Welcome to Experts Exchange

Add your voice to the tech community where 5M+ people, just like you, are talking about what matters.

  • Help others & share knowledge
  • Earn cash & points
  • Learn & ask questions
Solved

How to write LINQ query

Posted on 2014-12-22
4
88 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 33

Accepted Solution

by:
it_saige earned 500 total points
ID: 40513104
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
ID: 40513139
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
ID: 40513231
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
ID: 40514441
Thanks,
0

Featured Post

Networking for the Cloud Era

Join Microsoft and Riverbed for a discussion and demonstration of enhancements to SteelConnect:
-One-click orchestration and cloud connectivity in Azure environments
-Tight integration of SD-WAN and WAN optimization capabilities
-Scalability and resiliency equal to a data center

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

Suggested Solutions

Article by: Najam
Having new technologies does not mean they will completely replace old components.  Recently I had to create WCF that will be called by VB6 component.  Here I will describe what steps one should follow while doing so, please feel free to post any qu…
This article aims to explain the working of CircularLogArchiver. This tool was designed to solve the buildup of log file in cases where systems do not support circular logging or where circular logging is not enabled
The Email Laundry PDF encryption service allows companies to send confidential encrypted  emails to anybody. The PDF document can also contain attachments that are embedded in the encrypted PDF. The password is randomly generated by The Email Laundr…

808 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