Question

拥有一组对象。示意性地：

[
    { A = 1, B = 1 }
    { A = 1, B = 2 }
    { A = 2, B = 3 }
    { A = 2, B = 4 }
    { A = 1, B = 5 }
    { A = 3, B = 6 }
]

需要：

[
    { A = 1, Bs = [ 1, 2 ] }
    { A = 2, Bs = [ 3, 4 ] }
    { A = 1, Bs = [ 5 ] }
    { A = 3, Bs = [ 6 ] }
]

LINQ可以吗？

注意：订购非常重要。因此，Bs = [5]无法与Bs = [1, 2]

合并

Answer 1

鉴于这些简单的课程：

class C {
  public int A;
  public int B;
}
class R {
  public int A;
  public List<int> Bs = new List<int>();
}

你可以这样做：

var cs = new C[] {
  new C() { A = 1, B = 1 },
  new C() { A = 1, B = 2 },
  new C() { A = 2, B = 3 },
  new C() { A = 2, B = 4 },
  new C() { A = 1, B = 5 },
  new C() { A = 3, B = 6 }
};

var rs = cs.
  OrderBy(o => o.B).
  ThenBy(o => o.A).
  Aggregate(new List<R>(), (l, o) => {
    if (l.Count > 0 && l.Last().A == o.A) {
      l.Last().Bs.Add(o.B);
    }
    else {
      l.Add(new R { A = o.A, Bs = { o.B } });
    }
    return l;
  });

注意：在上面我假设必须对Bs和As进行排序。如果情况并非如此，那么删除排序说明就很简单了：

var rs = cs.
  Aggregate(new List<R>(), (l, o) => {
    if (l.Count > 0 && l.Last().A == o.A) {
      l.Last().Bs.Add(o.B);
    }
    else {
      l.Add(new R { A = o.A, Bs = { o.B } });
    }
    return l;
  });

Answer 2

所以基本上你想把具有相同A - 值的东西组合在一起并且是连续的。

您需要将对象列表转换为包含上一个/下一个元素的匿名类型。我使用了两个Selects来使它更加红色。然后你需要检查两个元素是否是连续的（相邻的索引）。现在，您拥有GroupBy所需的全部价值和bool。

你的对象：

var list = new System.Collections.Generic.List<Foo>(){
    new Foo(){ A = 1, B = 1 },
    new Foo(){ A = 1, B = 2 },
    new Foo(){ A = 2, B = 3 },
    new Foo(){ A = 2, B = 4 },
    new Foo(){ A = 1, B = 5 },
    new Foo(){ A = 3, B = 6 }
};

查询：

var groups = list
    .Select((f, i) => new
    {
        Obj = f,
        Next = list.ElementAtOrDefault(i + 1),
        Prev = list.ElementAtOrDefault(i - 1)
    })
    .Select(x => new
    {
        A = x.Obj.A,
        x.Obj,
        Consecutive = (x.Next != null && x.Next.A == x.Obj.A)
                   || (x.Prev != null && x.Prev.A == x.Obj.A)
    })
    .GroupBy(x => new { x.Consecutive, x.A });

输出结果：

foreach (var abGroup in groups)
{
    int aKey = abGroup.Key.A;
    var bList = string.Join(",", abGroup.Select(x => x.Obj.B));
    Console.WriteLine("A = {0}, Bs = [ {1} ] ", aKey, bList);
}

以下是有效的演示：http://ideone.com/fXgQ3

Answer 3

您可以使用The GroupAdjacent Extension Method。

然后，你只需要

var grps = objects.GroupAdjacent(p => new { p.A });

我认为这是实施它的最简单方法。

修改

这是我的测试代码。

class Program { static void Main(string[] args) { var ia = new Dummycls[] { new Dummycls{ A = 1, B = 1 }, new Dummycls{ A = 1, B = 2 }, new Dummycls{ A = 2, B = 3 }, new Dummycls{ A = 2, B = 4 }, new Dummycls{ A = 1, B = 5 }, new Dummycls{ A = 3, B = 6 }, }; var groups = ia.GroupAdjacent(i => i.A); foreach (var g in groups) { Console.WriteLine("Group {0}", g.Key); foreach (var i in g) Console.WriteLine(i.ToString()); Console.WriteLine(); } Console.ReadKey(); } } class Dummycls { public int A { get; set; } public int B { get; set; } public override string ToString() { return string.Format("A={0};B={1}" , A , B); } }

结果是

Group 1 A=1;B=1 A=1;B=2 Group 2 A=2;B=3 A=2;B=4 Group 1 A=1;B=5 Group 3 A=3;B=6

Answer 4

这是执行您想要的方法的结构：

public static IEnumerable<IGrouping<TKey, TElement>> GroupWithKeyBreaks<T, TKey, TElement>(IEnumerable<T> enumerable,
        Func<T, TKey> keySelector,
        Func<T, TElement> itemSelector)    
{
    // Error handling goes here
    TKey currentKey = default(TKey);
    List<TElement> elements = new List<TElement>();

    foreach (T element in enumerable)
    {
        TKey thisKey = keySelector(element);
        if (thisKey == null)
        {
            continue;
        }

        if (!thisKey.Equals(currentKey) && elements.Count > 0)
        {
            yield return new SimpleGrouping<TKey, TElement>(currentKey, elements);
            elements = new List<TElement>();
        }

        elements.Add(itemSelector(element));
        currentKey = thisKey;
     }

    // Add the "last" item
    if (elements.Count > 0)
    {
        yield return new SimpleGrouping<TKey, TElement>(currentKey, elements);
    }
}

它使用以下助手类：

private class SimpleGrouping<T, U> : IGrouping<T, U>
{
    private T key;
    private IEnumerable<U> grouping;

    T IGrouping<T, U>.Key
    {
        get { return key; }
    }

    IEnumerator<U> IEnumerable<U>.GetEnumerator()
    {
        return grouping.GetEnumerator();
    }

    System.Collections.IEnumerator System.Collections.IEnumerable.GetEnumerator()
    {
        return grouping.GetEnumerator();
    }

    public SimpleGrouping(T k, IEnumerable<U> g)
    {
        this.key = k;
        this.grouping = g;
    }
}

以下是一个示例用法：

foreach (var grouping in data.GroupWithKeyBreaks(x => x.A, x => x.B))
{
    Console.WriteLine("Key: " + grouping.Key);
    foreach (var element in grouping)
    {
        Console.Write(element);
    }
}

Answer 5

var result = list.ToKeyValuePairs(x => x.A)
                 .Select(x => new { A = x.Key, Bs = x.Value.Select(y => y.B) }); 

foreach (var item in result)
{
    Console.WriteLine("A = {0} Bs=[{1}]",item.A, String.Join(",",item.Bs));
}

-

public static class MyExtensions
{
    public static IEnumerable<KeyValuePair<S,IEnumerable<T>>> ToKeyValuePairs<T,S>(
            this IEnumerable<T> list, 
            Func<T,S> keySelector)
    {
        List<T> retList = new List<T>();
        S prev = keySelector(list.FirstOrDefault());
        foreach (T item in list)
        {
            if (keySelector(item).Equals(prev))
                retList.Add(item);
            else
            {
                yield return new KeyValuePair<S, IEnumerable<T>>(prev, retList);
                prev = keySelector(item);
                retList = new List<T>();
                retList.Add(item);
            }
        }
        if(retList.Count>0)
            yield return new KeyValuePair<S, IEnumerable<T>>(prev, retList);
    }
}

<强>输出：

A = 1 Bs=[1,2]
A = 2 Bs=[3,4]
A = 1 Bs=[5]
A = 3 Bs=[6]

Answer 6

var groupCounter = 0;
int? prevA = null;

collection
    .Select(item => { 
        var groupId = item.A == prevA ? groupCounter : ++groupCounter; 
        prevA = item.A;
        return new { groupId, item.A, item.B }; 
    })
    .GroupBy(item => item.groupId)
    .Select(grp => new { A = grp.First().A, Bs = grp.Select(g => g.B) });

Answer 7

如果您的收藏集在o，那么：

    var trans = o.Aggregate
    (
            new {
                List = new List<Tuple<int, List<int>>>(),
                            LastSeed = (int?)0
            },
            (acc, item) =>
            {
                if (acc.LastSeed == null || item.A != acc.LastSeed)
                    acc.List.Add(Tuple.Create(item.A, new List<int>()));
                acc.List[acc.List.Count - 1].Item2.Add(item.B);
                return new { List = acc.List, LastSeed = (int?)item.A};
            },
            acc => acc.List.Select(
                  z=>new {A = z.Item1,
                          B = z.Item2 as IEnumerable<int>
                         })
       );

这会产生IEnumerable<int, IEnumerable<int>>所需的表格。

转换集合

7 个答案: