Standardize data

The biggest issue with the computation above is the data isn't standardized. For example the distance between point variance and point average is likely going to be a lot higher, causing variance to have a lot higher weight in the distance computation. Looking at the Kristaps Porzingis vs Paul George example, the difference between average points (36.05-34.5947)=1.4553 and variance points (52.6875-61.93)=-9.2425 is a lot further off, especially after you square each number in the Euclidean formula. This will lead to variance having a lot higher influence in the overall formula.

To counter this, we can standardize the variables between 0 and 1:

Where X is a given variable for a player, and Min and Max are the minimum and maximum amounts of that particular variable within the dataset.
Example:
Average points max=58.4 (Russell Westbrook)
Average points min=26.555 (Serge Ibaka)
Paul George's average points in this standardization computation would be
(36.05-26.555)/(58.4-26.555)=0.298
This is a very basic standardization technique, but it will work for this example. Extreme outliers can greatly skew this computation so it is not ideal a lot of the time.

C# Example

Say we have a list named "tonight" of players playing tonight. We do not know how many points they will score. We can compute another list titled "previous" of players who played on a previous night. We know how many points players on previous nights scored. We also have average, variance, and opponent average for each player in both lists. To compute a list of distances we can do the following:

Dictionary<string,List<double>  player_Alldistances = new Dictionary<string,List<double>>();
foreach(player p in tonight){
    List<double> distances = new List<double>();
    foreach(player p2 in previous){
           distances.Add(calculateDistance(p, p2));
    }
    player_Alldistances.Add(p.Name, distances);          
}
public static double calculateDistance(player p, player p2)
{
    double mean = Math.Pow((p.meanPoints - meanMin) / (meanMax - meanMin) -
        (p2.meanPoints - meanMin) / (meanMax - meanMin), 2);

    double variationPoints = Math.Pow((p.variancePoints - variancePointsMin) / (variancePointsMax - variancePointsMin) -
        (p2.deviationPoints - variancePointsMin) / (variancePointsMax - variancePointsMin), 2);

    double opptTotalPointsAverage = Math.Pow((p.OpptTotalPointsAverage - OpptTotalPointsAverageMin) / (OpptTotalPointsAverageMax - OpptTotalPointsAverageMin) -
        (p2.OpptTotalPointsAverage - OpptTotalPointsAverageMin) / (OpptTotalPointsAverageMax - OpptTotalPointsAverageMin), 2);

    double distsum = mean + variationPoints + opptTotalPointsAverage;
    double distance= System.Math.Sqrt(Math.Abs(distance));

    return distance;
}

			

Now we have a list of distances for each player compared to players who played on previous nights. This is where "K" comes in. There is some theory on how to choose K, but I'll just take a percentile of the length of the list to compute a K. Below I'll take the closest 2% of the list of distances. From here, I'll take the output value (points on given night) of the top 2% of the nearest neighbors, average them, and this will be my output. Example:

foreach(KeyValuePair<string, List<double>> entry in player_Alldistances)
{
    //get index of top 2%
    int twoPercent = entry.Value.Count * .02;
    entry.Value=entry.Value.OrderByAscending(x => x).ToList();
    double total = 0;
    for(int i = 0;i(less than)twoPercent;i++){
        total+=entry.Value.ElementAt(i);
    }
    double average = total / twoPercent;
    Console.Write(entry.Key+" guessed value="+average);
}

			

PG-The Unibrow: sqrt((((36.05-26.555)/(58.4-26.55))-((55.1-26.55)/(58.4-26.55)))^2+(((52.68-52.68)/(321.98-52.68))-((321.98-52.6875)/(321.98-52.6875)))^2+(((185.019-185.019)/(209.71-185.019))-((203.155-185.019)/(209.71-185.019)))^2)=1.37
PG-Boogie: sqrt((((36.05-26.555)/(58.4-26.55))-((48.1-26.55)/(58.4-26.55)))^2+(((52.68-52.68)/(321.98-52.68))-((139.74-52.6875)/(321.98-52.6875)))^2+(((185.019-185.019)/(209.71-185.019))-((202.22-185.019)/(209.71-185.019)))^2)=0.856
PG-RW: sqrt((((36.05-26.555)/(58.4-26.555))-((58.4-26.555)/(58.4-26.555)))^2+(((52.6875-52.6875)/(321.98-52.6875))-((133.23-52.6875)/(321.98-52.6875)))^2+(((185.019-185.019)/(209.71-185.019))-((209.55-185.019)/(209.71-185.019)))^2)=1.25
PG-BG: sqrt((((36.05-26.555)/(58.4-26.555))-((40.095-26.555)/(58.4-26.555)))^2+(((52.6875-52.6875)/(321.98-52.6875))-((55.81-52.6875)/(321.98-52.6875)))^2+(((185.019-185.019)/(209.71-185.019))-((209.71-185.019)/(209.71-185.019)))^2)=1.008
PG-SI: sqrt((((36.05-26.555)/(58.4-26.555))-((26.555-26.555)/(58.4-26.555)))^2+(((52.6875-52.6875)/(321.98-52.6875))-((84.34-52.6875)/(321.98-52.6875)))^2+(((185.019-185.019)/(209.71-185.019))-((191.70-185.019)/(209.71-185.019)))^2)=0.4194
PG-RG: sqrt((((36.05-26.555)/(58.4-26.555))-((33.1278-26.555)/(58.4-26.555)))^2+(((185.019-52.6875)/(321.98-52.6875))-((97.208-52.6875)/(321.98-52.6875)))^2+(((185.019-185.019)/(209.71-185.019))-((202.22-185.019)/(209.71-185.019)))^2)=0.7746
PG-KPizzle: sqrt((((36.05-26.555)/(58.4-26.555))-((34.5947-26.555)/(58.4-26.555)))^2+(((52.6875-52.6875)/(321.98-52.6875))-((61.93-52.6875)/(321.98-52.6875)))^2+(((185.019-185.019)/(209.71-185.019))-((204.37-185.019)/(209.71-185.019)))^2)=0.7858

Another thing you could do is weight the distances by correlation. I won't go through the correlation computations but I'll include a formula. This is a method they use in gene expression microarray data. Here is the formula for the Pearson correlation coefficient: