How can I get a value in a List

Question

How can I get RSI_Value in the rsi_Wilder by the IdDate? I sort my rsi_Wilder, so the highest IdDate always will be the last index. Therefore I can do this:

class RSI
{
    public int IdDate { get; set; }
    public decimal RSI_Value { get; set; }
}

List<Datamodel.RSI> rsi_Wilder = BeregnRsi_Wilder(idVirksomhedensStamdata, BeregnAntalDage: 21);
var lastValue = rsi_Wilder[rsi_Wilder.Count - 1].RSI_Value;

It works fine, but I think there must be a better way/better performance to find to RSI_Value by the highest IdDate. What's the best way to do this?

EDIT: Example.

rsi_wilder.Add(new Datamodel.RSI{IdDate = 25, RSI_Value = 17});
rsi_wilder.Add(new Datamodel.RSI{IdDate = 26, RSI_Value = 26});
rsi_wilder.Add(new Datamodel.RSI{IdDate = 27, RSI_Value = 9});
rsi_wilder.Add(new Datamodel.RSI{IdDate = 31, RSI_Value = 70});
rsi_wilder.Add(new Datamodel.RSI{IdDate = 32, RSI_Value = 55});

I need to get the RSI_Value with the highest IdDate and I don't know the IdDate number. The result I seek is RSI_Value = 55. The code var lastValue = rsi_Wilder[rsi_Wilder.Count - 1].RSI_Value; do the job, because the IdDate in rsi_Wilder is sorted from lowest to highest number.

You mean, how could you get the highest value if `BeregnRsi_Wilder` happened to return the results in reverse or unsorted order? — Chris Sinclair, Jul 23 '15 at 18:46
"better way" does it mean by better performance? Do you get the `rsi_Wilder ` ordered? — Hamlet Hakobyan, Jul 23 '15 at 18:47
Maybe you could use a different data structure for better performance. SortedList or SortedSet for example, depends on the needs of your application. — James Buck, Jul 23 '15 at 18:49
You mean this? `var max = rsi_Wilder.Max(rsi => rsi.RSI_Value);` ?? — Lasse V. Karlsen, Jul 23 '15 at 18:51
Hanlet, with a better way i mean better performance. I get the rsi_Wilder ordered. I update my question in a minut. — MHP, Jul 23 '15 at 19:04

score 3 · Accepted Answer · edited May 23 '17 at 12:07

3

Ordering the entire sequence costs O(n*log(n)), and it's not really needed if all you want is the max of your collection.

You can find the max in O(n) just by iterating over the sequence, and keeping track of the maximum element that you have encountered. You might even create an extension method to do it for you, or just do it using LINQ:

var result = yourcollection.Aggregate((x,y) => x.IdDate > y.IdDate ? x : y);

you can find many more examples here

edited May 23 '17 at 12:07

Community

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answered Jul 23 '15 at 18:52

Save

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Was about to post an answer similar to this, due to that O(n) would be the most effective. Upvoted! – Bauss Jul 23 '15 at 18:55

score 2 · Answer 2 · answered Jul 23 '15 at 18:56

2

using System.Linq;

var lastValue = rsi_Wilder.Max(x => x.RSI_Value);

answered Jul 23 '15 at 18:56

Kevin

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According to his description, he wants the max RSI_Value by IdDate; this just returns the max RSI_Value. – Save Jul 23 '15 at 19:10
I can not use rsi_Wilder.Max(x => x.RSI_Value); it should be something like this rsi_Wilder.Max(x => x.Id_Date).select(x => x.RSI_Value); of course it is not going to work. – MHP Jul 23 '15 at 19:16

score -1 · Answer 3 · answered Jul 23 '15 at 18:49

-1

decimal rsiValueOfHighestId =
    rsi_Wilder.OrderByDescending(rsi => rsi.IdDate).First().RSI_Value;

answered Jul 23 '15 at 18:49

Jashaszun

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3 Answers3