Patrick Desjardins
Patrick Desjardins

Reputation: 140753

LINQ's Distinct() on a particular property

I am playing with LINQ to learn about it, but I can't figure out how to use Distinct when I do not have a simple list (a simple list of integers is pretty easy to do, this is not the question). What I if want to use Distinct on a List<TElement> on one or more properties of the TElement?

Example: If an object is Person, with property Id. How can I get all Person and use Distinct on them with the property Id of the object?

Person1: Id=1, Name="Test1"
Person2: Id=1, Name="Test1"
Person3: Id=2, Name="Test2"

How can I get just Person1 and Person3? Is that possible?

If it's not possible with LINQ, what would be the best way to have a list of Person depending on some of its properties?

Upvotes: 1505

Views: 1376691

Answers (23)

Contango
Contango

Reputation: 80192

The following code is functionally equivalent to Jon Skeet's answer.

Tested on .NET 4.5, should work on any earlier version of LINQ.

public static IEnumerable<TSource> DistinctBy<TSource, TKey>(
  this IEnumerable<TSource> source, Func<TSource, TKey> keySelector)
{
  HashSet<TKey> seenKeys = new HashSet<TKey>();
  return source.Where(element => seenKeys.Add(keySelector(element)));
}

Incidentially, check out Jon Skeet's latest version of DistinctBy.cs on Google Code.

Update 2022-04-03

Based on an comment by Andrew McClement, best to take John Skeet's answer over this one.

Update 2024-05-15

New home of the project MoreLINQ

Upvotes: 24

Saksham Chaudhary
Saksham Chaudhary

Reputation: 627

Definitely not the most efficient but for those, who are looking for a short and simple answer:

list.Select(x => x.Id).Distinct().Select(x => list.First(y => x == y.Id)).ToList();

Upvotes: 0

Magnetron
Magnetron

Reputation: 8543

Starting with .NET 6, there is new solution using the new DistinctBy() extension in Linq, so we can do:

var distinctPersonsById = personList.DistinctBy(x => x.Id);

The signature of the DistinctBy method:

// Returns distinct elements from a sequence according to a specified
// key selector function.
public static IEnumerable<TSource> DistinctBy<TSource, TKey> (
    this IEnumerable<TSource> source,
    Func<TSource, TKey> keySelector);

Upvotes: 100

karcsi
karcsi

Reputation: 1401

Use:

List<Person> pList = new List<Person>();
/* Fill list */

var result = pList.Where(p => p.Name != null).GroupBy(p => p.Id)
    .Select(grp => grp.FirstOrDefault());

The where helps you filter the entries (could be more complex) and the groupby and select perform the distinct function.

Upvotes: 129

cahit beyaz
cahit beyaz

Reputation: 5117

Solution first group by your fields then select FirstOrDefault item.

List<Person> distinctPeople = allPeople
.GroupBy(p => p.PersonId)
.Select(g => g.FirstOrDefault())
.ToList();

Upvotes: 70

Amy B
Amy B

Reputation: 110071

What if I want to obtain a distinct list based on one or more properties?

Simple! You want to group them and pick a winner out of the group.

List<Person> distinctPeople = allPeople
  .GroupBy(p => p.PersonId)
  .Select(g => g.First())
  .ToList();

If you want to define groups on multiple properties, here's how:

List<Person> distinctPeople = allPeople
  .GroupBy(p => new {p.PersonId, p.FavoriteColor} )
  .Select(g => g.First())
  .ToList();

Note: Certain query providers are unable to resolve that each group must have at least one element, and that First is the appropriate method to call in that situation. If you find yourself working with such a query provider, FirstOrDefault may help get your query through the query provider.

Note2: Consider this answer for an EF Core (prior to EF Core 6) compatible approach. https://stackoverflow.com/a/66529949/8155

Upvotes: 2446

TOL
TOL

Reputation: 470

If you use old .NET version, where the extension method is not built-in, then you may define your own extension method:

public static class EnumerableExtensions
{
    public static IEnumerable<T> DistinctBy<T, TKey>(this IEnumerable<T> enumerable, Func<T, TKey> keySelector)
    {
        return enumerable.GroupBy(keySelector).Select(grp => grp.First());
    }
}

Example of usage:

var personsDist = persons.DistinctBy(item => item.Name);

Upvotes: 0

Timothy Khouri
Timothy Khouri

Reputation: 31845

I've written an article that explains how to extend the Distinct function so that you can do as follows:

var people = new List<Person>();

people.Add(new Person(1, "a", "b"));
people.Add(new Person(2, "c", "d"));
people.Add(new Person(1, "a", "b"));

foreach (var person in people.Distinct(p => p.ID))
    // Do stuff with unique list here.

Here's the article (now in the Web Archive): Extending LINQ - Specifying a Property in the Distinct Function

Upvotes: 13

Harry .Naeem
Harry .Naeem

Reputation: 1301

You can use DistinctBy() for getting Distinct records by an object property. Just add the following statement before using it:

using Microsoft.Ajax.Utilities;

and then use it like following:

var listToReturn = responseList.DistinctBy(x => x.Index).ToList();

where 'Index' is the property on which i want the data to be distinct.

Upvotes: 10

Waldemar Gałęzinowski
Waldemar Gałęzinowski

Reputation: 1233

Override Equals(object obj) and GetHashCode() methods:

class Person
{
    public int Id { get; set; }
    public int Name { get; set; }

    public override bool Equals(object obj)
    {
        return ((Person)obj).Id == Id;
        // or: 
        // var o = (Person)obj;
        // return o.Id == Id && o.Name == Name;
    }
    public override int GetHashCode()
    {
        return Id.GetHashCode();
    }
}

and then just call:

List<Person> distinctList = new[] { person1, person2, person3 }.Distinct().ToList();

Upvotes: 4

Mohamed Hammam
Mohamed Hammam

Reputation: 11

Please give a try with below code.

var Item = GetAll().GroupBy(x => x .Id).ToList();

Upvotes: -4

Chuck Rostance
Chuck Rostance

Reputation: 7104

You could also use query syntax if you want it to look all LINQ-like:

var uniquePeople = from p in people
                   group p by new {p.ID} //or group by new {p.ID, p.Name, p.Whatever}
                   into mygroup
                   select mygroup.FirstOrDefault();

Upvotes: 89

Caspian Canuck
Caspian Canuck

Reputation: 1490

If you don't want to add the MoreLinq library to your project just to get the DistinctBy functionality then you can get the same end result using the overload of Linq's Distinct method that takes in an IEqualityComparer argument.

You begin by creating a generic custom equality comparer class that uses lambda syntax to perform custom comparison of two instances of a generic class:

public class CustomEqualityComparer<T> : IEqualityComparer<T>
{
    Func<T, T, bool> _comparison;
    Func<T, int> _hashCodeFactory;

    public CustomEqualityComparer(Func<T, T, bool> comparison, Func<T, int> hashCodeFactory)
    {
        _comparison = comparison;
        _hashCodeFactory = hashCodeFactory;
    }

    public bool Equals(T x, T y)
    {
        return _comparison(x, y);
    }

    public int GetHashCode(T obj)
    {
        return _hashCodeFactory(obj);
    }
}

Then in your main code you use it like so:

Func<Person, Person, bool> areEqual = (p1, p2) => int.Equals(p1.Id, p2.Id);

Func<Person, int> getHashCode = (p) => p.Id.GetHashCode();

var query = people.Distinct(new CustomEqualityComparer<Person>(areEqual, getHashCode));

Voila! :)

The above assumes the following:

  • Property Person.Id is of type int
  • The people collection does not contain any null elements

If the collection could contain nulls then simply rewrite the lambdas to check for null, e.g.:

Func<Person, Person, bool> areEqual = (p1, p2) => 
{
    return (p1 != null && p2 != null) ? int.Equals(p1.Id, p2.Id) : false;
};

EDIT

This approach is similar to the one in Vladimir Nesterovsky's answer but simpler.

It is also similar to the one in Joel's answer but allows for complex comparison logic involving multiple properties.

However, if your objects can only ever differ by Id then another user gave the correct answer that all you need to do is override the default implementations of GetHashCode() and Equals() in your Person class and then just use the out-of-the-box Distinct() method of Linq to filter out any duplicates.

Upvotes: 4

Arindam
Arindam

Reputation: 176

List<Person>lst=new List<Person>
        var result1 = lst.OrderByDescending(a => a.ID).Select(a =>new Player {ID=a.ID,Name=a.Name} ).Distinct();

Upvotes: 1

David Fahlander
David Fahlander

Reputation: 5646

You can do this with the standard Linq.ToLookup(). This will create a collection of values for each unique key. Just select the first item in the collection

Persons.ToLookup(p => p.Id).Select(coll => coll.First());

Upvotes: 35

Vladimir Nesterovsky
Vladimir Nesterovsky

Reputation: 649

When we faced such a task in our project we defined a small API to compose comparators.

So, the use case was like this:

var wordComparer = KeyEqualityComparer.Null<Word>().
    ThenBy(item => item.Text).
    ThenBy(item => item.LangID);
...
source.Select(...).Distinct(wordComparer);

And API itself looks like this:

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

public static class KeyEqualityComparer
{
    public static IEqualityComparer<T> Null<T>()
    {
        return null;
    }

    public static IEqualityComparer<T> EqualityComparerBy<T, K>(
        this IEnumerable<T> source,
        Func<T, K> keyFunc)
    {
        return new KeyEqualityComparer<T, K>(keyFunc);
    }

    public static KeyEqualityComparer<T, K> ThenBy<T, K>(
        this IEqualityComparer<T> equalityComparer,
        Func<T, K> keyFunc)
    {
        return new KeyEqualityComparer<T, K>(keyFunc, equalityComparer);
    }
}

public struct KeyEqualityComparer<T, K>: IEqualityComparer<T>
{
    public KeyEqualityComparer(
        Func<T, K> keyFunc,
        IEqualityComparer<T> equalityComparer = null)
    {
        KeyFunc = keyFunc;
        EqualityComparer = equalityComparer;
    }

    public bool Equals(T x, T y)
    {
        return ((EqualityComparer == null) || EqualityComparer.Equals(x, y)) &&
                EqualityComparer<K>.Default.Equals(KeyFunc(x), KeyFunc(y));
    }

    public int GetHashCode(T obj)
    {
        var hash = EqualityComparer<K>.Default.GetHashCode(KeyFunc(obj));

        if (EqualityComparer != null)
        {
            var hash2 = EqualityComparer.GetHashCode(obj);

            hash ^= (hash2 << 5) + hash2;
        }

        return hash;
    }

    public readonly Func<T, K> KeyFunc;
    public readonly IEqualityComparer<T> EqualityComparer;
}

More details is on our site: IEqualityComparer in LINQ.

Upvotes: 5

gcoleman0828
gcoleman0828

Reputation: 1542

The best way to do this that will be compatible with other .NET versions is to override Equals and GetHash to handle this (see Stack Overflow question This code returns distinct values. However, what I want is to return a strongly typed collection as opposed to an anonymous type), but if you need something that is generic throughout your code, the solutions in this article are great.

Upvotes: 2

Andrzej Gis
Andrzej Gis

Reputation: 14306

In case you need a Distinct method on multiple properties, you can check out my PowerfulExtensions library. Currently it's in a very young stage, but already you can use methods like Distinct, Union, Intersect, Except on any number of properties;

This is how you use it:

using PowerfulExtensions.Linq;
...
var distinct = myArray.Distinct(x => x.A, x => x.B);

Upvotes: 5

Jon Skeet
Jon Skeet

Reputation: 1499660

EDIT: This is now part of MoreLINQ.

What you need is a "distinct-by" effectively. I don't believe it's part of LINQ as it stands, although it's fairly easy to write:

public static IEnumerable<TSource> DistinctBy<TSource, TKey>
    (this IEnumerable<TSource> source, Func<TSource, TKey> keySelector)
{
    HashSet<TKey> seenKeys = new HashSet<TKey>();
    foreach (TSource element in source)
    {
        if (seenKeys.Add(keySelector(element)))
        {
            yield return element;
        }
    }
}

So to find the distinct values using just the Id property, you could use:

var query = people.DistinctBy(p => p.Id);

And to use multiple properties, you can use anonymous types, which implement equality appropriately:

var query = people.DistinctBy(p => new { p.Id, p.Name });

Untested, but it should work (and it now at least compiles).

It assumes the default comparer for the keys though - if you want to pass in an equality comparer, just pass it on to the HashSet constructor.

Upvotes: 1545

Joel
Joel

Reputation: 7569

Personally I use the following class:

public class LambdaEqualityComparer<TSource, TDest> : 
    IEqualityComparer<TSource>
{
    private Func<TSource, TDest> _selector;

    public LambdaEqualityComparer(Func<TSource, TDest> selector)
    {
        _selector = selector;
    }

    public bool Equals(TSource obj, TSource other)
    {
        return _selector(obj).Equals(_selector(other));
    }

    public int GetHashCode(TSource obj)
    {
        return _selector(obj).GetHashCode();
    }
}

Then, an extension method:

public static IEnumerable<TSource> Distinct<TSource, TCompare>(
    this IEnumerable<TSource> source, Func<TSource, TCompare> selector)
{
    return source.Distinct(new LambdaEqualityComparer<TSource, TCompare>(selector));
}

Finally, the intended usage:

var dates = new List<DateTime>() { /* ... */ }
var distinctYears = dates.Distinct(date => date.Year);

The advantage I found using this approach is the re-usage of LambdaEqualityComparer class for other methods that accept an IEqualityComparer. (Oh, and I leave the yield stuff to the original LINQ implementation...)

Upvotes: 10

Ivan
Ivan

Reputation: 1863

I think it is enough:

list.Select(s => s.MyField).Distinct();

Upvotes: 92

GWLlosa
GWLlosa

Reputation: 24403

You should be able to override Equals on person to actually do Equals on Person.id. This ought to result in the behavior you're after.

Upvotes: 0

mqp
mqp

Reputation: 71937

You can do it (albeit not lightning-quickly) like so:

people.Where(p => !people.Any(q => (p != q && p.Id == q.Id)));

That is, "select all people where there isn't another different person in the list with the same ID."

Mind you, in your example, that would just select person 3. I'm not sure how to tell which you want, out of the previous two.

Upvotes: 4

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