如何在C＃中執行Python的zip？

Question

Python的zip功能執行以下操作：

a = [1, 2, 3] 
b = [6, 7, 8] 
zipped = zip(a, b) 

結果

[[1, 6], [2, 7], [3, 8]] 

Answer 1

如何this？

C＃4.0 LINQ的新ZIP算

public static IEnumerable<TResult> Zip<TFirst, TSecond, TResult>(
     this IEnumerable<TFirst> first, 
     IEnumerable<TSecond> second, 
     Func<TFirst, TSecond, TResult> func); 

Answer 2

解決方案1：

IEnumerable<KeyValuePair<T1, T2>> Zip<T1, T2>(
    IEnumerable<T1> a, IEnumerable<T2> b) 
{ 
    var enumeratorA = a.GetEnumerator(); 
    var enumeratorB = b.GetEnumerator(); 
    while (enumeratorA.MoveNext()) 
    { 
     enumeratorB.MoveNext(); 
     yield return new KeyValuePair<T1, T2> 
     (
      enumeratorA.Current, 
      enumeratorB.Current 
     ); 
    } 
} 

Answer 3

解決方案2：以C＃4.0郵編相似，但你可以用它在C＃3.0

public static IEnumerable<TResult> Zip<TFirst, TSecond, TResult>(
     this IEnumerable<TFirst> first, 
     IEnumerable<TSecond> second, 
     Func<TFirst, TSecond, TResult> func) 
    { 
     using(var enumeratorA = first.GetEnumerator()) 
     using(var enumeratorB = second.GetEnumerator()) 
     { 
      while (enumeratorA.MoveNext()) 
      { 
       enumeratorB.MoveNext(); 
       yield return func(enumeratorA.Current, enumeratorB.Current); 
      } 
     } 
    } 

Answer 4

此外，在Cadenza看看它有所有各種漂亮的實用方法。

具體看一下這裏的郵編擴展方法： http://gitorious.org/cadenza/cadenza/blobs/master/src/Cadenza/Cadenza.Collections/Enumerable.cs#line1303

Answer 5

我只是遇到了同樣的問題。 .NET庫不提供解決方案，所以我自己做了。這是我的解決方案。

Pivot方法作爲IEnumerable<IEnumerable<T>>的擴展。 它要求所有序列的元素都是相同類型的T。

public static class LinqUtil 
{ 
    /// <summary> 
    /// From a number of input sequences makes a result sequence of sequences of elements 
    /// taken from the same position of each input sequence. 
    /// Example: ((1,2,3,4,5), (6,7,8,9,10), (11,12,13,14,15)) --> ((1,6,11), (2,7,12), (3,8,13), (4,9,14), (5,10,15)) 
    /// </summary> 
    /// <typeparam name="T">Type of sequence elements</typeparam> 
    /// <param name="source">source seq of seqs</param> 
    /// <param name="fillDefault"> 
    /// Defines how to handle situation when input sequences are of different length. 
    ///  false -- throw InvalidOperationException 
    ///  true -- fill missing values by the default values for the type T. 
    /// </param> 
    /// <returns>Pivoted sequence</returns> 
    public static IEnumerable<IEnumerable<T>> Pivot<T>(this IEnumerable<IEnumerable<T>> source, bool fillDefault = false) 
    { 
     IList<IEnumerator<T>> heads = new List<IEnumerator<T>>(); 

     foreach (IEnumerable<T> sourceSeq in source) 
     { 
      heads.Add(sourceSeq.GetEnumerator()); 
     } 

     while (MoveAllHeads(heads, fillDefault)) 
     { 
      yield return ReadHeads(heads); 
     } 
    } 

    private static IEnumerable<T> ReadHeads<T>(IEnumerable<IEnumerator<T>> heads) 
    { 
     foreach (IEnumerator<T> head in heads) 
     { 
      if (head == null) 
       yield return default(T); 
      else 
       yield return head.Current; 
     } 
    } 

    private static bool MoveAllHeads<T>(IList<IEnumerator<T>> heads, bool fillDefault) 
    { 
     bool any = false; 
     bool all = true; 

     for (int i = 0; i < heads.Count; ++i) 
     { 
      bool hasNext = false; 

      if(heads[i] != null) hasNext = heads[i].MoveNext(); 

      if (!hasNext) heads[i] = null; 

      any |= hasNext; 
      all &= hasNext; 
     } 

     if (any && !all && !fillDefault) 
      throw new InvalidOperationException("Input sequences are of different length"); 

     return any; 
    } 
}