代码之家 › 专栏 › 技术社区 › Roger Johansson

Linq中的按元素分组

group-by linq c#

Roger Johansson · 技术社区 · 14 年前

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");

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

ienumerable扩展中是否有任何东西可以让我这样做,或者我必须在这里按方法滚动我自己的组?

3 回复 | 直到 14 年前

Timwi 14 年前

为此,我编写了一个扩展方法,可以在任何 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,在您的案例中,您可以这样使用它:

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));

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

(包括结尾的空行,因为 "\r"

Ronald Wildenberg 14 年前

如果你想使用标准 IEnumerable 扩展方法,你必须使用 Aggregate (但这并不像Timwi的解决方案那样可重用):

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;
                         });

Community CDub 8 年前

看到这个了吗 nest yields to return IEnumerable<IEnumerable<T>> with lazy evaluation 我也是。你可以喝一杯 SplitBy 接受要拆分的谓词的扩展方法:

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;
}

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)
    {
    }