numl.Supervised.NaiveBayes.Measure.Normalize C# (CSharp) Method

        internal void Normalize()
        {
            double total = Probabilities.Select(p => p.Count).Sum();
            for (int i = 0; i < Probabilities.Length; i++)
                Probabilities[i].Probability = (double)Probabilities[i].Count / total;
        }