按两列分组并根据其中一列计算累计值

Question

请考虑此列表：

List<Data> lst = new List<Data>
{
    new Data() { Id = 1, Val1 = 100 },
    new Data() { Id = 1, Val1 = 200 },
    new Data() { Id = 1, Val1 = 300 },
    new Data() { Id = 2, Val1 = 100 },
    new Data() { Id = 2, Val1 = 200 },
    new Data() { Id = 3, Val1 = 300 },
    new Data() { Id = 3, Val1 = 300 },
    new Data() { Id = 3, Val1 = 300 },
    new Data() { Id = 1, Val1 = 200 },
    new Data() { Id = 1, Val1 = 200 },
    new Data() { Id = 1, Val1 = 200 },
    new Data() { Id = 2, Val1 = 200 },
    new Data() { Id = 3, Val1 = 100 },
    new Data() { Id = 3, Val1 = 100 },
};

然后是这段代码：

decimal Cumulative_Probability = 0;
var Result1 = (lst.OrderBy(o => o.Id).GroupBy(x => new { x.Val1 })
    .Select(y => new
    {
        y.Key.Val1,
        Probability = (Convert.ToDecimal(y.Count()) / lst.Count),
        Cumulative_Probability = (Cumulative_Probability = 
            Cumulative_Probability + 
            (Convert.ToDecimal(y.Count()) / lst.Count))
    })).OrderBy(o => o.Val1).ToList();

此代码运行良好，Cumulative_Probability 计算正确。

现在请考虑这段代码：

decimal Cumulative_Probability2 = 0;
var Result2 = (lst.OrderBy(o => o.Id).GroupBy(x => new { x.Id, x.Val1 })
    .Select(y => new
    {
        y.Key.Id,
        y.Key.Val1,
        Probability = (Convert.ToDecimal(y.Count()) 
            / lst.Where(o => o.Id == y.Key.Id).Count()),
        Cumulative_Probability = (Cumulative_Probability2 = 
            Cumulative_Probability2 + 
            (Convert.ToDecimal(y.Count()) / 
            lst.Where(o => o.Id == y.Key.Id).Count()))
    })).OrderBy(o => o.Id).ThenBy(o => o.Val1).ToList();

此代码生成此结果：

如您所见，Probability 在每个组中计算正确，但 Cumulative_Probability 不正确。我想在每个 Id 组中计算 Cumulative_Probability（组记录首先根据 Id 然后 Val1）并且 Cumulative_Probability2 不会重置在每个组中。如何计算每个组中的 Cumulative_Probability？

谢谢

编辑 1)

我想要这个结果：

 Id             Val1             Probability       Cumulative_Probability 
 -------------------------------------------------------------------------
 1              100                0.16                 0.16
 1              200                0.66                 0.82
 1              300                0.16                 0.98
 2              100                0.33                 0.33
 2              200                0.66                 0.66
 ...

Answer 1

我设法在扩展方法的帮助下做到了这一点，该方法累积了累积概率，以及一些嵌套的 GroupBy。我敢肯定一定有更简单的方法，但我正在摸索着寻找它。

分机号是：

public static class EnumerableExtensions
{
    public static IEnumerable<TResult> Accumulate<TSource, TAccumulate, TResult>(
        this IEnumerable<TSource> source, 
        TAccumulate seed, 
        Func<TAccumulate, TSource, (TAccumulate,TResult)> accumulator)
    {
        var acc = seed;
        foreach(TSource value in source)
        {
            var (newSeed, newSource) = accumulator.Invoke(acc, value);
            yield return newSource;
            acc = newSeed;
        }
    }
}

完成的代码如下所示：

var result = lst.GroupBy( x => x.Id)
        .SelectMany( (grpId,i) =>  grpId.GroupBy(x => x.Val1)
                    .Accumulate(0M, (acc,grpVal) => (acc + (decimal)grpVal.Count()/grpId.Count(), new {
                        Id = grpId.Key,
                        Val1 = grpVal.Key,
                        Probability = (decimal)grpVal.Count()/grpId.Count(),
                        Cumulative_Probability = acc + ((decimal)grpVal.Count()/grpId.Count())
                    }))
                   )
        .OrderBy(x => x.Id);

实例：https://dotnetfiddle.net/dvW1qo

Answer 2

此代码有效：

var Result2 = (from a in lst.OrderBy(o => o.Id)
               group a by new { a.Id, a.Val1 } into grp
               select new
               {
                   grp.Key.Id,
                   grp.Key.Val1,
                   Probability = (Convert.ToDecimal(grp.Count()) / lst.Where(o => o.Id == grp.Key.Id).Count()),
                   Cumulative_Probability = (from b in lst.Where(o => o.Id == grp.Key.Id && o.Val1 <= grp.Key.Val1)
                                             group b by new { b.Val1 } into grp2
                                             select new
                                             {
                                                 Probability2 = (Convert.ToDecimal(grp2.Count()) / lst.Where(o => o.Id == grp.Key.Id).Count())
                                             }).Sum(o => o.Probability2)
                }).OrderBy(o => o.Id).ThenBy(o => o.Val1).ToList();

按两列分组并根据其中一列计算累计值

Group by two columns and calculate cumulative value based on one of them

c#

linq

lambda

group-by

c#-4.0