Solved

# How to write LINQ query

Posted on 2014-12-22
82 Views
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
}
}
}
``````
0
Question by:John Bolter
• 2

LVL 32

Accepted Solution

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

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));
}
}
}
``````

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

-saige-
0

Author Closing Comment

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

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());
}
}
``````

Output:
``````totalNumberOfDistinctValue = 5
value = 1, Count = 2
value = 1.1, Count = 2
value = 2, Count = 3
value = 3, Count = 1
value = 4, Count = 1
``````
0

Author Comment

Thanks,
0

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