我需要帮助 我有桌子：
Id Day Group Value
-------------------------------------------------
1 2016-10-11 1 10.5
2 2016-10-11 1 10.8
3 2016-10-11 1 10.7
4 2016-10-11 1 10.6
5 2016-10-11 1 10.5
6 2016-10-11 1 10.4
7 2016-10-11 1 10.8
8 2016-10-11 1 10.2
9 2016-10-11 1 10.0
10 2016-10-11 1 10.9
11 2016-10-11 2 10.1
12 2016-10-11 2 10.0
13 2016-10-11 2 10.1
14 2016-10-11 2 10.6
15 2016-10-11 2 10.7
16 2016-10-11 2 10.2
17 2016-10-11 2 10.0
18 2016-10-11 2 10.5
19 2016-10-11 2 10.5
20 2016-10-11 2 10.8
21 2016-10-12 1 11.1
22 2016-10-12 1 11.7
23 2016-10-12 1 11.0
24 2016-10-12 1 11.4
25 2016-10-12 1 11.7
26 2016-10-12 1 11.8
27 2016-10-12 1 11.1
28 2016-10-12 1 11.1
29 2016-10-12 1 11.4
30 2016-10-12 1 11.6
31 2016-10-12 2 11.9
32 2016-10-12 2 11.6
...
我想要
[[[10.5,10.8],[10.0,10.1]],[[11.1,11.7],[11.6,11.9]]]（示例）
我需要更多地使用这个值，当我获得价值时，我会在Day获得更多价值。然后我将这批次分成4批，然后我从每个组中选择第一个值。因此，我将为每个组提供4个值，对于1天，我将有2个批次，每个批次将具有4个值。并且因为我需要在每个批次中都有5个值，如最后一个值，我找到Max Value并将其添加到具有4个值的批处理中。我不知道你是否理解我的代码：
var valueList = await Task.Factory.StartNew(() =>
(
from pv in db.Values
where DbFunctions.TruncateTime(pv.Day) >= startDate
&& DbFunctions.TruncateTime(pv.Day) <= endDate
orderby pv.Value
//group pv by pv.Day into gpv
select pv
));
var group1 = await Task.Factory.StartNew(() => (
(
from vl in valueList
where vl.Group == 1
select vl)
));
var group2 = await Task.Factory.StartNew(() =>
(
from vl in valueList
where vl.Group == 2
select vl
));
var group3 = await Task.Factory.StartNew(() =>
(
from vl in valueList
where vl.Group == 3
select vl
));
var group1Split = group1.Split(group1.Count() / 4 + 1).Select(x => x.First());
var group1Max = group1.Max();
var group2Split = group2.Split(group2.Count() / 4 + 1).Select(x => x.First());
var group2Max = group2.Max();
var group3Split = group3.Split(group3.Count() / 4 + 1).Select(x => x.First());
var group3Max = group3.Max();
var group1Values = group1Split.Concat(group1Max);
var group2Values = group2Split.Concat(group2Max);
var group3Values = group3Split.Concat(group3Max);
var groupAllValues = group1Values.Concat(group2Values).Concat(group3Values);
var groupSplitAllValues = groupAllValues.Split(5);
return Request.CreateResponse(HttpStatusCode.OK, groupSplitAllValues.ToList());
更新
这项工作：
var result = db.Values
.Where(x => (DbFunctions.TruncateTime(x.Day)) >= startDate
&& (DbFunctions.TruncateTime(x.Day)) <= endDate)
.GroupBy(x => new { Day = x.Day, Group = x.Group })
.Select(x => x.Select(y => y.Value).OrderBy(z => z).ToList()).ToList();
我明白了：
[
[10.0, 10.2, 10.4, 10.5, 10.5, 10.6, 10.7, 10.8, 10.8, 10.9],
[10.0, 10.0, 10.1, 10.1, 10.2, 10.5, 10.5, 10.6, 10.7, 10.8],
[11.0, 11.1, 11.1, 11.1, 11.4, 11.4, 11.6, 11.7, 11.7, 11.8],
[...]
]
但是我需要将每个批次拆分为2个部分并从该部分中选择第一个值。所以我将得到2个值，然后我想从这个批次得到concat到2个值的最大值，所以对于这个例子，我的最终结果将是：
[
[10.0, 10.6, 10.9],
[10.0, 10.2, 10.8],
[11.0, 11.4, 11.8],
[...]
]
对于拆分我使用这种方法：
public static IEnumerable<IEnumerable<T>> Split<T>(this IEnumerable<T> source, int length)
{
if (length <= 0)
throw new ArgumentOutOfRangeException("length");
var section = new List<T>(length);
foreach (var item in source)
{
section.Add(item);
if (section.Count == length)
{
yield return section.AsReadOnly();
section = new List<T>(length);
}
}
if (section.Count > 0)
yield return section.AsReadOnly();
}
答案 0 :(得分：1)
试试这个
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答案 1 :(得分：0)
以下是我如何做到这一点的示例，我确信必须有一种更优雅的方式来生成输出字符串
static void Main(string[] args)
{
List<Example> samples = new List<Example>();
samples.Add(new Example(1, DateTime.Parse("2016-10-11"), 1, 10.5));
samples.Add(new Example(2, DateTime.Parse("2016-10-11"), 1, 10.8));
samples.Add(new Example(3, DateTime.Parse("2016-10-11"), 2, 10.1));
samples.Add(new Example(4, DateTime.Parse("2016-10-11"), 2, 10.0));
samples.Add(new Example(5, DateTime.Parse("2016-10-12"), 1, 11.1));
samples.Add(new Example(6, DateTime.Parse("2016-10-12"), 1, 11.7));
samples.Add(new Example(7, DateTime.Parse("2016-10-12"), 2, 11.9));
samples.Add(new Example(8, DateTime.Parse("2016-10-12"), 2, 11.6));
var daily_results = samples.GroupBy(g => new { g.group, g.date }).GroupBy(g => g.Key.date);
StringBuilder sb = new StringBuilder();
sb.Append("[");
foreach (var group_result in daily_results)
{
sb.Append("[");
foreach(var result in group_result)
{
sb.Append($"[{string.Join(",", result.OrderBy(r => r.value).Select(r => r.value))}]");
}
sb.Append("]");
}
sb.Append("]");
Console.WriteLine(sb);
Console.Read();
}
}
class Example
{
public int id;
public DateTime date;
public int group;
public double value;
public Example(int i, DateTime d, int g, double v)
{
id = i;
date = d;
group = g;
value = v;
}
}
更新：输出代码可以简化如下：
var daily_results = samples.GroupBy(g => new { g.group, g.date }).GroupBy(g => g.Key.date);
string output = string.Empty;
daily_results.ToList().ForEach(grp =>
{
output += "[";
grp.ToList().ForEach(res =>
{
output += $"[{string.Join(",", res.OrderBy(r => r.value).Select(r => r.value))}],";
});
output = output.TrimEnd(',') + "],";
});
Console.WriteLine($"[{output.TrimEnd(',')}]");
Console.Read();