在 LINQ 中获取多个集合的平均值答案

【问题标题】：Getting an Average over Multiple Collections in LINQ在 LINQ 中获取多个集合的平均值
【发布时间】：2008-11-25 19:19:36
【问题描述】：

我有一个集合集合，所有子集合都有相同数量的元素。父集合除了保存子集合之外什么都不做。

[0] [Child_0] [ID: 1]
[0] [Child_0] [Amount: 4]
[0] [Child_1] [ID: 2]
[0] [Child_1] [Amount: 7]
[1] [Child_0] [ID: 1]
[1] [Child_0] [Amount: 2]
[1] [Child_1] [ID: 2]
[1] [Child_1] [Amount: 4]
[2] [Child_0] [ID: 1]
[2] [Child_0] [Amount: 5]
[2] [Child_1] [ID: 2]
[2] [Child_1] [Amount: 3]

对于我的输出，我不关心父集合。我只想要一个匿名类型的 ID 和平均金额，所以对于上述情况，它会是

ID     Avg
1      3.66
2      4.66

回复的语言无关紧要。

谢谢。

【问题讨论】：

标签： c# vb.net linq

【解决方案1】：

听起来像一个简单的 selectmany，group 和 average 应该可以完成这项工作；这里我使用匿名类型和数组纯粹是为了方便输入...

        // the data
        var outer = new[] {
            new[] {
                new {ID=1,Amount=4}, // [0] [Child_0] [ID: 1, Amount: 4]
                new {ID=2,Amount=7}  // [0] [Child_1] [ID: 2, Amount: 7]
            },
            new[] {
                new {ID=1, Amount=2}, // [1] [Child_0] [ID: 1, Amount: 2]
                new {ID=2, Amount=4}  // [1] [Child_1] [ID: 2, Amount: 4]
            },
            new[] {
                new {ID=1, Amount=5}, // [2] [Child_0] [ID: 1, Amount: 5]
                new {ID=2, Amount=3}  // [2] [Child_1] [ID: 2, Amount: 3]
            }
        };
        var qry = from x in outer
                  from y in x
                  group y by y.ID into grp
                  select new { Id = grp.Key, Avg = grp.Average(z => z.Amount) };

        foreach (var item in qry)
        {
            Console.WriteLine("{0}: {1}", item.Id, item.Avg);
        }

【讨论】：

【解决方案2】：

您可以以此为指导

        var items = new[] 
        { 
            new { ID = 1, Amount = 4 }, 
            new { ID = 1, Amount = 5 },
            new { ID = 2, Amount = 5 },
            new { ID = 2, Amount = 3 },
        };

        var results = from item in items group item by item.ID into g select new { ID = g.Key, Avg = g.Average(item => item.Amount) };

        foreach (var result in results)
        {
            Console.WriteLine("{0} - {1}", result.ID, result.Avg);
        }

【讨论】：

比我的回答好多了。 +1