如何避免列表重复

Question

I'm trying to get a distinct list to my view.I need to select records from a list randomly and put it in to another list.The following code works..But it contain duplication records..How can I overcome this problem?

Note: the variable "budget" is a parameter passed in to the controller and "model1" is a List of PlanObjectsViewModel

int count = 0;
foreach (var item in model1) { count++; }

List<PlanObjectsViewModel> result = new List<PlanObjectsViewModel>();

Random rand = new Random();
double? temp=0;

while(budget>temp)
{

    int randi = rand.Next(0, count);
    var nthItem = model1.OrderBy(p => p.Id).Skip(randi).First();
    temp += nthItem.Price;

    if (!result.Contains(nthItem)) // Think this is the wrong point
    {
       result.Add(nthItem);
    }


}

Answer 1

Use a HashSet<PlanObjectsViewModel>

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

class Program
{
    static void Main()
    {
    // Input array that contains three duplicate strings.
    string[] array1 = { "cat", "dog", "cat", "leopard", "tiger", "cat" };

    // Display the array.
    Console.WriteLine(string.Join(",", array1));

    // Use HashSet constructor to ensure unique strings.
    var hash = new HashSet<string>(array1);

    // Convert to array of strings again.
    string[] array2 = hash.ToArray();

    // Display the resulting array.
    Console.WriteLine(string.Join(",", array2));
    }
}

Output:

cat,dog,cat,leopard,tiger,cat
cat,dog,leopard,tiger

Answer 2

there are two ways to do this, use a hashset instead of list for your result, or use Distinct()

HashSet<PlanObjectsViewModel> result

or

return result.Distinct();

You will have have to implement the Equals() method for this to work with objects, a which point your current code should work too.

Answer 3

Actually you have made it the correct way. For me it looks like you didnt implemented Equals and GetHashCode which are used by List.Contains to compare objects. Well basically GetHashCode is not mandatory but its a good design if you implemented the one to implement the other one.

But ofcourse you can use HashSet as pointed in the other answeres.