Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
719 views
in Technique[技术] by (71.8m points)

linq - Good way to get the key of the highest value of a Dictionary in C#

I'm trying to get the key of the maximum value in the Dictionary<string, double> results.

This is what I have so far:

double max = results.Max(kvp => kvp.Value);
return results.Where(kvp => kvp.Value == max).Select(kvp => kvp.Key).First();

However, since this seems a little inefficient, I was wondering whether there was a better way to do this.

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

I think this is the most readable O(n) answer using standard LINQ.

var max = results.Aggregate((l, r) => l.Value > r.Value ? l : r).Key;

edit: explanation for CoffeeAddict

Aggregate is the LINQ name for the commonly known functional concept Fold

It loops over each element of the set and applies whatever function you provide. Here, the function I provide is a comparison function that returns the bigger value. While looping, Aggregate remembers the return result from the last time it called my function. It feeds this into my comparison function as variable l. The variable r is the currently selected element.

So after aggregate has looped over the entire set, it returns the result from the very last time it called my comparison function. Then I read the .Key member from it because I know it's a dictionary entry

Here is a different way to look at it [I don't guarantee that this compiles ;) ]

var l = results[0];
for(int i=1; i<results.Count(); ++i)
{
    var r = results[i];
    if(r.Value > l.Value)
        l = r;        
}
var max = l.Key;

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...