是否有任何方法可以将List<SomeObject>分离为SomeObject的几个单独的列表，使用项目索引作为每个分割的分隔符?
让我举个例子:
我有一个List<SomeObject>，我需要一个List<List<SomeObject>>或List<SomeObject>[]，这样每个结果列表将包含一组原始列表的3个项目(依次)。
eg.:
原始列表:[a, g, e, w, p, s, q, f, x, y, i, m, c] 结果列表:[a、g e], [w、p, s], [q, f, x]、[y,我,m], [c]
我还需要结果列表的大小是这个函数的参数。
当前回答
看看这个!我有一个序列计数器和日期的元素列表。每次序列重新启动时，我都想创建一个新列表。
例:消息列表。
List<dynamic> messages = new List<dynamic>
{
new { FcntUp = 101, CommTimestamp = "2019-01-01 00:00:01" },
new { FcntUp = 102, CommTimestamp = "2019-01-01 00:00:02" },
new { FcntUp = 103, CommTimestamp = "2019-01-01 00:00:03" },
//restart of sequence
new { FcntUp = 1, CommTimestamp = "2019-01-01 00:00:04" },
new { FcntUp = 2, CommTimestamp = "2019-01-01 00:00:05" },
new { FcntUp = 3, CommTimestamp = "2019-01-01 00:00:06" },
//restart of sequence
new { FcntUp = 1, CommTimestamp = "2019-01-01 00:00:07" },
new { FcntUp = 2, CommTimestamp = "2019-01-01 00:00:08" },
new { FcntUp = 3, CommTimestamp = "2019-01-01 00:00:09" }
};
我想在计数器重新启动时将列表拆分为单独的列表。代码如下:
var arraylist = new List<List<dynamic>>();
List<dynamic> messages = new List<dynamic>
{
new { FcntUp = 101, CommTimestamp = "2019-01-01 00:00:01" },
new { FcntUp = 102, CommTimestamp = "2019-01-01 00:00:02" },
new { FcntUp = 103, CommTimestamp = "2019-01-01 00:00:03" },
//restart of sequence
new { FcntUp = 1, CommTimestamp = "2019-01-01 00:00:04" },
new { FcntUp = 2, CommTimestamp = "2019-01-01 00:00:05" },
new { FcntUp = 3, CommTimestamp = "2019-01-01 00:00:06" },
//restart of sequence
new { FcntUp = 1, CommTimestamp = "2019-01-01 00:00:07" },
new { FcntUp = 2, CommTimestamp = "2019-01-01 00:00:08" },
new { FcntUp = 3, CommTimestamp = "2019-01-01 00:00:09" }
};
//group by FcntUp and CommTimestamp
var query = messages.GroupBy(x => new { x.FcntUp, x.CommTimestamp });
//declare the current item
dynamic currentItem = null;
//declare the list of ranges
List<dynamic> range = null;
//loop through the sorted list
foreach (var item in query)
{
//check if start of new range
if (currentItem == null || item.Key.FcntUp < currentItem.Key.FcntUp)
{
//create a new list if the FcntUp starts on a new range
range = new List<dynamic>();
//add the list to the parent list
arraylist.Add(range);
}
//add the item to the sublist
range.Add(item);
//set the current item
currentItem = item;
}
其他回答
几年前我写了一个集群扩展方法。工作得很好，是这里最快的实现。: P
/// <summary>
/// Clumps items into same size lots.
/// </summary>
/// <typeparam name="T"></typeparam>
/// <param name="source">The source list of items.</param>
/// <param name="size">The maximum size of the clumps to make.</param>
/// <returns>A list of list of items, where each list of items is no bigger than the size given.</returns>
public static IEnumerable<IEnumerable<T>> Clump<T>(this IEnumerable<T> source, int size)
{
if (source == null)
throw new ArgumentNullException("source");
if (size < 1)
throw new ArgumentOutOfRangeException("size", "size must be greater than 0");
return ClumpIterator<T>(source, size);
}
private static IEnumerable<IEnumerable<T>> ClumpIterator<T>(IEnumerable<T> source, int size)
{
Debug.Assert(source != null, "source is null.");
T[] items = new T[size];
int count = 0;
foreach (var item in source)
{
items[count] = item;
count++;
if (count == size)
{
yield return items;
items = new T[size];
count = 0;
}
}
if (count > 0)
{
if (count == size)
yield return items;
else
{
T[] tempItems = new T[count];
Array.Copy(items, tempItems, count);
yield return tempItems;
}
}
}
看看这个!我有一个序列计数器和日期的元素列表。每次序列重新启动时，我都想创建一个新列表。
例:消息列表。
List<dynamic> messages = new List<dynamic>
{
new { FcntUp = 101, CommTimestamp = "2019-01-01 00:00:01" },
new { FcntUp = 102, CommTimestamp = "2019-01-01 00:00:02" },
new { FcntUp = 103, CommTimestamp = "2019-01-01 00:00:03" },
//restart of sequence
new { FcntUp = 1, CommTimestamp = "2019-01-01 00:00:04" },
new { FcntUp = 2, CommTimestamp = "2019-01-01 00:00:05" },
new { FcntUp = 3, CommTimestamp = "2019-01-01 00:00:06" },
//restart of sequence
new { FcntUp = 1, CommTimestamp = "2019-01-01 00:00:07" },
new { FcntUp = 2, CommTimestamp = "2019-01-01 00:00:08" },
new { FcntUp = 3, CommTimestamp = "2019-01-01 00:00:09" }
};
我想在计数器重新启动时将列表拆分为单独的列表。代码如下:
var arraylist = new List<List<dynamic>>();
List<dynamic> messages = new List<dynamic>
{
new { FcntUp = 101, CommTimestamp = "2019-01-01 00:00:01" },
new { FcntUp = 102, CommTimestamp = "2019-01-01 00:00:02" },
new { FcntUp = 103, CommTimestamp = "2019-01-01 00:00:03" },
//restart of sequence
new { FcntUp = 1, CommTimestamp = "2019-01-01 00:00:04" },
new { FcntUp = 2, CommTimestamp = "2019-01-01 00:00:05" },
new { FcntUp = 3, CommTimestamp = "2019-01-01 00:00:06" },
//restart of sequence
new { FcntUp = 1, CommTimestamp = "2019-01-01 00:00:07" },
new { FcntUp = 2, CommTimestamp = "2019-01-01 00:00:08" },
new { FcntUp = 3, CommTimestamp = "2019-01-01 00:00:09" }
};
//group by FcntUp and CommTimestamp
var query = messages.GroupBy(x => new { x.FcntUp, x.CommTimestamp });
//declare the current item
dynamic currentItem = null;
//declare the list of ranges
List<dynamic> range = null;
//loop through the sorted list
foreach (var item in query)
{
//check if start of new range
if (currentItem == null || item.Key.FcntUp < currentItem.Key.FcntUp)
{
//create a new list if the FcntUp starts on a new range
range = new List<dynamic>();
//add the list to the parent list
arraylist.Add(range);
}
//add the item to the sublist
range.Add(item);
//set the current item
currentItem = item;
}
如果列表的类型为system.collections.generic，则可以使用“CopyTo”方法将数组中的元素复制到其他子数组中。您可以指定开始元素和要复制的元素数量。
你也可以对你的原始列表做3个克隆，并在每个列表上使用“RemoveRange”将列表缩小到你想要的大小。
或者只是创建一个helper方法来为您做这件事。
可与无限发电机工作:
a.Zip(a.Skip(1), (x, y) => Enumerable.Repeat(x, 1).Concat(Enumerable.Repeat(y, 1)))
.Zip(a.Skip(2), (xy, z) => xy.Concat(Enumerable.Repeat(z, 1)))
.Where((x, i) => i % 3 == 0)
演示代码:https://ideone.com/GKmL7M
using System;
using System.Collections.Generic;
using System.Linq;
public class Test
{
private static void DoIt(IEnumerable<int> a)
{
Console.WriteLine(String.Join(" ", a));
foreach (var x in a.Zip(a.Skip(1), (x, y) => Enumerable.Repeat(x, 1).Concat(Enumerable.Repeat(y, 1))).Zip(a.Skip(2), (xy, z) => xy.Concat(Enumerable.Repeat(z, 1))).Where((x, i) => i % 3 == 0))
Console.WriteLine(String.Join(" ", x));
Console.WriteLine();
}
public static void Main()
{
DoIt(new int[] {1});
DoIt(new int[] {1, 2});
DoIt(new int[] {1, 2, 3});
DoIt(new int[] {1, 2, 3, 4});
DoIt(new int[] {1, 2, 3, 4, 5});
DoIt(new int[] {1, 2, 3, 4, 5, 6});
}
}
1
1 2
1 2 3
1 2 3
1 2 3 4
1 2 3
1 2 3 4 5
1 2 3
1 2 3 4 5 6
1 2 3
4 5 6
但实际上我更喜欢写相应的方法没有linq。
我们可以改进@JaredPar的解决方案来做真正的惰性求值。我们使用GroupAdjacentBy方法生成一组具有相同键的连续元素:
sequence
.Select((x, i) => new { Value = x, Index = i })
.GroupAdjacentBy(x=>x.Index/3)
.Select(g=>g.Select(x=>x.Value))
因为基团是一个接一个地产生的，所以这个解决方案对长序列或无限序列有效。
