20

Let's take a class called Cls:

public class Cls
{
    public int SequenceNumber { get; set; }
    public int Value { get; set; }
}

Now, let's populate some collection with following elements:

Sequence
Number      Value
========    =====
1           9
2           9
3           15
4           15
5           15
6           30
7           9

What I need to do, is to enumerate over Sequence Numbers and check if the next element has the same value. If yes, values are aggregated and so, desired output is as following:

Sequence    Sequence
Number      Number
From        To          Value
========    ========    =====
1           2           9
3           5           15
6           6           30
7           7           9

How can I perform this operation using LINQ query?

4

7 回答 7

18

您可以在修改后的版本中使用 Linq GroupBy，该版本仅在两个项目相邻时才分组，那么它很简单：

var result = classes
    .GroupAdjacent(c => c.Value)
    .Select(g => new { 
        SequenceNumFrom = g.Min(c => c.SequenceNumber),
        SequenceNumTo = g.Max(c => c.SequenceNumber),  
        Value = g.Key
    });

foreach (var x in result)
    Console.WriteLine("SequenceNumFrom:{0} SequenceNumTo:{1} Value:{2}", x.SequenceNumFrom, x.SequenceNumTo, x.Value);

演示

结果：

SequenceNumFrom:1  SequenceNumTo:2  Value:9
SequenceNumFrom:3  SequenceNumTo:5  Value:15
SequenceNumFrom:6  SequenceNumTo:6  Value:30
SequenceNumFrom:7  SequenceNumTo:7  Value:9

这是对相邻项目进行分组的扩展方法：

public static IEnumerable<IGrouping<TKey, TSource>> GroupAdjacent<TSource, TKey>(
        this IEnumerable<TSource> source,
        Func<TSource, TKey> keySelector)
    {
        TKey last = default(TKey);
        bool haveLast = false;
        List<TSource> list = new List<TSource>();
        foreach (TSource s in source)
        {
            TKey k = keySelector(s);
            if (haveLast)
            {
                if (!k.Equals(last))
                {
                    yield return new GroupOfAdjacent<TSource, TKey>(list, last);
                    list = new List<TSource>();
                    list.Add(s);
                    last = k;
                }
                else
                {
                    list.Add(s);
                    last = k;
                }
            }
            else
            {
                list.Add(s);
                last = k;
                haveLast = true;
            }
        }
        if (haveLast)
            yield return new GroupOfAdjacent<TSource, TKey>(list, last);
    }
}

和使用的类：

public class GroupOfAdjacent<TSource, TKey> : IEnumerable<TSource>, IGrouping<TKey, TSource>
{
    public TKey Key { get; set; }
    private List<TSource> GroupList { get; set; }
    System.Collections.IEnumerator System.Collections.IEnumerable.GetEnumerator()
    {
        return ((System.Collections.Generic.IEnumerable<TSource>)this).GetEnumerator();
    }
    System.Collections.Generic.IEnumerator<TSource> System.Collections.Generic.IEnumerable<TSource>.GetEnumerator()
    {
        foreach (var s in GroupList)
            yield return s;
    }
    public GroupOfAdjacent(List<TSource> source, TKey key)
    {
        GroupList = source;
        Key = key;
    }
}
于 2013-02-14T16:32:51.847 回答
3

您可以使用此 linq 查询

演示

var values = (new[] { 9, 9, 15, 15, 15, 30, 9 }).Select((x, i) => new { x, i });

var query = from v in values
            let firstNonValue = values.Where(v2 => v2.i >= v.i && v2.x != v.x).FirstOrDefault()
            let grouping = firstNonValue == null ? int.MaxValue : firstNonValue.i
            group v by grouping into v
            select new
            {
              From = v.Min(y => y.i) + 1,
              To = v.Max(y => y.i) + 1,
              Value = v.Min(y => y.x)
            };
于 2013-02-14T17:17:29.117 回答
3

MoreLinq 开箱即用地提供了此功能

它被调用GroupAdjacent并作为扩展方法实现IEnumerable：

根据指定的键选择器函数对序列的相邻元素进行分组。

enumerable.GroupAdjacent(e => e.Key)

如果您不想引入额外的二进制 Nuget 包，甚至还有一个仅包含该方法的Nuget“源”包。

该方法返回一个IEnumerable<IGrouping<TKey, TValue>>，因此可以以与输出相同的方式处理其输出GroupBy。

于 2015-04-24T06:12:28.970 回答
2

你可以这样做：

var all = new [] {
    new Cls(1, 9)
,   new Cls(2, 9)
,   new Cls(3, 15)
,   new Cls(4, 15)
,   new Cls(5, 15)
,   new Cls(6, 30)
,   new Cls(7, 9)
};
var f = all.First();
var res = all.Skip(1).Aggregate(
    new List<Run> {new Run {From = f.SequenceNumber, To = f.SequenceNumber, Value = f.Value} }
,   (p, v) => {
    if (v.Value == p.Last().Value) {
        p.Last().To = v.SequenceNumber;
    } else {
        p.Add(new Run {From = v.SequenceNumber, To = v.SequenceNumber, Value = v.Value});
    }
    return p;
});
foreach (var r in res) {
    Console.WriteLine("{0} - {1} : {2}", r.From, r.To, r.Value);
}

这个想法是Aggregate创造性地使用：从一个由单个 组成的列表开始Run，检查到目前为止我们在聚合的每个阶段获得的列表的内容（iflambda 中的语句）。根据最后一个值，要么继续旧运行，要么开始新运行。

这是关于 ideone 的演示。

于 2013-02-14T16:32:31.887 回答
2

我能够通过创建自定义扩展方法来完成它。

static class Extensions {
  internal static IEnumerable<Tuple<int, int, int>> GroupAdj(this IEnumerable<Cls> enumerable) {
    Cls start = null;
    Cls end = null;
    int value = Int32.MinValue;

    foreach (Cls cls in enumerable) {
      if (start == null) {
        start = cls;
        end = cls;
        continue;
      }

      if (start.Value == cls.Value) {
        end = cls;
        continue;
      }

      yield return Tuple.Create(start.SequenceNumber, end.SequenceNumber, start.Value);
      start = cls;
      end = cls;
    }

    yield return Tuple.Create(start.SequenceNumber, end.SequenceNumber, start.Value);
  }
}

这是实现：

static void Main() {
  List<Cls> items = new List<Cls> {
    new Cls { SequenceNumber = 1, Value = 9 },
    new Cls { SequenceNumber = 2, Value = 9 },
    new Cls { SequenceNumber = 3, Value = 15 },
    new Cls { SequenceNumber = 4, Value = 15 },
    new Cls { SequenceNumber = 5, Value = 15 },
    new Cls { SequenceNumber = 6, Value = 30 },
    new Cls { SequenceNumber = 7, Value = 9 }
  };

  Console.WriteLine("From  To    Value");
  Console.WriteLine("===== ===== =====");
  foreach (var item in items.OrderBy(i => i.SequenceNumber).GroupAdj()) {
    Console.WriteLine("{0,-5} {1,-5} {2,-5}", item.Item1, item.Item2, item.Item3);
  }
}

和预期的输出：

From  To    Value
===== ===== =====
1     2     9
3     5     15
6     6     30
7     7     9
于 2013-02-14T16:38:42.023 回答
2

这是一个没有任何辅助方法的实现：

var grp = 0;
var results =
from i
in
input.Zip(
    input.Skip(1).Concat(new [] {input.Last ()}),
    (n1, n2) => Tuple.Create(
        n1, (n2.Value == n1.Value) ? grp : grp++
    )
)
group i by i.Item2 into gp
select new {SequenceNumFrom = gp.Min(x => x.Item1.SequenceNumber),SequenceNumTo = gp.Max(x => x.Item1.SequenceNumber), Value = gp.Min(x => x.Item1.Value)};

这个想法是：

  • 跟踪您自己的分组指标 grp。
  • 将集合中的每个项目连接到集合中的下一个项目（通过 Skip(1) 和 Zip）。
  • 如果值匹配，则它们在同一组中；否则，递增 grp 以指示下一组的开始。
于 2013-02-14T17:24:04.847 回答
1

未经测试的黑魔法随之而来。在这种情况下，命令式版本似乎会更容易。

IEnumerable<Cls> data = ...;
var query = data
    .GroupBy(x => x.Value)
    .Select(g => new
    {
        Value = g.Key,
        Sequences = g
            .OrderBy(x => x.SequenceNumber)
            .Select((x,i) => new
            {
                x.SequenceNumber,
                OffsetSequenceNumber = x.SequenceNumber - i
            })
            .GroupBy(x => x.OffsetSequenceNumber)
            .Select(g => g
                .Select(x => x.SequenceNumber)
                .OrderBy(x => x)
                .ToList())
            .ToList()
    })
    .SelectMany(x => x.Sequences
        .Select(s => new { First = s.First(), Last = s.Last(), x.Value }))
    .OrderBy(x => x.First)
    .ToList();
于 2013-02-14T16:51:18.960 回答