最小最大DataPoint归一化

Question

我有一个DataPoint列表，如

List<DataPoint> newpoints=new List<DataPoint>(); 

其中DataPoint是一个类，由A到I的九个双重特性组成，

newpoints.count=100000 double points (i.e each point consists of nine double features from A to I)

我需要应用赋范的列表新点使用最小-最大的去甲方法和scale_range之间的0和1。

到目前为止，我已经执行了以下步骤

1. 每个DataPoints特征都分配给一维数组。例如，特性A的代码 对于所有九个双重特性(int = 0；i< newpoints.Count；i++) { array_Ai = newpointsi.A;}等等
2. 我采用了最大最小归一化法.例如，特性A的代码： normilized_featureA= (array_Ai- array_A.Min()) *(1-0))/ (array_A.Max() -array_A.Min()+0；

该方法是成功完成的，但需要更多的时间(即3分45秒)。

如何使用Max_min ( C#中的LINQ代码)来将我的时间减少到几秒钟？我在Stackoverflow How to normalize a list of int values中发现了这个问题，但我的问题是

double valueMax = list.Max(); // I need Max point for feature A  for all 100000
double valueMin = list.Min(); //I need Min point for feature A  for all 100000

诸如此类，对于所有其他九个功能，您的帮助将受到高度赞赏。

Answer 1

作为将您的9个特性建模为类" datapoint“上的双重属性的替代方法，您还可以将一个9倍的数据点建模为一个数组，其好处是您可以再次使用LINQ一次完成所有9个计算：

var newpoints = new List<double[]>
{
    new []{1.23, 2.34, 3.45, 4.56, 5.67, 6.78, 7.89, 8.90, 9.12},
    new []{2.34, 3.45, 4.56, 5.67, 6.78, 7.89, 8.90, 9.12, 12.23},
    new []{3.45, 4.56, 5.67, 6.78, 7.89, 8.90, 9.12, 12.23, 13.34},
    new []{4.56, 5.67, 6.78, 7.89, 8.90, 9.12, 12.23, 13.34, 15.32}
};

var featureStats = newpoints
// We make the assumption that all 9 data points are present on each row.
.First()
// 2 Anon Projections - first to determine min / max as a function of column
.Select((np, idx) => new
{ 
   Idx = idx,
   Max = newpoints.Max(x => x[idx]),
   Min = newpoints.Min(x => x[idx])
})
// Second to add in the dynamic Range
.Select(x => new {
  x.Idx,
  x.Max,
  x.Min,
  Range = x.Max - x.Min
})
// Back to array for O(1) lookups.
.ToArray();

// Do the normalizaton for the columns, for each row.
var normalizedFeatures = newpoints
   .Select(np => np.Select(
      (i, idx) => (i - featureStats[idx].Min) / featureStats[idx].Range));

foreach(var datapoint in normalizedFeatures)
{
  Console.WriteLine(string.Join(",", datapoint.Select(x => x.ToString("0.00"))));
}

结果：

0.00,0.00,0.00,0.00,0.00,0.00,0.00,0.00,0.00
0.33,0.33,0.33,0.33,0.34,0.47,0.23,0.05,0.50
0.67,0.67,0.67,0.67,0.69,0.91,0.28,0.75,0.68
1.00,1.00,1.00,1.00,1.00,1.00,1.00,1.00,1.00

Answer 2

停止一次又一次地重新计算最大值/最小值，它不会改变。

double maxInFeatureA = array_A.Max();
double minInFeatureA = array_A.Min();

// somewher in the loop:
normilized_featureA= (((array_A[i] - minInFeatureA ) * (1 - 0)) / 
                  (maxInFeatureA  - minInFeatureA ))+0;

当在foreach/for中使用许多元素时，Max/Min对于数组来说是非常昂贵的。

我建议你接受以下代码：Array data normalization

并把它当作

var normalizedPoints = newPoints.Select(x => x.A)
            .NormalizeData(1, 1)
            .ToList(); 

Answer 3

double min = newpoints.Min(p => p.A);
double max = newpoints.Max(p => p.A);
double readonly normalizer = 1 / (max - min);

var normalizedFeatureA = newpoints.Select(p => (p.A - min) * normalizer);