Alternative to Random.Next method [duplicate] - vb.net

Question

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How to Choose Random number from given list of numbers in VB.Net
(4 answers)
Closed 4 years ago.
In visual basic, is there a method that will allow me to select a random number from a list of numbers, rather than selecting a random number between two numbers?
In other words, I want something like Random().Next(1, 3, 4, 6, 7) where it will select a random number from those numbers (so selecting 2 is not an option)
I've looked here, but can't seem to find anything.

Answer 1

Try the following:
Imports System
Public Module Module1
Private Function NextRandom(ParamArray numbers() As Integer) As Integer
Dim Result As Integer = Integer.MinValue
If numbers.Length > 0 Then
Dim rnd As New Random
Dim i As Integer = rnd.Next(0, numbers.Length)
Result = numbers(i)
End If
Return Result
End Function
Public Sub Main()
Console.writeline("Your next number is: {0}", NextRandom(1, 3, 5, 6, 7, 8, 9))
End Sub
End Module
More info and examples here
Working example of the code here
Hope this helps.

Answer 2

If you want each value in your list only once then just shuffle your list.
Private Sub Button1_Click(sender As Object, e As EventArgs) Handles Button1.Click
Dim lst As New List(Of Integer) From {1, 3, 4, 6, 7}
Dim shuffledList As List(Of Integer) = lst.OrderBy(Function(i) Rnd.Next()).ToList()
For Each i In shuffledList
Debug.Print(i.ToString)
Next
End Sub
You get a different order each time you click the button.

Answer 3

Define your list as an array:
Dim List = New Integer() {Your numbers here}
Then use the standard random.next to generate an index for the array:
Dim generator As New Random
Dim index As Integer
index = generator.Next(0, List.length)
Then define your random number as the number that corresponds to the index in the array:
Dim randNumber as Integer
randNumber = List(index)