Question

让我们假设我们有以下数组

var arr = new string[] {"foo","bar","jar","\r","a","b,"c","\r","x","y","z","\r");

还要忽略这是字符串的事实，所以请不要使用字符串黑客解决方案。

我想按序列中的每个“\ r”对这些元素进行分组。也就是说，我想要一个数组/可枚举“foo”，“bar”，“jar”和另一个“a”，“b”，“c”等。

在可以使用的扩展名中是否有任何内容可以让我这样做，或者我必须按方法滚动我自己的小组吗？

Answer 1

我为此目的编写了一个扩展方法，适用于任何IEnumerable<T>。

/// <summary>
/// Splits the specified IEnumerable at every element that satisfies a
/// specified predicate and returns a collection containing each sequence
/// of elements in between each pair of such elements. The elements
/// satisfying the predicate are not included.
/// </summary>
/// <param name="splitWhat">The collection to be split.</param>
/// <param name="splitWhere">A predicate that determines which elements
/// constitute the separators.</param>
/// <returns>A collection containing the individual pieces taken from the
/// original collection.</returns>
public static IEnumerable<IEnumerable<T>> Split<T>(
        this IEnumerable<T> splitWhat, Func<T, bool> splitWhere)
{
    if (splitWhat == null)
        throw new ArgumentNullException("splitWhat");
    if (splitWhere == null)
        throw new ArgumentNullException("splitWhere");
    return splitIterator(splitWhat, splitWhere);
}
private static IEnumerable<IEnumerable<T>> splitIterator<T>(
        IEnumerable<T> splitWhat, Func<T, bool> splitWhere)
{
    int prevIndex = 0;
    foreach (var index in splitWhat
        .Select((elem, ind) => new { e = elem, i = ind })
        .Where(x => splitWhere(x.e)))
    {
        yield return splitWhat.Skip(prevIndex).Take(index.i - prevIndex);
        prevIndex = index.i + 1;
    }
    yield return splitWhat.Skip(prevIndex);
}

例如，在您的情况下，您可以像这样使用它：

var arr = new string[] { "foo", "bar", "jar", "\r", "a", "b", "c", "\r", "x", "y", "z", "\r" };
var results = arr.Split(elem => elem == "\r");

foreach (var result in results)
    Console.WriteLine(string.Join(", ", result));

这将打印：

foo, bar, jar
a, b, c
x, y, z

（最后包括一个空行，因为收藏品末尾有"\r"。

Answer 2

如果您想使用标准IEnumerable扩展方法，则必须使用Aggregate（但这不像Timwi的解决方案那样可重复使用）：

var list = new[] { "foo","bar","jar","\r","a","b","c","\r","x","y","z","\r" };
var res = list.Aggregate(new List<List<string>>(),
                         (l, s) =>
                         {
                             if (s == "\r")
                             {
                                 l.Add(new List<string>());
                             }
                             else
                             {
                                 if (!l.Any())
                                 {
                                     l.Add(new List<string>());
                                 }
                                 l.Last().Add(s);
                             }
                             return l;
                         });

Answer 3

也请参阅此nest yields to return IEnumerable<IEnumerable<T>> with lazy evaluation。您可以使用SplitBy扩展方法接受要拆分的谓词：

public static IEnumerable<IList<T>> SplitBy<T>(this IEnumerable<T> source, 
                                              Func<T, bool> separatorPredicate,
                                              bool includeEmptyEntries = false, 
                                              bool includeSeparators = false)
{
    var l = new List<T>();
    foreach (var x in source)
    {
        if (!separatorPredicate(x))
            l.Add(x);
        else
        {
            if (includeEmptyEntries || l.Count != 0)
            {
                if (includeSeparators)
                    l.Add(x);

                yield return l;
            }

            l = new List<T>();
        }
    }

    if (l.Count != 0)
        yield return l;
}

所以在你的情况下：

var arr = new string[] {"foo","bar","jar","\r","a","b,"c","\r","x","y","z","\r");
foreach (var items in arr.SplitBy(x => x == "\r"))
    foreach (var item in items)
    {
    }

与Timwi相同，实施方式不同。没有错误检查，这就是你。由于您只遍历列表一次，因此速度会更快。

Linq中的元素分组

3 个答案: