in both List A and List B and counter of that Id is more than 1 in List A or List B
I was looking to get items from ListA, where the value of Id is same in both of the lists, and the count of Id must be more than 1 in list A or list B我想从 ListA 中获取项目,其中两个列表中的 Id 值相同,并且 Id 的计数在列表 A 或列表 B 中必须大于 1
var items = itemsA.Where(x => itemsB.Select(y => y.Id == x.Id).Count() > 1);
This gives me the result where same Ids in itemsB is more then 1, I want to use a or condition to check for the same counter in itemsA这给我的结果是 itemsB 中的相同 ID 大于 1,我想使用 or 条件来检查 itemsA 中的相同计数器
Eg 1:
ListA=[{"id"=1,"name="abc"},{"id=1, "name"="def"}]
ListB=[{"id=2","name="xyz"}, {"id=1, "name"="mno"}]
Should return [{"id"=1,"name="abc"},{"id=1, "name"="def"}] because id =1 exists in listB and the count of id with value 1 in listA is more then 1.
Eg 2:
ListA=[{"id"=2,"name="abc"},{"id=1, "name"="def"}]
ListB=[{"id=1","name="xyz"}, {"id=1, "name"="mno"}]
should return {"id=1, "name"="def"} because common id in both list is 1 and the count of id with value 1 in ListB is more then 1.
I am not certain this is the best solution, but as far as I've understood the question, it should be a solution.我不确定这是最好的解决方案,但据我了解这个问题,它应该是一个解决方案。
Assuming you have an Item
class as follows:假设你有一个
Item
class 如下:
public class Item
{
public int Id { get; set; }
public string Name { get; set; }
}
and define itemsA
and itemsB
as List<Item>
s, you can first find all Id
s that are present in both lists, then select the applicable items from itemsA
based on occurrence of each Id
in either list:并将
itemsA
和itemsB
定义为List<Item>
s,您可以首先找到两个列表中都存在的所有Id
s,然后根据任一列表中每个Id
的出现从itemsA
中找到适用的项目 select:
IEnumerable<int> idsInBothItemLists = itemsA
.Select(a => a.Id)
.Intersect(itemsB.Select(b => b.Id))
.Distinct();
List<Item> items = itemsA
.Where(a => idsInBothItemLists.Contains(a.Id))
.GroupBy(a => a.Id)
.Where(gr =>
gr.Skip(1).Any() ||
itemsB.Where(b => b.Id == gr.Key).Skip(1).Any())
.SelectMany(gr => gr.Select(item => item))
.ToList();
( .Skip(1).Any()
serves the same purpose as .Count() > 1
in your original code; it simply checks whether there are any items left after skipping the first item.) (
.Skip(1).Any()
.Count() > 1
具有相同的目的;它只是检查在跳过第一个项目后是否还有剩余的项目。)
Printing the output from the suggested population of itemsA
and itemsB
从建议的项目 A 和
itemsB
中打印itemsA
foreach (var entry in items)
{
Console.WriteLine(entry.Id + " " + entry.Name);
}
eg for input例如输入
var itemsA = new List<Item>
{
new Item { Id = 1, Name = "abc" },
new Item { Id = 3, Name = "def" },
new Item { Id = 1, Name = "ghi" },
new Item { Id = 2, Name = "jkl" }
};
var itemsB = new List<Item>
{
new Item { Id = 2, Name = "xyz" },
new Item { Id = 2, Name = "jkl" },
new Item { Id = 1, Name = "mno" },
new Item { Id = 3, Name = "pqr" }
};
gives给
1 abc
1 个字母
1 ghi1 吉
2 jkl2 jkl
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.