Weighted Random Number Generation in C#

问题:

Question

How can I randomly generate one of two states, with the probability of 'red' being generated 10% of the time, and 'green' being generated 90% of the time?

Background

Every 2 second either a green or a red light will blink.

This sequence will continue for 5 minutes.

The total number of occurrences of a blinking light should be 300.

回答1:

Random.NextDouble returns a number between 0 and 1, so the following should work:

if (random.NextDouble() < 0.90)
{
    BlinkGreen();
}
else
{
    BlinkRed();
}

回答2:

Either

Random rg = new Random();

int n = rg.Next(10); 
if(n == 0) {
    // blink red
}
else {
    // blink green
}

or

Random rg = new Random();

double value = rg.NextDouble();
if(value < 0.1) {
    // blink red
}
else {
    // blink green
}

This works because Random.Next(int maxValue) returns a uniformly distributed integer in [0, maxValue) and Random.NextDouble returns a uniformly distributed double in [0, 1).

回答3:

public class NewRandom
{
    private static Random _rnd = new Random();
    public static bool PercentChance(int percent)
    {
        double d = (double)percent / 100.0;
        return (_rnd.NextDouble() <= d);
    }
}

To use:

if (NewRandom.PercentChance(10))
{
    // blink red
}
else
{
    // blink green
}

回答4:

The other answers will definitely work if you need a random distribution that favors 90% green.

However, if you need a precise distribution, something like this will work:

void Main()
{
    Light[] lights = new Light[300];
    int i=0;
    Random rand = new Random();
    while(i<270)
    {
        int tryIndex = rand.Next(300);
        if(lights[tryIndex] == Light.NotSet)
        {
            lights[tryIndex] = Light.Green;
            i++;
        }
    }
    for(i=0;i<300;i++)
    {
        if(lights[i] == Light.NotSet)
        {
            lights[i] = Light.Red;
        }
    }

    //iterate over lights and do what you will
}


enum Light
{
    NotSet,
    Green,
    Red
}

回答5:

Building on Michaels answer, but adding further context from the question:

public static void PerformBlinks()
{
    var random = new Random();
    for (int i = 0; i < 300; i++)
    {
        if (random.Next(10) == 0)
        {
            BlinkGreen();
        }
        else
        {
            BlinkRed();
        }
        // Pause the thread for 2 seconds.
        Thread.Sleep(2000);
    }
}

回答6:

I'm guessing you have the timing part down (so this code doesn't address that). Assuming "nice" division, this will generate 10% reds and 90% greens. If the exactness isn't important, Michael's answer already has my vote.

static void Main(string[] args)
{
    int blinkCount = 300, redPercent = 10, greenPercent = 90;
    List<BlinkObject> blinks = new List<BlinkObject>(300);

    for (int i = 0; i < (blinkCount * redPercent / 100); i++)
    {
        blinks.Add(new BlinkObject("red"));
    }

    for (int i = 0; i < (blinkCount * greenPercent / 100); i++)
    {
        blinks.Add(new BlinkObject("green"));
    }

    blinks.Sort();

    foreach (BlinkObject b in blinks)
    {
        Console.WriteLine(b);
    }
}

class BlinkObject : IComparable<BlinkObject>
{
    object Color { get; set; }
    Guid Order { get; set; }

    public BlinkObject(object color)
    {
        Color = color;
        Order = Guid.NewGuid();
    }

    public int CompareTo(BlinkObject obj)
    {
        return Order.CompareTo(obj.Order);
    }

    public override string ToString()
    {
        return Color.ToString();
    }
}

回答7:

var random = new Random();
for (var i = 0; i < 150; ++i) {
  var red = random.Next(10) == 0;
  if (red)
    // ..
  else
    // Green
}

random.Next(10) will randomly return the numbers 0..9 and there is 10% chance of it returning 0.

回答8:

If you want these just to look random, you might want to implement shuffle bag

http://web.archive.org/web/20111203113141/http://kaioa.com:80/node/53

and

Need for predictable random generator

This way the blinking period should look more natural and you can simply implement the restricted number of blinks.