如何在字典列表上动态构建分组依据 [英] How to build group-by dynamically on a list of dictionaries
问题描述
我正在尝试对 IEnumerable 执行 groupby.问题是我在编译时不知道我想分组哪些字段.我在堆栈上找到了另一篇文章,它解释了当类已知时如何执行此操作并具有属性,但在我的情况下,我正在处理字典,并且键也仅在运行时才知道.
I am trying to perform a groupby on an IEnumerable. The problem is that I do not know at compile-time which fields i want to groupby. I have found another post on stack that explains how to do this when the class is known and has properties, but in my case i am dealing with a dictionary and the keys are also only known at run-time.
我的代码类似于这样(我知道这不会编译...):
My code would resemble something like this (i know this doesn't compile...):
private object GetValuesGroupedBy(List<string> groupbyNames, List<string> summableNames)
{
// get the list of items in the grid
var listOfDicos = grid.AllItems;
return listOfDicos
.GroupBy(x => new { x[groupbyNames[0]],
x[groupbyNames[1]],
x[groupbyNames[2]] })
.Select(group => new { group.Key,
group.Sum(x => x[summableNames[0]]),
group.Sum(x => x[summableNames[1]]) });
}
有什么想法吗?我已经开始研究动态 LINQ,但被卡住了(因为我没有使用属性,而是使用键/值集合)...
Any ideas? I've started looking into Dynamic LINQ but am stuck (because I am not using properties but a Key/Value collection)...
谢谢大家!!
肖恩
推荐答案
所以我能够让 groupby 工作......(select 语句是另一个问题).感谢 c0d1ng 让我走上正确的道路.语法不是那么简单,因为我使用的是索引器而不是属性...
So I am able to get the groupby to work... (the select statement is another question). Thanks to c0d1ng for putting me on the right path. The syntax was not so trivial because I am using indexers and not properties...
下面是我的代码:
private void GetValuesGroupedBy(List<Dictionary<string, object>> list, List<string> groupbyNames, List<string> summableNames)
{
// build the groupby string
StringBuilder groupBySB = new StringBuilder();
groupBySB.Append("new ( ");
bool useComma = false;
foreach (var name in groupbyNames)
{
if (useComma)
groupBySB.Append(", ");
else
useComma = true;
groupBySB.Append("it["");
groupBySB.Append(name);
groupBySB.Append(""]");
groupBySB.Append(" as ");
groupBySB.Append(name);
}
groupBySB.Append(" )");
var groupby = list.GroupBy(groupBySB.ToString(), "it");
}
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