我需要在中点将一个不确定大小的数组拆分成两个独立的数组。
使用ToArray()从字符串列表生成数组。
public void AddToList ()
{
bool loop = true;
string a = "";
Console.WriteLine("Enter a string value and press enter to add it to the list");
while (loop == true)
{
a = Console.ReadLine();
if (a != "")
{
mylist.Add(a);
}
else
{
loop = false;
}
}
}
public void ReturnList()
{
string x = "";
foreach (string number in mylist)
{
x = x + number + " ";
}
Console.WriteLine(x);
Console.ReadLine();
}
}
class SplitList
{
public string[] sTop;
public string[] sBottom;
public void Split(ref UList list)
{
string[] s = list.mylist.ToArray();
//split the array into top and bottom halfs
}
}
static void Main(string[] args)
{
UList list = new UList();
SplitList split = new SplitList();
list.AddToList();
list.ReturnList();
split.Split(ref list);
}
}
}
答案 0 :(得分:47)
您可以使用以下方法将数组拆分为2个单独的数组
public void Split<T>(T[] array, int index, out T[] first, out T[] second) {
first = array.Take(index).ToArray();
second = array.Skip(index).ToArray();
}
public void SplitMidPoint<T>(T[] array, out T[] first, out T[] second) {
Split(array, array.Length / 2, out first, out second);
}
答案 1 :(得分:9)
使用通用拆分方法:
public static void Split<T>(T[] source, int index, out T[] first, out T last)
{
int len2 = source.Length - index;
first = new T[index];
last = new T[len2];
Array.Copy(source, 0, first, 0, index);
Array.Copy(source, index, last, 0, len2);
}
答案 2 :(得分:6)
我还想添加一个解决方案,将一个数组拆分成几个包含确定数量单元格的较小数组。
一种不错的方法是创建一个通用/扩展方法来拆分任何数组。这是我的:
/// <summary>
/// Splits an array into several smaller arrays.
/// </summary>
/// <typeparam name="T">The type of the array.</typeparam>
/// <param name="array">The array to split.</param>
/// <param name="size">The size of the smaller arrays.</param>
/// <returns>An array containing smaller arrays.</returns>
public static IEnumerable<IEnumerable<T>> Split<T>(this T[] array, int size)
{
for (var i = 0; i < (float)array.Length / size; i++)
{
yield return array.Skip(i * size).Take(size);
}
}
此外,该解决方案被推迟。然后,只需在您的阵列上调用split(size)
。
var array = new byte[] {10, 20, 30, 40, 50};
var splitArray = array.Split(2);
玩得开心:)
答案 3 :(得分:5)
在处理具有大量元素(即字节数组)的数组时,我遇到了Linq的Skip()和Take()函数的问题,其中元素数量是数百万。
这种方法大大减少了我的分割执行时间。
public static IEnumerable<IEnumerable<T>> Split<T>(this ICollection<T> self, int chunkSize)
{
var splitList = new List<List<T>>();
var chunkCount = (int)Math.Ceiling((double)self.Count / (double)chunkSize);
for(int c = 0; c < chunkCount; c++)
{
var skip = c * chunkSize;
var take = skip + chunkSize;
var chunk = new List<T>(chunkSize);
for(int e = skip; e < take && e < self.Count; e++)
{
chunk.Add(self.ElementAt(e));
}
splitList.Add(chunk);
}
return splitList;
}
答案 4 :(得分:1)
如果您没有Linq,可以使用Array.Copy:
public void Split(ref UList list)
{
string[] s = list.mylist.ToArray();
//split the array into top and bottom halfs
string[] top = new string[s.Length / 2];
string[] bottom = new string[s.Length - s.Length / 2];
Array.Copy(s, top, top.Length);
Array.Copy(s, top.Length, bottom, 0, bottom.Length);
Console.WriteLine("Top: ");
foreach (string item in top) Console.WriteLine(item);
Console.WriteLine("Bottom: ");
foreach (string item in bottom) Console.WriteLine(item);
}
答案 5 :(得分:0)
为什么不分配两个数组并复制内容?
编辑:你走了: String[] origin = new String[4];
origin[0] = "zero";
origin[1] = "one";
origin[2] = "two";
origin[3] = "three";
Int32 topSize = origin.Length / 2;
Int32 bottomSize = origin.Length - topSize;
String[] sTop = new String[topSize];
String[] sBottom = new String[bottomSize];
Array.Copy(origin, sTop, topSize);
Array.Copy(origin, topSize , sBottom, 0, bottomSize);
答案 6 :(得分:0)
为什么要将UList
作为参考号传递?似乎没有必要这样做。
如果我需要这样做,我会使用通用Split
方法:
public void Split<T>(T[] array, out T[] left, out T[] right)
{
left = new T[array.Length / 2];
right = new T[array.Length - left.Length];
Array.Copy(array, left, left.Length);
Array.Copy(array, left.Length, right, 0, right.Length);
}
答案 7 :(得分:0)
我认为你要找的是Array
类,特别是Array.Copy
静态方法。如果C#数组有方法,你可以认为该类包含的方法是数组的实例方法。
答案 8 :(得分:0)
如果需要考虑功能范例,这可能会有所帮助:
public static IEnumerable<IEnumerable<T>> Split<T>(this IEnumerable<T> seq, Int32 sizeSplits) {
Int32 numSplits = (seq.Count() / sizeSplits) + 1;
foreach ( Int32 ns in Enumerable.Range(start: 1, count: numSplits) ) {
(Int32 start, Int32 end) = GetIndexes(ns);
yield return seq.Where((_, i) => (start <= i && i <= end));
}
(Int32 start, Int32 end) GetIndexes(Int32 numSplit) {
Int32 indBase1 = numSplit * sizeSplits;
Int32 start = indBase1 - sizeSplits;
Int32 end = indBase1 - 1;
return (start, end);
}
}