Question

请考虑以下列表：

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 },
};

然后是此代码：

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();

此代码可以正常工作，并且Cumulative_Probability的计算正确。

现在请考虑以下代码：

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();

此代码生成以下结果：

您可以看到在每个组中正确计算出Probability，但没有Cumulative_Probability。我想计算每个 Id 组中的Cumulative_Probability（组记录首先按Id，然后按Val1），并且Cumulative_Probability2不在每个组中重置。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

我设法借助扩展方法来做到这一点，该方法累积了累积概率以及一些嵌套的next-redux-wrapper。我敢肯定一定有一种更简单的方法，但是我正在努力寻找答案。

扩展名是：

GroupBy

完成的代码如下：

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;
        }
    }
}

实时示例：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

2 个答案: