Python's zip
function does the following:
a = [1, 2, 3]
b = [6, 7, 8]
zipped = zip(a, b)
result
[[1, 6], [2, 7], [3, 8]]
Python's zip
function does the following:
a = [1, 2, 3]
b = [6, 7, 8]
zipped = zip(a, b)
result
[[1, 6], [2, 7], [3, 8]]
How about this?
C# 4.0 LINQ'S NEW ZIP OPERATOR
public static IEnumerable<TResult> Zip<TFirst, TSecond, TResult>(
this IEnumerable<TFirst> first,
IEnumerable<TSecond> second,
Func<TFirst, TSecond, TResult> func);
Solution 2: Similar to C# 4.0 Zip, but you can use it in 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);
}
}
}
Solution 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
);
}
}
Also take a look at Cadenza which has all sorts of nifty utility methods.
Specifically look at the Zip extension methods under Cadenza.Collections.EnumerableCoda
.
Here's a more modern rendition of Python's zip:
public static IEnumerable<(T1, T2)> Zip<T1, T2>(this IEnumerable<T1> t1, IEnumerable<T2> t2)
{
using var t1e = t1.GetEnumerator();
using var t2e = t2.GetEnumerator();
while (t1e.MoveNext() && t2e.MoveNext())
yield return (t1e.Current, t2e.Current);
}
public static IEnumerable<(T1, T2, T3)> Zip<T1, T2, T3>(this IEnumerable<T1> t1, IEnumerable<T2> t2, IEnumerable<T3> t3)
{
using var t1e = t1.GetEnumerator();
using var t2e = t2.GetEnumerator();
using var t3e = t3.GetEnumerator();
while (t1e.MoveNext() && t2e.MoveNext() && t3e.MoveNext())
yield return (t1e.Current, t2e.Current, t3e.Current);
}
It's easy to make more extensions that handle more IEnumerable
s at once, I've kept it to the 2 and 3 versions for brevity.
I just have come across the same problem. .NET library does not offer the solution, so I made it by myself. Here is my solution.
The Pivot
method is made as extension to IEnumerable<IEnumerable<T>>
.
It requires all the sequences' elements to be of the same type 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;
}
}