斯普利特不同日期的列表 [英] Split a list by distinct date

问题描述

另一个容易的希望。

让我们说我有一个集合是这样的：

 列表与LT; DateTime的> allDates; 
  

我想将其转换成

 列表<名单，LT; DateTime的>> dividedDates; 
  

其中'dividedDates每个列表包含属于一个独立的allDates所有日期一年。

有一点LINQ挂羊头卖狗肉，我的疲惫的心灵不能挑出现在呢？

解决方案

接受的答案是正确的。

谢谢，我不ŧ想我是知道GROUPBY的进位的，我试图用.GroupBy（）之类的方法，而不是像语法的SQL。并感谢确认了ToList（）修正案，并把它包含在接受的答案： - ）

解决方案

 变种q =从日期allDates 
组由日期到date.Year datesByYear 
选择datesByYear.ToList（）; 
 q.ToList（）; //返回列表<名单，LT; DateTime的>> 
  

Another easy one hopefully.

Let's say I have a collection like this:

List<DateTime> allDates;

I want to turn that into

List<List<DateTime>> dividedDates;

where each List in 'dividedDates' contains all of the dates in 'allDates' that belong to a distinct year.

Is there a bit of LINQ trickery that my tired mind can't pick out right now?

Solution

The Accepted Answer is correct.

Thanks, I don't think I was aware of the 'into' bit of GroupBy and I was trying to use the .GroupBy() sort of methods rather than the SQL like syntax. And thanks for confirming the ToList() amendment and including it in the Accepted Answer :-)

解决方案

var q  = from date in allDates 
         group date by date.Year into datesByYear
         select datesByYear.ToList();
q.ToList(); //returns List<List<DateTime>>

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