为IList中的每个字符串构建中值<IDictionary<string,double>>

事实上,我有点脑筋急转弯,只是找不到正确的方向让它工作:

给定是一个 IList<IDictionary<string, double>>

Name|Value
----+-----
"x" | 3.8
"y" | 4.2
"z" | 1.5
----+-----
"x" | 7.2
"y" | 2.9
"z" | 1.3
----+-----
... | ...

var list = CreateRandomPoints(new string[] { "x", "y", "z" }, 20);

其工作原理如下:

private IList<IDictionary<string, double>> CreateRandomPoints(string[] variableNames, int count)
{
    var list = new List<IDictionary<string, double>>(count);
    list.AddRange(CreateRandomPoints(variableNames).Take(count));

    return list;
}

private IEnumerable<IDictionary<string, double>> CreateRandomPoints(string[] variableNames)
{
    while (true)
        yield return CreateRandomLine(variableNames);
}

private IDictionary<string, double> CreateRandomLine(string[] variableNames)
{
    var dict = new Dictionary<string, double>(variableNames.Length);

    foreach (var variable in variableNames)
    {
        dict.Add(variable, _Rand.NextDouble() * 10);
    }

    return dict;
}

我还可以说,已经确保 列表中的每个字典都包含相同的键 (但是从一个列表到另一个列表,键的名称和计数可以更改)。

所以这就是我得到的。现在来谈谈我需要的东西:

我想得到所有字典中每个键的中位数(或任何其他数学聚合运算),以便调用的函数类似于:

IDictionary<string, double> GetMedianOfRows(this IList<IDictionary<string, double>> list)

最好是给函数提供某种聚合操作作为参数,使其更通用(不知道func是否有正确的参数,但应该想象一下我想做什么):

private IDictionary<string, double> Aggregate(this IList<IDictionary<string, double>> list, Func<IEnumerable<double>, double> aggregator)

如果列表中有20个变量和1000个值,我不想在列表中重复20次。相反,我会在列表上遍历一次,然后一次计算所有20个变量(这样做的最大优点是也可以将其用作 IEnumerable<T> 列表的任何部分)。

public static IDictionary<string, double> GetMedianOfRows(this IList<IDictionary<string, double>> list)
{
    //Check of parameters is left out!

    //Get first item for initialization of result dictionary
    var firstItem = list[0];

    //Create result dictionary and fill in variable names
    var dict = new Dictionary<string, double>(firstItem.Count);

    //Iterate over the whole list
    foreach (IDictionary<string, double> row in list)
    {
        //Iterate over each key/value pair within the list
        foreach (var kvp in row)
        {
            //How to determine median of all values?
        }
    }

    return dict;
}

只是想确定一下这里有一个关于 the Median .

解决方案

有了它,我们现在可以做这样美丽的陈述:

var result0 = list.Cast<IEnumerable<KeyValuePair<string, double>>>()
                  .Transpose()
                  .ToDictionary(column => column.First().Key,
                                column => column.Select(kvp => kvp.Value)
                                                .Sum());

或

var result1 = list.Cast<IEnumerable<KeyValuePair<string, double>>>()
                  .Transpose()
                  .ToDictionary(column => column.First().Key,
                                          //Performing my own function on the IEnumerable<double>
                                column => GetSumOfElements(column.Select(kvp => kvp.Value)));

private double GetSumOfElements(IEnumerable<double> elements)
{
    double result = 0;

    foreach (var element in elements)
    {
        result += element;
    }

    return result;
}