How to Choose Random number from given list of num

问题:

I want to create a random number generator in VB.NET But from my own given list of numbers

Like Chose random numbers from [1,2,3,4,5,6] e.t.c

回答1:

This is how you get a random natural number in the interval of [0, n - 1]:

CInt(Rnd() * n)

Let's suppose you have a List of n elements. This is how you get a random element from it:

MyList(CInt(Rnd() * n))

回答2:

Already built into .NET base of 'Random' and then extending that into your existing choices. This is NOT the same as generating the number from a Random as you are specifying YOUR OWN list first and then merely getting positioning with the help of a new Rand and using your length as a ceiling for it.

 Sub Main()
    'Say you have four items in your list
    Dim ls = New List(Of Integer)({1, 4, 8, 20})
    'I can find the 'position' of where the count of my array could be
    Dim rand = New Random().Next(0, ls.Count)
    'This will give a different 'position' every time.
    Console.WriteLine(ls(rand))

    Console.ReadLine()
  End Sub

回答3:

I would create a random number generator to generate a random number in the range of the list/array length, then use the result to point to the index of your number list.

Dim numbers As Integer() = New Integer() {1,2,5,6,7,8,12,43,56,67}

Dim randomKey = numbers(CInt(Rnd() * numbers.length))

*Edited based on Lajos Arpad's answer of how to get the random number

回答4:

Here is the function you can try, see more here Random integer in VB.NET

Public Function GetRandom(ByVal Min As Integer, ByVal Max As Integer) As    Integer
Dim Generator As System.Random = New System.Random()
Return Generator.Next(Min, Max + 1)
End Function