假设我有一个字符串:
string str = "1111222233334444";
我如何把这个字符串分成一定大小的块?
例如，将它分解为4的大小将返回字符串:
"1111"
"2222"
"3333"
"4444"
当前回答
我不记得是谁给我的了，但它很好用。我快速测试了几种将枚举类型分成组的方法。用法是这样的…
List<string> Divided = Source3.Chunk(24).Select(Piece => string.Concat<char>(Piece)).ToList();
扩展代码看起来像这样…
#region Chunk Logic
private class ChunkedEnumerable<T> : IEnumerable<T>
{
class ChildEnumerator : IEnumerator<T>
{
ChunkedEnumerable<T> parent;
int position;
bool done = false;
T current;
public ChildEnumerator(ChunkedEnumerable<T> parent)
{
this.parent = parent;
position = -1;
parent.wrapper.AddRef();
}
public T Current
{
get
{
if (position == -1 || done)
{
throw new InvalidOperationException();
}
return current;
}
}
public void Dispose()
{
if (!done)
{
done = true;
parent.wrapper.RemoveRef();
}
}
object System.Collections.IEnumerator.Current
{
get { return Current; }
}
public bool MoveNext()
{
position++;
if (position + 1 > parent.chunkSize)
{
done = true;
}
if (!done)
{
done = !parent.wrapper.Get(position + parent.start, out current);
}
return !done;
}
public void Reset()
{
// per http://msdn.microsoft.com/en-us/library/system.collections.ienumerator.reset.aspx
throw new NotSupportedException();
}
}
EnumeratorWrapper<T> wrapper;
int chunkSize;
int start;
public ChunkedEnumerable(EnumeratorWrapper<T> wrapper, int chunkSize, int start)
{
this.wrapper = wrapper;
this.chunkSize = chunkSize;
this.start = start;
}
public IEnumerator<T> GetEnumerator()
{
return new ChildEnumerator(this);
}
System.Collections.IEnumerator System.Collections.IEnumerable.GetEnumerator()
{
return GetEnumerator();
}
}
private class EnumeratorWrapper<T>
{
public EnumeratorWrapper(IEnumerable<T> source)
{
SourceEumerable = source;
}
IEnumerable<T> SourceEumerable { get; set; }
Enumeration currentEnumeration;
class Enumeration
{
public IEnumerator<T> Source { get; set; }
public int Position { get; set; }
public bool AtEnd { get; set; }
}
public bool Get(int pos, out T item)
{
if (currentEnumeration != null && currentEnumeration.Position > pos)
{
currentEnumeration.Source.Dispose();
currentEnumeration = null;
}
if (currentEnumeration == null)
{
currentEnumeration = new Enumeration { Position = -1, Source = SourceEumerable.GetEnumerator(), AtEnd = false };
}
item = default(T);
if (currentEnumeration.AtEnd)
{
return false;
}
while (currentEnumeration.Position < pos)
{
currentEnumeration.AtEnd = !currentEnumeration.Source.MoveNext();
currentEnumeration.Position++;
if (currentEnumeration.AtEnd)
{
return false;
}
}
item = currentEnumeration.Source.Current;
return true;
}
int refs = 0;
// needed for dispose semantics
public void AddRef()
{
refs++;
}
public void RemoveRef()
{
refs--;
if (refs == 0 && currentEnumeration != null)
{
var copy = currentEnumeration;
currentEnumeration = null;
copy.Source.Dispose();
}
}
}
/// <summary>Speed Checked. Works Great!</summary>
public static IEnumerable<IEnumerable<T>> Chunk<T>(this IEnumerable<T> source, int chunksize)
{
if (chunksize < 1) throw new InvalidOperationException();
var wrapper = new EnumeratorWrapper<T>(source);
int currentPos = 0;
T ignore;
try
{
wrapper.AddRef();
while (wrapper.Get(currentPos, out ignore))
{
yield return new ChunkedEnumerable<T>(wrapper, chunksize, currentPos);
currentPos += chunksize;
}
}
finally
{
wrapper.RemoveRef();
}
}
#endregion
其他回答
六年后o_O
仅仅因为
public static IEnumerable<string> Split(this string str, int chunkSize, bool remainingInFront)
{
var count = (int) Math.Ceiling(str.Length/(double) chunkSize);
Func<int, int> start = index => remainingInFront ? str.Length - (count - index)*chunkSize : index*chunkSize;
Func<int, int> end = index => Math.Min(str.Length - Math.Max(start(index), 0), Math.Min(start(index) + chunkSize - Math.Max(start(index), 0), chunkSize));
return Enumerable.Range(0, count).Select(i => str.Substring(Math.Max(start(i), 0),end(i)));
}
or
private static Func<bool, int, int, int, int, int> start = (remainingInFront, length, count, index, size) =>
remainingInFront ? length - (count - index) * size : index * size;
private static Func<bool, int, int, int, int, int, int> end = (remainingInFront, length, count, index, size, start) =>
Math.Min(length - Math.Max(start, 0), Math.Min(start + size - Math.Max(start, 0), size));
public static IEnumerable<string> Split(this string str, int chunkSize, bool remainingInFront)
{
var count = (int)Math.Ceiling(str.Length / (double)chunkSize);
return Enumerable.Range(0, count).Select(i => str.Substring(
Math.Max(start(remainingInFront, str.Length, count, i, chunkSize), 0),
end(remainingInFront, str.Length, count, i, chunkSize, start(remainingInFront, str.Length, count, i, chunkSize))
));
}
AFAIK所有的边缘情况都处理好了。
Console.WriteLine(string.Join(" ", "abc".Split(2, false))); // ab c
Console.WriteLine(string.Join(" ", "abc".Split(2, true))); // a bc
Console.WriteLine(string.Join(" ", "a".Split(2, true))); // a
Console.WriteLine(string.Join(" ", "a".Split(2, false))); // a
使用正则表达式和Linq:
List<string> groups = (from Match m in Regex.Matches(str, @"\d{4}")
select m.Value).ToList();
我觉得这样更有可读性，但这只是个人观点。它也可以是一行代码:)。
它不是很漂亮，也不是很快，但它是有效的，它是一行程序，它是LINQy:
List<string> a = text.Select((c, i) => new { Char = c, Index = i }).GroupBy(o => o.Index / 4).Select(g => new String(g.Select(o => o.Char).ToArray())).ToList();
最近我不得不在工作中写一些东西来完成这个任务，所以我想我会把我对这个问题的解决方案贴出来。作为一个额外的好处，这个解决方案的功能提供了一种方法，以相反的方向分割字符串，并且它正确地处理上面Marvin Pinto提到的unicode字符。所以，就是这样:
using System;
using Extensions;
namespace TestCSharp
{
class Program
{
static void Main(string[] args)
{
string asciiStr = "This is a string.";
string unicodeStr = "これは文字列です。";
string[] array1 = asciiStr.Split(4);
string[] array2 = asciiStr.Split(-4);
string[] array3 = asciiStr.Split(7);
string[] array4 = asciiStr.Split(-7);
string[] array5 = unicodeStr.Split(5);
string[] array6 = unicodeStr.Split(-5);
}
}
}
namespace Extensions
{
public static class StringExtensions
{
/// <summary>Returns a string array that contains the substrings in this string that are seperated a given fixed length.</summary>
/// <param name="s">This string object.</param>
/// <param name="length">Size of each substring.
/// <para>CASE: length > 0 , RESULT: String is split from left to right.</para>
/// <para>CASE: length == 0 , RESULT: String is returned as the only entry in the array.</para>
/// <para>CASE: length < 0 , RESULT: String is split from right to left.</para>
/// </param>
/// <returns>String array that has been split into substrings of equal length.</returns>
/// <example>
/// <code>
/// string s = "1234567890";
/// string[] a = s.Split(4); // a == { "1234", "5678", "90" }
/// </code>
/// </example>
public static string[] Split(this string s, int length)
{
System.Globalization.StringInfo str = new System.Globalization.StringInfo(s);
int lengthAbs = Math.Abs(length);
if (str == null || str.LengthInTextElements == 0 || lengthAbs == 0 || str.LengthInTextElements <= lengthAbs)
return new string[] { str.ToString() };
string[] array = new string[(str.LengthInTextElements % lengthAbs == 0 ? str.LengthInTextElements / lengthAbs: (str.LengthInTextElements / lengthAbs) + 1)];
if (length > 0)
for (int iStr = 0, iArray = 0; iStr < str.LengthInTextElements && iArray < array.Length; iStr += lengthAbs, iArray++)
array[iArray] = str.SubstringByTextElements(iStr, (str.LengthInTextElements - iStr < lengthAbs ? str.LengthInTextElements - iStr : lengthAbs));
else // if (length < 0)
for (int iStr = str.LengthInTextElements - 1, iArray = array.Length - 1; iStr >= 0 && iArray >= 0; iStr -= lengthAbs, iArray--)
array[iArray] = str.SubstringByTextElements((iStr - lengthAbs < 0 ? 0 : iStr - lengthAbs + 1), (iStr - lengthAbs < 0 ? iStr + 1 : lengthAbs));
return array;
}
}
}
此外，这里还有一个运行此代码结果的图像链接:http://i.imgur.com/16Iih.png
class StringHelper
{
static void Main(string[] args)
{
string str = "Hi my name is vikas bansal and my email id is bansal.vks@gmail.com";
int offSet = 10;
List<string> chunks = chunkMyStr(str, offSet);
Console.Read();
}
static List<string> chunkMyStr(string str, int offSet)
{
List<string> resultChunks = new List<string>();
for (int i = 0; i < str.Length; i += offSet)
{
string temp = str.Substring(i, (str.Length - i) > offSet ? offSet : (str.Length - i));
Console.WriteLine(temp);
resultChunks.Add(temp);
}
return resultChunks;
}
}
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