How to get random value from property from one List?

Question

I have a List of 10,000 entries of type Element:

public class Element
{
    public Element()
    {
        this.Id = Guid.NewGuid();
    }

    public Guid Id { get; set };
}

And I have another List of 5,000 entries of type Link:

public class Link
{
    public Link(Guid ElementOne, Guid ElementTwo)
    {
        this.ElementOne = ElementOne;
        this.ElementTwo = ElementTwo;
    }

    public Guid ElementOne { get; set; }
    public Guid ElementTwo { get; set; }
}

I am populating my Lists here:

for (int i = 0; i < 10,000; i++)
    this.ListOfElements.Add(new Element());

for (int i = 0; i < 5,000; i++)
{
    this.ListOfLinks.Add(new Link(need ElementOne, need ElementTwo));
}

I am not sure what to pass for ElementOne and ElementTwo. I want to grab a random Id from the Element List (Element.Id) for both parameters and ensure they're unique (ElementOne could never be ElementTwo).

You need to keep track of elements you are adding or quering ListOfLinks everytime you want to add something, and adding a method to give you next one to add — Reza, May 11 '18 at 15:47
So, can ElementOne and ElementTwo NEVER repeat? Or do you just want to avoid linking 1 - 1. Is linking 1-2 and then 1-3 ok? Do you need to avoid linking 1-2, then later linking 1-2 again? — Tyler Lee, May 11 '18 at 15:52
Oh sorry, I wasn't clear - I just don't want 1 - 1. 1 - 2 repeated is fine. — , May 11 '18 at 15:53

score 2 · Accepted Answer · answered May 11 '18 at 15:56

Sounds like you just need two random numbers. You can compare and repoll if you happen to get a duplicate.

for (int i = 0; i < 5,000; i++)
{
    int ele1 = Random.NextInt(10000);
    int ele2 = Random.NextInt(10000);

    while(ele1 == ele2){
        ele2 = Random.NextInt(10000);
    }
    this.ListOfLinks.Add(new Link(ListOfElements[ele1], ListOfElements[ele2]));
}

score 0 · Answer 2 · answered May 11 '18 at 16:04

The code below will give a random array of the index to your list. So you can take randIndex[0] and randIndex[1] to get two random items.

            Random rand = new Random();
            List<string> myList = new List<string>() { "a","b","c"};
            int[] randIndex = myList.Select((x,i) => new {i = i, rand = rand.Next()}).OrderBy(x => x.rand).Select(x => x.i).ToArray();

Reza · Answer 3 · 2018-05-11T18:00:53.157

You need to find next element which is not used, also using .OrderBy(i => Guid.NewGuid()) to get random element

private Element GetNextAvaibaleElement() {
  var usedElementIds = listOfLinks.Select(i => i.ElementOne).Union(listOfLinks.Select(i => i.ElementTwo));
  return listOfElements.Where(i => !usedElementIds.Contains(i.Id)).OrderBy(i => Guid.NewGuid()).FirstOrDefault();
}

and then get two available elements

for (int i = 0; i < 5,000; i++)
{
    this.ListOfLinks.Add(new Link(GetNextAvaibaleElement().Id,GetNextAvaibaleElement().Id));
}

See this dotnet fiddle here

score -1 · Answer 4 · answered May 11 '18 at 15:58

Create an instance of the Random class somewhere.

E.g.

static Random rdm = new Random();

Then, use this rdm instance to generate a random integer, delete that element from the list (so as not to reuse it), and keep going until you deplete your list.

List<Element> listOfElements; // list of 10k Elements

for (int i = 0; i < 5,000; i++)
{
    int r = rdm.Next(listOfElements.Count);
    Element elementOne = listOfElements[r]; 
    listOfElements.RemoveAt(r);
    r = rdm.Next(listOfElements.Count);
    Element elementTwo = listOfElements[r]; 
    this.ListOfLinks.Add(new Link(elementOne, elementTwo));
}

score -1 · Answer 5 · answered May 11 '18 at 16:10

One approach to solving this is to create a list of integers representing the indexes into the ListOfElements, then order that list in a random way. Then you can simply reference items in the ListOfElements by index, sequentially walking through the randomized index list:

private static void Main()
{
    for (int i = 0; i < 10000; i++) ListOfElements.Add(new Element());

    // Create a randomly-ordered list of indexes to choose from
    var rnd = new Random();
    var randomIndexes = Enumerable.Range(0, ListOfElements.Count)
        .OrderBy(i => rnd.NextDouble()).ToList();

    // Now use our random indexes to pull unique items from the Element list
    for (int i = 0; i < 10000; i += 2)
        ListOfLinks.Add(new Link(ListOfElements[randomIndexes[i]].Id,
            ListOfElements[randomIndexes[i + 1]].Id));

    GetKeyFromUser("\nDone! Press any key to exit...");
}

How to get random value from property from one List?

5 Answers5