Getting random numbers from a list of integers

Question

If I have a list of integers:

List<int> myValues = new List<int>(new int[] { 1, 2, 3, 4, 5, 6 } ); 

How do I get 3 random numbers from this list?

Answer 1

One easy way:

 Random r = new Random(); IEnumerable<int> threeRandom = myValues.OrderBy(x => r.Next()).Take(3); 

The best way: Fisher-Yates shuffle :

 public static class EnumerableExtensions { public static IEnumerable<T> Shuffle<T>(this IEnumerable<T> source) { return source.Shuffle(new Random()); } public static IEnumerable<T> Shuffle<T>(this IEnumerable<T> source, Random rng) { if (source == null) throw new ArgumentNullException("source"); if (rng == null) throw new ArgumentNullException("rng"); return source.ShuffleIterator(rng); } private static IEnumerable<T> ShuffleIterator<T>( this IEnumerable<T> source, Random rng) { List<T> buffer = source.ToList(); for (int i = 0; i < buffer.Count; i++) { int j = rng.Next(i, buffer.Count); yield return buffer[j]; buffer[j] = buffer[i]; } } } 

how do you use it:

 IEnumerable<int> threeRandom = myValues.Shuffle().Take(3); 

Answer 2

The easiest way is something like this:

 var r = new Random(); var myValues = new int[] { 1, 2, 3, 4, 5, 6 }; // Will work with array or list var randomValues = Enumerable.Range(0, 3) .Select(e => myValues[r.Next(myValues.Length)]); 

But the best method, if you want to avoid duplicates, you need to use a shuffle algorithm, such as the Fisher-Yates algorithm, then take the first 3 elements:

 public static T[] Shuffle<T>(IEnumerable<T> items) { var result = items.ToArray(); var r = new Random(); for (int i = items.Length; i > 1; i--) { int j = r.Next(i); var t = result[j]; result[j] = result[i - 1]; result[i - 1] = t; } return result; } var myValues = new int[] { 1, 2, 3, 4, 5, 6 }; // Will work with any enumerable var randomValues = myValues.Shuffle().Take(3); 

Answer 3

Combining another answer with this answer can lead you to the following:

 var rand = new Random(); var numbers = Enumerable.Range(1, 6).OrderBy(i => rand.Next()).ToList(); 

In this case, 1 is the initial value (inclusive), and 6 is the number of integers that need to be generated.

Answer 4

There are ways to do it! Simple Google can get you hundreds of answers. However you can do it!

 myList.OrderBy(x => rnd.Next()).Take(3) 

Answer 5

Use the code below to get the number:

 int k = 3; // items to select var items = new List<int>(new[] { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 }); var selected = new List<int>(); var neededItem = k; var availableItem = items.Count; var rand = new Random(); while (selected.Count < k) { if( rand.NextDouble() < neededItem / availableItem ) { selected.Add(items[availableItem-1]) neededItem--; } availableItem--; } 

Answer 6

or that:

 myList.OrderBy(x => Guid.newGuid()).Take(3) 

Answer 7

 int[] x = {1,2,3,4}; string result = Convert.ToString(x[(new Random()).Next(4)]);