Question

下午我需要找出将数组分成较小“块”的最佳方法是什么。

我正在传递大约1200个项目，并且需要将这些项目拆分为更容易处理的100个组，然后我需要将它们传递给已处理的项目。

有人可以提出一些建议吗？

Answer 1

Array.Copy自1.1以来一直存在，并且在分块数组方面表现非常出色。

string[] buffer;

for(int i = 0; i < source.Length; i+=100)
{
    buffer = new string[100];
    Array.Copy(source, i, buffer, 0, 100);
    // process array
}

并为其做出扩展：

public static class Extensions
{
    public static T[] Slice<T>(this T[] source, int index, int length)
    {       
        T[] slice = new T[length];
        Array.Copy(source, index, slice, 0, length);
        return slice;
    }
}

并使用扩展程序：

string[] source = new string[] { 1200 items here };

// get the first 100
string[] slice = source.Slice(0, 100);

更新：我认为你可能想要ArraySegment<>不需要性能检查，因为它只是使用原始数组作为源，并维护一个Offset和Count属性来确定'segment'。不幸的是，没有办法将段作为一个数组检索，所以有些人为它编写了包装器，如下所示：ArraySegment - Returning the actual segment C#

ArraySegment<string> segment;

for (int i = 0; i < source.Length; i += 100)
{
    segment = new ArraySegment<string>(source, i, 100);

    // and to loop through the segment
    for (int s = segment.Offset; s < segment.Array.Length; s++)
    {
        Console.WriteLine(segment.Array[s]);
    }
}

Array.Copy与Skip / Take vs LINQ的性能

测试方法（在发布模式下）：

static void Main(string[] args)
{
    string[] source = new string[1000000];
    for (int i = 0; i < source.Length; i++)
    {
        source[i] = "string " + i.ToString();
    }

    string[] buffer;

    Console.WriteLine("Starting stop watch");

    Stopwatch sw = new Stopwatch();

    for (int n = 0; n < 5; n++)
    {
        sw.Reset();
        sw.Start();
        for (int i = 0; i < source.Length; i += 100)
        {
            buffer = new string[100];
            Array.Copy(source, i, buffer, 0, 100);
        }

        sw.Stop();
        Console.WriteLine("Array.Copy: " + sw.ElapsedMilliseconds.ToString());

        sw.Reset();
        sw.Start();
        for (int i = 0; i < source.Length; i += 100)
        {
            buffer = new string[100];
            buffer = source.Skip(i).Take(100).ToArray();
        }
        sw.Stop();
        Console.WriteLine("Skip/Take: " + sw.ElapsedMilliseconds.ToString());

        sw.Reset();
        sw.Start();
        String[][] chunks = source                            
            .Select((s, i) => new { Value = s, Index = i })                            
            .GroupBy(x => x.Index / 100)                            
            .Select(grp => grp.Select(x => x.Value).ToArray())                            
            .ToArray();
        sw.Stop();
        Console.WriteLine("LINQ: " + sw.ElapsedMilliseconds.ToString());
    }
    Console.ReadLine();
}

结果（以毫秒为单位）：

Array.Copy:    15
Skip/Take:  42464
LINQ:         881

Array.Copy:    21
Skip/Take:  42284
LINQ:         585

Array.Copy:    11
Skip/Take:  43223
LINQ:         760

Array.Copy:     9
Skip/Take:  42842
LINQ:         525

Array.Copy:    24
Skip/Take:  43134
LINQ:         638

Answer 2

您可以使用LINQ按块大小对所有项目进行分组，然后创建新的数组。

// build sample data with 1200 Strings
string[] items = Enumerable.Range(1, 1200).Select(i => "Item" + i).ToArray();
// split on groups with each 100 items
String[][] chunks = items
                    .Select((s, i) => new { Value = s, Index = i })
                    .GroupBy(x => x.Index / 100)
                    .Select(grp => grp.Select(x => x.Value).ToArray())
                    .ToArray();

for (int i = 0; i < chunks.Length; i++)
{
    foreach (var item in chunks[i])
        Console.WriteLine("chunk:{0} {1}", i, item);
}

请注意，没有必要创建新数组（需要cpu周期和内存）。当您省略两个IEnumerable<IEnumerable<String>>时，您也可以使用ToArrays。

以下是正在运行的代码：http://ideone.com/K7Hn2

Answer 3

您可以使用Skip()和Take()

string[] items = new string[]{ "a", "b", "c"};
string[] chunk = items.Skip(1).Take(1).ToArray();

Answer 4

here我找到了另一个linq解决方案：

int[] source = new[] { 1, 2, 3, 4, 5, 6, 7, 8, 9 };
int i = 0;
int chunkSize = 3;
int[][] result = source.GroupBy(s => i++ / chunkSize).Select(g => g.ToArray()).ToArray();

//result = [1,2,3][4,5,6][7,8,9]

Answer 5

    string[]  amzProductAsins = GetProductAsin();;
    List<string[]> chunks = new List<string[]>();
    for (int i = 0; i < amzProductAsins.Count; i += 100)
    {
        chunks.Add(amzProductAsins.Skip(i).Take(100).ToArray());
    }

Answer 6

您可以使用List.GetRange：

for(var i = 0; i < source.Count; i += chunkSize)
{
    List<string> items = source.GetRange(i, Math.Min(chunkSize, source.Count - i));
}

虽然没有像Array.Copy一样快，但我觉得它看起来更干净：

var list = Enumerable.Range(0, 723748).ToList();

var stopwatch = new Stopwatch();

for (int n = 0; n < 5; n++)
{
    stopwatch.Reset();
    stopwatch.Start();
    for(int i = 0; i < list.Count; i += 100)
    {
        List<int> c = list.GetRange(i, Math.Min(100, list.Count - i));
    }
    stopwatch.Stop();
    Console.WriteLine("List<T>.GetRange: " + stopwatch.ElapsedMilliseconds.ToString());

    stopwatch.Reset();
    stopwatch.Start();
    for (int i = 0; i < list.Count; i += 100)
    {
        List<int> c = list.Skip(i).Take(100).ToList();
    }
    stopwatch.Stop();
    Console.WriteLine("Skip/Take: " + stopwatch.ElapsedMilliseconds.ToString());

    stopwatch.Reset();
    stopwatch.Start();
    var test = list.ToArray();
    for (int i = 0; i < list.Count; i += 100)
    {
        int length = Math.Min(100, list.Count - i);
        int[] c = new int[length];
        Array.Copy(test, i, c, 0, length);
    }
    stopwatch.Stop();
    Console.WriteLine("Array.Copy: " + stopwatch.ElapsedMilliseconds.ToString());

    stopwatch.Reset();
    stopwatch.Start();
    List<List<int>> chunks = list
        .Select((s, i) => new { Value = s, Index = i })
        .GroupBy(x => x.Index / 100)
        .Select(grp => grp.Select(x => x.Value).ToList())
        .ToList();
    stopwatch.Stop();
    Console.WriteLine("LINQ: " + stopwatch.ElapsedMilliseconds.ToString());
}

以毫秒为单位的结果：

List<T>.GetRange: 1
Skip/Take: 9820
Array.Copy: 1
LINQ: 161

List<T>.GetRange: 9
Skip/Take: 9237
Array.Copy: 1
LINQ: 148

List<T>.GetRange: 5
Skip/Take: 9470
Array.Copy: 1
LINQ: 186

List<T>.GetRange: 0
Skip/Take: 9498
Array.Copy: 1
LINQ: 110

List<T>.GetRange: 8
Skip/Take: 9717
Array.Copy: 1
LINQ: 148

Answer 7

使用LINQ，您可以使用Take（）和Skip（）函数

Answer 8

一般递归扩展方法：

    public static IEnumerable<IEnumerable<T>> SplitList<T>(this IEnumerable<T> source, int maxPerList)
    {
        var enumerable = source as IList<T> ?? source.ToList();
        if (!enumerable.Any())
        {
            return new List<IEnumerable<T>>();
        }
        return (new List<IEnumerable<T>>() { enumerable.Take(maxPerList) }).Concat(enumerable.Skip(maxPerList).SplitList<T>(maxPerList));
    }

Answer 9

如果你有一个要分割的数组，但是除了这个简单的解决方案之外，你可以将缺少的元素分享到各种＆＃34;块＆＃34;相等。在我的解决方案中，块被链接为一个长字符串，但您可以轻松地更改它。

    public static string[] SplitArrey(string[] ArrInput, int n_column)
    {

        string[] OutPut = new string[n_column];
        int NItem = ArrInput.Length; // total elements
        int ItemsForColum = NItem / n_column; // n elements for each chunk
        int _total = ItemsForColum * n_column; // Count the equal elements
        int MissElement = NItem - _total; // Count missing element

        int[] _Arr = new int[n_column];
        for (int i = 0; i < n_column; i++)
        {
            int AddOne = (i < MissElement) ? 1 : 0;
            _Arr[i] = ItemsForColum + AddOne;
        }

        int offset = 0;
        for (int Row = 0; Row < n_column; Row++)
        {
            for (int i = 0; i < _Arr[Row]; i++)
            {
                OutPut[Row] += ArrInput[i + offset] + " "; // <- Here to change how the strings are linked 
            }
            offset += _Arr[Row];
        }
        return OutPut;
    }

拆分阵列的最佳方法

9 个答案:

Array.Copy与Skip / Take vs LINQ的性能