Solved

How to write LINQ query

Posted on 2014-12-22
4
82 Views
Last Modified: 2016-02-15
Hi, yet another LINQ query. My apologies I cannot get the hang of this. This should be simple. My code is below.
Thank you
John

using System;
using System.Collections.Generic;
using System.Linq;

namespace ConsoleApplication5
{
    class Program
    {
        static void Main(string[] args)
        {
            Dictionary<int, List<Single>> test = new Dictionary<int, List<Single>>();

            test.Add(1, new List<Single> { 1f, 2f, 3f, 4f });
            test.Add(15, new List<Single> { 1.1f, 2f });
            test.Add(99, new List<Single> { 1.1f, 1f, 2f });

            int totalNumberOfValues = test.Sum(n => n.Value.Count());   //returns 9 which is correct

            //the distinct values are 1f, 2f, 3f, 4f, 1.1f, counting them there are 5 values
            //I want to count them
            int totalNumberOfDistinctValue = what goes here;      //I want this to be 5
        }
    }
}

Open in new window

0
Comment
Question by:John Bolter
  • 2
4 Comments
 
LVL 32

Accepted Solution

by:
it_saige earned 500 total points
Comment Utility
You would use:
int totalNumberOfDistinctValue = test.SelectMany(n => n.Value).Distinct().Count();

Open in new window


Example -
using System;
using System.Collections.Generic;
using System.Linq;

namespace EE_Q28585282
{
	class Program
	{
		static void Main(string[] args)
		{
			Dictionary<int, List<Single>> test = new Dictionary<int, List<Single>>();

			test.Add(1, new List<Single> { 1f, 2f, 3f, 4f });
			test.Add(15, new List<Single> { 1.1f, 2f });
			test.Add(99, new List<Single> { 1.1f, 1f, 2f });

			int totalNumberOfValues = test.Sum(n => n.Value.Count());   //returns 9 which is correct
			Console.WriteLine(string.Format("Total number of values: {0}", totalNumberOfValues));

			//the distinct values are 1f, 2f, 3f, 4f, 1.1f, counting them there are 5 values
			//I want to count them
			int totalNumberOfDistinctValue = test.SelectMany(n => n.Value).Distinct().Count();      //I want this to be 5
			Console.WriteLine(string.Format("Total number of distinct values: {0}", totalNumberOfDistinctValue));
			Console.ReadLine();
		}
	}
}

Open in new window


Produces the following output -Capture.JPG
More on SelectMany - http://msdn.microsoft.com/en-us/library/system.linq.enumerable.selectmany%28v=vs.100%29.aspx

-saige-
0
 

Author Closing Comment

by:John Bolter
Comment Utility
Oh of course.

I even wrote

test.Select(n => n.Value).Distinct().Count();

that didn't work. I wasn't even away of SelectMany.

Thank you!
0
 
LVL 85

Expert Comment

by:Mike Tomlinson
Comment Utility
If you're interested in what those distinct values are, and how many of them there were, then:
        static void Main(string[] args)
        {
            Dictionary<int, List<Single>> test = new Dictionary<int, List<Single>>();

            test.Add(1, new List<Single> { 1f, 2f, 3f, 4f });
            test.Add(15, new List<Single> { 1.1f, 2f });
            test.Add(99, new List<Single> { 1.1f, 1f, 2f });

            int totalNumberOfValues = test.Sum(n => n.Value.Count());  

            var distinctValuesWithCount = test.SelectMany(x => x.Value) // flatten the dictionary values to a single List<Single>
                .GroupBy(x => x) // grouped by each Single in the List
                .Select(group => new // create an anonymous type with the value and number of occurrences of that Single
                {
                    Value = group.Key,
                    Count = group.Count()
                })
                .OrderBy(x => x.Value); // sorted ascending by the Single

            int totalNumberOfDistinctValue = distinctValuesWithCount.Count();

            Console.WriteLine("totalNumberOfDistinctValue = " + totalNumberOfDistinctValue.ToString());
            foreach(var g in distinctValuesWithCount)
            {
                Console.WriteLine("value = " + g.Value.ToString() + ", Count = " + g.Count.ToString());
            }
            Console.ReadLine();
        }

Open in new window


Output:
totalNumberOfDistinctValue = 5
value = 1, Count = 2
value = 1.1, Count = 2
value = 2, Count = 3
value = 3, Count = 1
value = 4, Count = 1

Open in new window

0
 

Author Comment

by:John Bolter
Comment Utility
Thanks,
0

Featured Post

IT, Stop Being Called Into Every Meeting

Highfive is so simple that setting up every meeting room takes just minutes and every employee will be able to start or join a call from any room with ease. Never be called into a meeting just to get it started again. This is how video conferencing should work!

Join & Write a Comment

Introduction Hi all and welcome to my first article on Experts Exchange. A while ago, someone asked me if i could do some tutorials on object oriented programming. I decided to do them on C#. Now you may ask me, why's that? Well, one of the re…
This article is for Object-Oriented Programming (OOP) beginners. An Interface contains declarations of events, indexers, methods and/or properties. Any class which implements the Interface should provide the concrete implementation for each Inter…
Access reports are powerful and flexible. Learn how to create a query and then a grouped report using the wizard. Modify the report design after the wizard is done to make it look better. There will be another video to explain how to put the final p…
In this tutorial you'll learn about bandwidth monitoring with flows and packet sniffing with our network monitoring solution PRTG Network Monitor (https://www.paessler.com/prtg). If you're interested in additional methods for monitoring bandwidt…

728 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question

Need Help in Real-Time?

Connect with top rated Experts

9 Experts available now in Live!

Get 1:1 Help Now