如何对一个集合的连续类似项目进行分组？

Question

考虑以下集合。

• 真正
• 假
• 假
• 假
• 真正
• 真正
• 假
• 假

我希望以结构化的方式显示它，例如，在TreeView中。我希望能够围绕整个团体等绘制边框。

• 真实集团 真正
• 假组 假 假 假
• 真实集团 真正 真正
• 假组 假 假

如何使用尽可能少的程序代码完成此操作？

Answer 1

这就是您正在寻找的并且是通用的：

private static IEnumerable<IGrouping<int, T>> GroupConsecutive<T>(this IEnumerable<T> set, Func<T, T, bool> predicate)
{
    var i = 0;
    var k = 0;
    var ranges = from e in set
                 let idx = ++i
                 let next = set.ElementAtOrDefault(idx)
                 let key = (predicate(e, next)) ? k : k++
                 group e by key into g
                 select g;
    return ranges;
}

用法：

var set = new List<bool>
            {
                true,
                false,
                false,
                false,
                true,
                true,
                false,
                false,
            };
var groups = set.GroupConsecutive((b1, b2) => (b1 == b2));
foreach (var g in groups)
{
    Console.WriteLine(g.Key);
    foreach (var b in g)
        Console.WriteLine("\t{0}", b);
}

输出：

0
        True
1
        False
        False
        False
2
        True
        True
3
        False
        False

Answer 2

虽然接受的答案中的代码满足原始问题的需要，但在处理更复杂对象的IEnumerables时它将会失败（因为谓词在将枚举中的最后一项与“下一个”项目进行比较时会抛出异常[根据定义，它总是为空]）。

此版本处理更复杂的对象：

   public static IEnumerable<IGrouping<int, T>> GroupConsecutive<T>(this IEnumerable<T> set, Func<T, T, bool> predicate)
    {
        var i = 0;
        var k = 0;
        var ranges = from e in set
                     let idx = ++i
                     let next = set.ElementAtOrDefault(idx)
                     let key = next == null ? k : (predicate(e, next)) ? k : k++
                     group e by key into g
                     select g;
        return ranges;
    } 

Answer 3

last = null;
foreach (var option in list)
{
   if (last != option)
      newlist.Add(new Group(option, new[]));
   newlist.Last().Add(option);
   last = option;
}

Answer 4

public class GroupConsecutiveEqualItemsConverter : IValueConverter
{
    static readonly object UnsetValue = new object();

    public object Convert(object value, Type targetType, object parameter, CultureInfo culture)
    {
        IEnumerable source = value as IEnumerable;
        if (source == null) return DependencyProperty.UnsetValue;
        string propertyName = parameter as string;
        var result = new ObservableCollection<List<object>>();

        var notify = value as INotifyCollectionChanged;
        if (notify != null) notify.CollectionChanged += delegate { Reload(result, source, propertyName); };

        Reload(result, source, propertyName);
        return result;
    }

    public object ConvertBack(object value, Type targetType, object parameter, CultureInfo culture)
    {
        throw new NotSupportedException();
    }

    void Reload(ObservableCollection<List<object>> result, IEnumerable source, string propertyName)
    {
        result.Clear();
        object previous = UnsetValue;
        List<object> group = null;
        foreach (object i in source)
        {
            object current = UnsetValue;
            if (propertyName == null)
            {
                current = i;
            }
            else
            {
                try
                {
                    var property = i.GetType().GetProperty(propertyName);
                    if (property != null) current = property.GetValue(i, null);
                }
                catch (AmbiguousMatchException) { }
            }
            if (!object.Equals(previous, current))
            {
                if (group != null) result.Add(group);
                group = new List<object>();
            }
            group.Add(i);
            previous = current;
        }
        if (group != null && group.Count > 0) result.Add(group);
    }
}

Answer 5

你可能想在TreeGrid看看Delay's blog上的伪http://blogs.msdn.com/delay/archive/2009/09/23/if-it-walks-like-a-duck-and-talks-like-a-duck-it-must-be-a-treegrid-a-simple-xaml-only-treegrid-ui-for-wpf.aspx

（来源：msdn.com）