Remove duplicates from a List<T> in C#

Question

Anyone have a quick method for de-duplicating a generic List in C#?

Answer 1

Perhaps you should consider using a HashSet.

From the MSDN link:

using System;
using System.Collections.Generic;

class Program
{
    static void Main()
    {
        HashSet<int> evenNumbers = new HashSet<int>();
        HashSet<int> oddNumbers = new HashSet<int>();

        for (int i = 0; i < 5; i++)
        {
            // Populate numbers with just even numbers.
            evenNumbers.Add(i * 2);

            // Populate oddNumbers with just odd numbers.
            oddNumbers.Add((i * 2) + 1);
        }

        Console.Write("evenNumbers contains {0} elements: ", evenNumbers.Count);
        DisplaySet(evenNumbers);

        Console.Write("oddNumbers contains {0} elements: ", oddNumbers.Count);
        DisplaySet(oddNumbers);

        // Create a new HashSet populated with even numbers.
        HashSet<int> numbers = new HashSet<int>(evenNumbers);
        Console.WriteLine("numbers UnionWith oddNumbers...");
        numbers.UnionWith(oddNumbers);

        Console.Write("numbers contains {0} elements: ", numbers.Count);
        DisplaySet(numbers);
    }

    private static void DisplaySet(HashSet<int> set)
    {
        Console.Write("{");
        foreach (int i in set)
        {
            Console.Write(" {0}", i);
        }
        Console.WriteLine(" }");
    }
}

/* This example produces output similar to the following:
 * evenNumbers contains 5 elements: { 0 2 4 6 8 }
 * oddNumbers contains 5 elements: { 1 3 5 7 9 }
 * numbers UnionWith oddNumbers...
 * numbers contains 10 elements: { 0 2 4 6 8 1 3 5 7 9 }
 */

Answer 2

If you're using .Net 3+, you can use Linq.

List<T> withDupes = LoadSomeData();
List<T> noDupes = withDupes.Distinct().ToList();

Answer 3

How about:-

var noDupes = list.Distinct().ToList();

In .net 3.5?

Answer 4

Simply initialize a HashSet with a List of the same type:

var noDupes = new HashSet<T>(withDupes);

Or, if you want a List returned:

var noDupsList = new HashSet<T>(withDupes).ToList();

Answer 5

Sort it, then check two and two next to each others, as the duplicates will clump together.

Something like this:

list.Sort();
Int32 index = list.Count - 1;
while (index > 0)
{
    if (list[index] == list[index - 1])
    {
        if (index < list.Count - 1)
            (list[index], list[list.Count - 1]) = (list[list.Count - 1], list[index]);
        list.RemoveAt(list.Count - 1);
        index--;
    }
    else
        index--;
}

Notes:

• Comparison is done from back to front, to avoid having to resort list after each removal
• This example now uses C# Value Tuples to do the swapping, substitute with appropriate code if you can't use that
• The end-result is no longer sorted

Answer 6

It worked for me. simply use

List<Type> liIDs = liIDs.Distinct().ToList<Type>();

Replace "Type" with your desired type e.g. int.

Answer 7

I like to use this command:

List<Store> myStoreList = Service.GetStoreListbyProvince(provinceId)
                                                 .GroupBy(s => s.City)
                                                 .Select(grp => grp.FirstOrDefault())
                                                 .OrderBy(s => s.City)
                                                 .ToList();

I have these fields in my list: Id, StoreName, City, PostalCode I wanted to show list of cities in a dropdown which has duplicate values. solution: Group by city then pick the first one for the list.

I hope it helps :)

Answer 8

As kronoz said in .Net 3.5 you can use Distinct().

In .Net 2 you could mimic it:

public IEnumerable<T> DedupCollection<T> (IEnumerable<T> input) 
{
    var passedValues = new HashSet<T>();

    // Relatively simple dupe check alg used as example
    foreach(T item in input)
        if(passedValues.Add(item)) // True if item is new
            yield return item;
}

This could be used to dedupe any collection and will return the values in the original order.

It's normally much quicker to filter a collection (as both Distinct() and this sample does) than it would be to remove items from it.

Answer 9

An extension method might be a decent way to go... something like this:

public static List<T> Deduplicate<T>(this List<T> listToDeduplicate)
{
    return listToDeduplicate.Distinct().ToList();
}

And then call like this, for example:

List<int> myFilteredList = unfilteredList.Deduplicate();

Answer 10

In Java (I assume C# is more or less identical):

list = new ArrayList<T>(new HashSet<T>(list))

If you really wanted to mutate the original list:

List<T> noDupes = new ArrayList<T>(new HashSet<T>(list));
list.clear();
list.addAll(noDupes);

To preserve order, simply replace HashSet with LinkedHashSet.

Answer 11

Use Linq's Union method.

Note: This solution requires no knowledge of Linq, aside from that it exists.

Code

Begin by adding the following to the top of your class file:

using System.Linq;

Now, you can use the following to remove duplicates from an object called, obj1:

obj1 = obj1.Union(obj1).ToList();

Note: Rename obj1 to the name of your object.

How it works

1. The Union command lists one of each entry of two source objects. Since obj1 is both source objects, this reduces obj1 to one of each entry.

2. The ToList() returns a new List. This is necessary, because Linq commands like Union returns the result as an IEnumerable result instead of modifying the original List or returning a new List.

Answer 12

If you don't care about the order you can just shove the items into a HashSet, if you do want to maintain the order you can do something like this:

var unique = new List<T>();
var hs = new HashSet<T>();
foreach (T t in list)
    if (hs.Add(t))
        unique.Add(t);

Or the Linq way:

var hs = new HashSet<T>();
list.All( x =>  hs.Add(x) );

Edit: The HashSet method is O(N) time and O(N) space while sorting and then making unique (as suggested by @lassevk and others) is O(N*lgN) time and O(1) space so it's not so clear to me (as it was at first glance) that the sorting way is inferior (my apologies for the temporary down vote...)

Answer 13

Here's an extension method for removing adjacent duplicates in-situ. Call Sort() first and pass in the same IComparer. This should be more efficient than Lasse V. Karlsen's version which calls RemoveAt repeatedly (resulting in multiple block memory moves).

public static void RemoveAdjacentDuplicates<T>(this List<T> List, IComparer<T> Comparer)
{
    int NumUnique = 0;
    for (int i = 0; i < List.Count; i++)
        if ((i == 0) || (Comparer.Compare(List[NumUnique - 1], List[i]) != 0))
            List[NumUnique++] = List[i];
    List.RemoveRange(NumUnique, List.Count - NumUnique);
}

Answer 14

As a helper method (without Linq):

public static List<T> Distinct<T>(this List<T> list)
{
    return (new HashSet<T>(list)).ToList();
}

Answer 15

Installing the MoreLINQ package via Nuget, you can easily distinct object list by a property

IEnumerable<Catalogue> distinctCatalogues = catalogues.DistinctBy(c => c.CatalogueCode); 

Answer 16

This takes distinct (the elements without duplicating elements) and convert it into a list again:

List<type> myNoneDuplicateValue = listValueWithDuplicate.Distinct().ToList();

Answer 17

Might be easier to simply make sure that duplicates are not added to the list.

if(items.IndexOf(new_item) < 0) 
    items.add(new_item)

Answer 18

Another way in .Net 2.0

    static void Main(string[] args)
    {
        List<string> alpha = new List<string>();

        for(char a = 'a'; a <= 'd'; a++)
        {
            alpha.Add(a.ToString());
            alpha.Add(a.ToString());
        }

        Console.WriteLine("Data :");
        alpha.ForEach(delegate(string t) { Console.WriteLine(t); });

        alpha.ForEach(delegate (string v)
                          {
                              if (alpha.FindAll(delegate(string t) { return t == v; }).Count > 1)
                                  alpha.Remove(v);
                          });

        Console.WriteLine("Unique Result :");
        alpha.ForEach(delegate(string t) { Console.WriteLine(t);});
        Console.ReadKey();
    }

Answer 19

There are many ways to solve - the duplicates issue in the List, below is one of them:

List<Container> containerList = LoadContainer();//Assume it has duplicates
List<Container> filteredList = new  List<Container>();
foreach (var container in containerList)
{ 
  Container duplicateContainer = containerList.Find(delegate(Container checkContainer)
  { return (checkContainer.UniqueId == container.UniqueId); });
   //Assume 'UniqueId' is the property of the Container class on which u r making a search

    if(!containerList.Contains(duplicateContainer) //Add object when not found in the new class object
      {
        filteredList.Add(container);
       }
  }

Cheers Ravi Ganesan

Answer 20

Here's a simple solution that doesn't require any hard-to-read LINQ or any prior sorting of the list.

   private static void CheckForDuplicateItems(List<string> items)
    {
        if (items == null ||
            items.Count == 0)
            return;

        for (int outerIndex = 0; outerIndex < items.Count; outerIndex++)
        {
            for (int innerIndex = 0; innerIndex < items.Count; innerIndex++)
            {
                if (innerIndex == outerIndex) continue;
                if (items[outerIndex].Equals(items[innerIndex]))
                {
                    // Duplicate Found
                }
            }
        }
    }

Answer 21

David J.'s answer is a good method, no need for extra objects, sorting, etc. It can be improved on however:

for (int innerIndex = items.Count - 1; innerIndex > outerIndex ; innerIndex--)

So the outer loop goes top bottom for the entire list, but the inner loop goes bottom "until the outer loop position is reached".

The outer loop makes sure the entire list is processed, the inner loop finds the actual duplicates, those can only happen in the part that the outer loop hasn't processed yet.

Or if you don't want to do bottom up for the inner loop you could have the inner loop start at outerIndex + 1.

Answer 22

You can use Union

obj2 = obj1.Union(obj1).ToList();

Answer 23

If you have tow classes Product and Customer and we want to remove duplicate items from their list

public class Product
{
    public int Id { get; set; }
    public string ProductName { get; set; }

}

public class Customer
{
    public int Id { get; set; }
    public string CustomerName { get; set; }

}

You must define a generic class in the form below

public class ItemEqualityComparer<T> : IEqualityComparer<T> where T : class
{
    private readonly PropertyInfo _propertyInfo;

    public ItemEqualityComparer(string keyItem)
    {
        _propertyInfo = typeof(T).GetProperty(keyItem, BindingFlags.GetProperty | BindingFlags.Instance | BindingFlags.Public);
    }

    public bool Equals(T x, T y)
    {
        var xValue = _propertyInfo?.GetValue(x, null);
        var yValue = _propertyInfo?.GetValue(y, null);
        return xValue != null && yValue != null && xValue.Equals(yValue);
    }

    public int GetHashCode(T obj)
    {
        var propertyValue = _propertyInfo.GetValue(obj, null);
        return propertyValue == null ? 0 : propertyValue.GetHashCode();
    }
}

then, You can remove duplicate items in your list.

var products = new List<Product>
            {
                new Product{ProductName = "product 1" ,Id = 1,},
                new Product{ProductName = "product 2" ,Id = 2,},
                new Product{ProductName = "product 2" ,Id = 4,},
                new Product{ProductName = "product 2" ,Id = 4,},
            };
var productList = products.Distinct(new ItemEqualityComparer<Product>(nameof(Product.Id))).ToList();

var customers = new List<Customer>
            {
                new Customer{CustomerName = "Customer 1" ,Id = 5,},
                new Customer{CustomerName = "Customer 2" ,Id = 5,},
                new Customer{CustomerName = "Customer 2" ,Id = 5,},
                new Customer{CustomerName = "Customer 2" ,Id = 5,},
            };
var customerList = customers.Distinct(new ItemEqualityComparer<Customer>(nameof(Customer.Id))).ToList();

this code remove duplicate items by Id if you want remove duplicate items by other property, you can change nameof(YourClass.DuplicateProperty) same nameof(Customer.CustomerName) then remove duplicate items by CustomerName Property.

Answer 24

  public static void RemoveDuplicates<T>(IList<T> list )
  {
     if (list == null)
     {
        return;
     }
     int i = 1;
     while(i<list.Count)
     {
        int j = 0;
        bool remove = false;
        while (j < i && !remove)
        {
           if (list[i].Equals(list[j]))
           {
              remove = true;
           }
           j++;
        }
        if (remove)
        {
           list.RemoveAt(i);
        }
        else
        {
           i++;
        }
     }  
  }

Answer 25

A simple intuitive implementation:

public static List<PointF> RemoveDuplicates(List<PointF> listPoints)
{
    List<PointF> result = new List<PointF>();

    for (int i = 0; i < listPoints.Count; i++)
    {
        if (!result.Contains(listPoints[i]))
            result.Add(listPoints[i]);
        }

        return result;
    }

Answer 26

All answers copy lists, or create a new list, or use slow functions, or are just painfully slow.

To my understanding, this is the fastest and cheapest method I know (also, backed by a very experienced programmer specialized on real-time physics optimization).

// Duplicates will be noticed after a sort O(nLogn)
list.Sort();

// Store the current and last items. Current item declaration is not really needed, and probably optimized by the compiler, but in case it's not...
int lastItem = -1;
int currItem = -1;

int size = list.Count;

// Store the index pointing to the last item we want to keep in the list
int last = size - 1;

// Travel the items from last to first O(n)
for (int i = last; i >= 0; --i)
{
    currItem = list[i];

    // If this item was the same as the previous one, we don't want it
    if (currItem == lastItem)
    {
        // Overwrite last in current place. It is a swap but we don't need the last
       list[i] = list[last];

        // Reduce the last index, we don't want that one anymore
        last--;
    }

    // A new item, we store it and continue
    else
        lastItem = currItem;
}

// We now have an unsorted list with the duplicates at the end.

// Remove the last items just once
list.RemoveRange(last + 1, size - last - 1);

// Sort again O(n logn)
list.Sort();

Final cost is:

nlogn + n + nlogn = n + 2nlogn = O(nlogn) which is pretty nice.

Note about RemoveRange: Since we cannot set the count of the list and avoid using the Remove funcions, I don't know exactly the speed of this operation but I guess it is the fastest way.