Longest list in SortedList of Lists
c#, linq, sortedlist
Solution
There's maybe a more efficient way, but you can order by count (of value) descending, and take first.
myList.OrderByDescending(m => m.Value.Count()).First().Key;
of course, if you want all the keys with highest count (they may be multiple values with same length), you should do a group by count.
Something like that.
myList.GroupBy(m => m.Value.Count())
.OrderByDescending(m => m.Key)//I'm the key of the group by
.First()
.Select(g => g.Key);//I'm the key of the SortedList
So if you add to your sample an item with same list length
myList.Add(24, new List<string>());
myList[24].AddRange(new[] {"a", "b", "c", "d", "e"});
you will get 23 And 24.
same could be achieved with
from item in myList
let maxCount = myList.Max(x => x.Value.Count())
where item.Value.Count() == maxCount
select item.Key;
Problem
I have a SortedList of Lists and I am interested in finding the KEY that corresponds to the longest list (list with the most items in it). In code, that looks like: ``` // how the list is defined: var myList = new SortedList<long, List<string>>(); // EXAMPLE data only: myList.Add(0, new List<string>()); myList[0].AddRange(new []{"a", "b", "c"}); myList.Add(8, new List<string>()); myList[8].AddRange(new []{"1", "2"}); myList.Add(23, new List<string>()); myList[23].AddRange(new []{"c", "d", "e", "f", "g"}); ``` In the above example the result should be "23" since that is the key that goes with the longest list. I know how to write this with a for loop, but I think this should be a simple to do with LINQ. That said, I can't seem to get the syntax quite right! Any help is appreciated!