Is there a nice way to split a collection into n parts with LINQ?
Not necessarily evenly of course.
That is, I want to divide the collection into sub-collections, which each contains a subset of the elements, where the last collection can be ragged.
A pure linq and the simplest solution is as shown below.
static class LinqExtensions
{
public static IEnumerable<IEnumerable<T>> Split<T>(this IEnumerable<T> list, int parts)
{
int i = 0;
var splits = from item in list
group item by i++ % parts into part
select part.AsEnumerable();
return splits;
}
}
.AsEnumerable() is not necessary, IGrouping<T> is already an IEnumerable<T>. –
Corm IEnumerable<IEnumerable<T>> an implicit cast is taking place when you return part. No? –
Corm Dim i As Integer = 0 Dim splits As IEnumerable(Of IEnumerable(Of T)) = list.GroupBy(Function(item) Dim groupRemainder As Integer = i Mod parts i += 1 Return groupRemainder End Function).Select(Function(group) group.AsEnumerable()) –
Mafalda EDIT: Okay, it looks like I misread the question. I read it as "pieces of length n" rather than "n pieces". Doh! Considering deleting answer...
(Original answer)
I don't believe there's a built-in way of partitioning, although I intend to write one in my set of additions to LINQ to Objects. Marc Gravell has an implementation here although I would probably modify it to return a read-only view:
public static IEnumerable<IEnumerable<T>> Partition<T>
(this IEnumerable<T> source, int size)
{
T[] array = null;
int count = 0;
foreach (T item in source)
{
if (array == null)
{
array = new T[size];
}
array[count] = item;
count++;
if (count == size)
{
yield return new ReadOnlyCollection<T>(array);
array = null;
count = 0;
}
}
if (array != null)
{
Array.Resize(ref array, count);
yield return new ReadOnlyCollection<T>(array);
}
}
ReadOnlyCollection to a read-only array in this case? –
Mclane ReadOnlyCollection - the elements could be replaced. Thanks for your reply. –
Mclane yield return. It requires one batch to be in memory at a time, but that's all. –
Biyearly yield return and of course my comment "needs to traverse the complete list" is plainly wrong. Sorry. What happened was this: I was trying to use the method in a Parallel.ForEach in order to partition my data, which exploded in memory usage. I guess that the TPL spawns too many Tasks because the yield return only comes after filling a partition and so the first phase will be just preparing partitions in parallel. I guess what I need is a proper implementation of msdn.microsoft.com/en-us/library/dd381768(VS.100).aspx. –
Autobahn ToList in yield return GetNextBatch(enumerator).ToList();. Batching is necessary as otherwise you end up with very odd results if you skip one partition. –
Biyearly static class LinqExtensions
{
public static IEnumerable<IEnumerable<T>> Split<T>(this IEnumerable<T> list, int parts)
{
return list.Select((item, index) => new {index, item})
.GroupBy(x => x.index % parts)
.Select(x => x.Select(y => y.item));
}
}
var dept = {1,2,3,4,5} . After splitting the result is like dept1 = {1,3,5} and dept2 = { 2,4 } where parts = 2. But the result i need is dept1 = {1,2,3} and dept2 = {4,5} –
Subantarctic int columnLength = (int)Math.Ceiling((decimal)(list.Count()) / parts); then did division with .GroupBy(x => x.index / columnLength). One downside is Count() enumerates the list. –
Selfindulgent Ok, I'll throw my hat in the ring. The advantages of my algorithm:
The code:
public static IEnumerable<IEnumerable<T>>
Section<T>(this IEnumerable<T> source, int length)
{
if (length <= 0)
throw new ArgumentOutOfRangeException("length");
var section = new List<T>(length);
foreach (var item in source)
{
section.Add(item);
if (section.Count == length)
{
yield return section.AsReadOnly();
section = new List<T>(length);
}
}
if (section.Count > 0)
yield return section.AsReadOnly();
}
As pointed out in the comments below, this approach doesn't actually address the original question which asked for a fixed number of sections of approximately equal length. That said, you can still use my approach to solve the original question by calling it this way:
myEnum.Section(myEnum.Count() / number_of_sections + 1)
When used in this manner, the approach is no longer O(1) as the Count() operation is O(N).
myEnum.Section(myEnum.Count() / number_of_sections + 1) enumerates the first collection (and for the given question can be inefficient), so pls edit the advantages you have mentioned for your approach. –
Beguine This is same as the accepted answer, but a much simpler representation:
public static IEnumerable<IEnumerable<T>> Split<T>(this IEnumerable<T> items,
int numOfParts)
{
int i = 0;
return items.GroupBy(x => i++ % numOfParts);
}
The above method splits an IEnumerable<T> into N number of chunks of equal sizes or close to equal sizes.
public static IEnumerable<IEnumerable<T>> Partition<T>(this IEnumerable<T> items,
int partitionSize)
{
int i = 0;
return items.GroupBy(x => i++ / partitionSize).ToArray();
}
The above method splits an IEnumerable<T> into chunks of desired fixed size with total number of chunks being unimportant - which is not what the question is about.
The problem with the Split method, besides being slower, is that it scrambles the output in the sense that the grouping will be done on the basis of i'th multiple of N for each position, or in other words you don't get the chunks in the original order.
Almost every answer here either doesn't preserve order, or is about partitioning and not splitting, or is plainly wrong. Try this which is faster, preserves order but a lil' more verbose:
public static IEnumerable<IEnumerable<T>> Split<T>(this ICollection<T> items,
int numberOfChunks)
{
if (numberOfChunks <= 0 || numberOfChunks > items.Count)
throw new ArgumentOutOfRangeException("numberOfChunks");
int sizePerPacket = items.Count / numberOfChunks;
int extra = items.Count % numberOfChunks;
for (int i = 0; i < numberOfChunks - extra; i++)
yield return items.Skip(i * sizePerPacket).Take(sizePerPacket);
int alreadyReturnedCount = (numberOfChunks - extra) * sizePerPacket;
int toReturnCount = extra == 0 ? 0 : (items.Count - numberOfChunks) / extra + 1;
for (int i = 0; i < extra; i++)
yield return items.Skip(alreadyReturnedCount + i * toReturnCount).Take(toReturnCount);
}
The equivalent method for a Partition operation here
I have been using the Partition function I posted earlier quite often. The only bad thing about it was that is wasn't completely streaming. This is not a problem if you work with few elements in your sequence. I needed a new solution when i started working with 100.000+ elements in my sequence.
The following solution is a lot more complex (and more code!), but it is very efficient.
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Collections;
namespace LuvDaSun.Linq
{
public static class EnumerableExtensions
{
public static IEnumerable<IEnumerable<T>> Partition<T>(this IEnumerable<T> enumerable, int partitionSize)
{
/*
return enumerable
.Select((item, index) => new { Item = item, Index = index, })
.GroupBy(item => item.Index / partitionSize)
.Select(group => group.Select(item => item.Item) )
;
*/
return new PartitioningEnumerable<T>(enumerable, partitionSize);
}
}
class PartitioningEnumerable<T> : IEnumerable<IEnumerable<T>>
{
IEnumerable<T> _enumerable;
int _partitionSize;
public PartitioningEnumerable(IEnumerable<T> enumerable, int partitionSize)
{
_enumerable = enumerable;
_partitionSize = partitionSize;
}
public IEnumerator<IEnumerable<T>> GetEnumerator()
{
return new PartitioningEnumerator<T>(_enumerable.GetEnumerator(), _partitionSize);
}
IEnumerator IEnumerable.GetEnumerator()
{
return GetEnumerator();
}
}
class PartitioningEnumerator<T> : IEnumerator<IEnumerable<T>>
{
IEnumerator<T> _enumerator;
int _partitionSize;
public PartitioningEnumerator(IEnumerator<T> enumerator, int partitionSize)
{
_enumerator = enumerator;
_partitionSize = partitionSize;
}
public void Dispose()
{
_enumerator.Dispose();
}
IEnumerable<T> _current;
public IEnumerable<T> Current
{
get { return _current; }
}
object IEnumerator.Current
{
get { return _current; }
}
public void Reset()
{
_current = null;
_enumerator.Reset();
}
public bool MoveNext()
{
bool result;
if (_enumerator.MoveNext())
{
_current = new PartitionEnumerable<T>(_enumerator, _partitionSize);
result = true;
}
else
{
_current = null;
result = false;
}
return result;
}
}
class PartitionEnumerable<T> : IEnumerable<T>
{
IEnumerator<T> _enumerator;
int _partitionSize;
public PartitionEnumerable(IEnumerator<T> enumerator, int partitionSize)
{
_enumerator = enumerator;
_partitionSize = partitionSize;
}
public IEnumerator<T> GetEnumerator()
{
return new PartitionEnumerator<T>(_enumerator, _partitionSize);
}
IEnumerator IEnumerable.GetEnumerator()
{
return GetEnumerator();
}
}
class PartitionEnumerator<T> : IEnumerator<T>
{
IEnumerator<T> _enumerator;
int _partitionSize;
int _count;
public PartitionEnumerator(IEnumerator<T> enumerator, int partitionSize)
{
_enumerator = enumerator;
_partitionSize = partitionSize;
}
public void Dispose()
{
}
public T Current
{
get { return _enumerator.Current; }
}
object IEnumerator.Current
{
get { return _enumerator.Current; }
}
public void Reset()
{
if (_count > 0) throw new InvalidOperationException();
}
public bool MoveNext()
{
bool result;
if (_count < _partitionSize)
{
if (_count > 0)
{
result = _enumerator.MoveNext();
}
else
{
result = true;
}
_count++;
}
else
{
result = false;
}
return result;
}
}
}
Enjoy!
Interesting thread. To get a streaming version of Split/Partition, one can use enumerators and yield sequences from the enumerator using extension methods. Converting imperative code to functional code using yield is a very powerful technique indeed.
First an enumerator extension that turns a count of elements into a lazy sequence:
public static IEnumerable<T> TakeFromCurrent<T>(this IEnumerator<T> enumerator, int count)
{
while (count > 0)
{
yield return enumerator.Current;
if (--count > 0 && !enumerator.MoveNext()) yield break;
}
}
And then an enumerable extension that partitions a sequence:
public static IEnumerable<IEnumerable<T>> Partition<T>(this IEnumerable<T> seq, int partitionSize)
{
var enumerator = seq.GetEnumerator();
while (enumerator.MoveNext())
{
yield return enumerator.TakeFromCurrent(partitionSize);
}
}
The end result is a highly efficient, streaming and lazy implementation that relies on very simple code.
Enjoy!
I use this:
public static IEnumerable<IEnumerable<T>> Partition<T>(this IEnumerable<T> instance, int partitionSize)
{
return instance
.Select((value, index) => new { Index = index, Value = value })
.GroupBy(i => i.Index / partitionSize)
.Select(i => i.Select(i2 => i2.Value));
}
As of .NET 6 you can use Enumerable.Chunk<TSource>(IEnumerable<TSource>, Int32).
This is memory efficient and defers execution as much as possible (per batch) and operates in linear time O(n)
public static IEnumerable<IEnumerable<T>> InBatchesOf<T>(this IEnumerable<T> items, int batchSize)
{
List<T> batch = new List<T>(batchSize);
foreach (var item in items)
{
batch.Add(item);
if (batch.Count >= batchSize)
{
yield return batch;
batch = new List<T>();
}
}
if (batch.Count != 0)
{
//can't be batch size or would've yielded above
batch.TrimExcess();
yield return batch;
}
}
There are lots of great answers for this question (and its cousins). I needed this myself and had created a solution that is designed to be efficient and error tolerant in a scenario where the source collection can be treated as a list. It does not use any lazy iteration so it may not be suitable for collections of unknown size that may apply memory pressure.
static public IList<T[]> GetChunks<T>(this IEnumerable<T> source, int batchsize)
{
IList<T[]> result = null;
if (source != null && batchsize > 0)
{
var list = source as List<T> ?? source.ToList();
if (list.Count > 0)
{
result = new List<T[]>();
for (var index = 0; index < list.Count; index += batchsize)
{
var rangesize = Math.Min(batchsize, list.Count - index);
result.Add(list.GetRange(index, rangesize).ToArray());
}
}
}
return result ?? Enumerable.Empty<T[]>().ToList();
}
static public void TestGetChunks()
{
var ids = Enumerable.Range(1, 163).Select(i => i.ToString());
foreach (var chunk in ids.GetChunks(20))
{
Console.WriteLine("[{0}]", String.Join(",", chunk));
}
}
I have seen a few answers across this family of questions that use GetRange and Math.Min. But I believe that overall this is a more complete solution in terms of error checking and efficiency.
protected List<List<int>> MySplit(int MaxNumber, int Divider)
{
List<List<int>> lst = new List<List<int>>();
int ListCount = 0;
int d = MaxNumber / Divider;
lst.Add(new List<int>());
for (int i = 1; i <= MaxNumber; i++)
{
lst[ListCount].Add(i);
if (i != 0 && i % d == 0)
{
ListCount++;
d += MaxNumber / Divider;
lst.Add(new List<int>());
}
}
return lst;
}
Great Answers, for my scenario i tested the accepted answer , and it seems it does not keep order. there is also great answer by Nawfal that keeps order. But in my scenario i wanted to split the remainder in a normalized way, all answers i saw spread the remainder or at the beginning or at the end.
My answer also takes the remainder spreading in more normalized way.
static class Program
{
static void Main(string[] args)
{
var input = new List<String>();
for (int k = 0; k < 18; ++k)
{
input.Add(k.ToString());
}
var result = splitListIntoSmallerLists(input, 15);
int i = 0;
foreach(var resul in result){
Console.WriteLine("------Segment:" + i.ToString() + "--------");
foreach(var res in resul){
Console.WriteLine(res);
}
i++;
}
Console.ReadLine();
}
private static List<List<T>> splitListIntoSmallerLists<T>(List<T> i_bigList,int i_numberOfSmallerLists)
{
if (i_numberOfSmallerLists <= 0)
throw new ArgumentOutOfRangeException("Illegal value of numberOfSmallLists");
int normalizedSpreadRemainderCounter = 0;
int normalizedSpreadNumber = 0;
//e.g 7 /5 > 0 ==> output size is 5 , 2 /5 < 0 ==> output is 2
int minimumNumberOfPartsInEachSmallerList = i_bigList.Count / i_numberOfSmallerLists;
int remainder = i_bigList.Count % i_numberOfSmallerLists;
int outputSize = minimumNumberOfPartsInEachSmallerList > 0 ? i_numberOfSmallerLists : remainder;
//In case remainder > 0 we want to spread the remainder equally between the others
if (remainder > 0)
{
if (minimumNumberOfPartsInEachSmallerList > 0)
{
normalizedSpreadNumber = (int)Math.Floor((double)i_numberOfSmallerLists / remainder);
}
else
{
normalizedSpreadNumber = 1;
}
}
List<List<T>> retVal = new List<List<T>>(outputSize);
int inputIndex = 0;
for (int i = 0; i < outputSize; ++i)
{
retVal.Add(new List<T>());
if (minimumNumberOfPartsInEachSmallerList > 0)
{
retVal[i].AddRange(i_bigList.GetRange(inputIndex, minimumNumberOfPartsInEachSmallerList));
inputIndex += minimumNumberOfPartsInEachSmallerList;
}
//If we have remainder take one from it, if our counter is equal to normalizedSpreadNumber.
if (remainder > 0)
{
if (normalizedSpreadRemainderCounter == normalizedSpreadNumber-1)
{
retVal[i].Add(i_bigList[inputIndex]);
remainder--;
inputIndex++;
normalizedSpreadRemainderCounter=0;
}
else
{
normalizedSpreadRemainderCounter++;
}
}
}
return retVal;
}
}
If order in these parts is not very important you can try this:
int[] array = new int[] { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 };
int n = 3;
var result =
array.Select((value, index) => new { Value = value, Index = index }).GroupBy(i => i.Index % n, i => i.Value);
// or
var result2 =
from i in array.Select((value, index) => new { Value = value, Index = index })
group i.Value by i.Index % n into g
select g;
However these can't be cast to IEnumerable<IEnumerable<int>> by some reason...
This is my code, nice and short.
<Extension()> Public Function Chunk(Of T)(ByVal this As IList(Of T), ByVal size As Integer) As List(Of List(Of T))
Dim result As New List(Of List(Of T))
For i = 0 To CInt(Math.Ceiling(this.Count / size)) - 1
result.Add(New List(Of T)(this.GetRange(i * size, Math.Min(size, this.Count - (i * size)))))
Next
Return result
End Function
This is my way, listing items and breaking row by columns
int repat_count=4;
arrItems.ForEach((x, i) => {
if (i % repat_count == 0)
row = tbo.NewElement(el_tr, cls_min_height);
var td = row.NewElement(el_td);
td.innerHTML = x.Name;
});
I was looking for a split like the one with string, so the whole List is splitted according to some rule, not only the first part, this is my solution
List<int> sequence = new List<int>();
for (int i = 0; i < 2000; i++)
{
sequence.Add(i);
}
int splitIndex = 900;
List<List<int>> splitted = new List<List<int>>();
while (sequence.Count != 0)
{
splitted.Add(sequence.Take(splitIndex).ToList() );
sequence.RemoveRange(0, Math.Min(splitIndex, sequence.Count));
}
Here is a little tweak for the number of items instead of the number of parts:
public static class MiscExctensions
{
public static IEnumerable<IEnumerable<T>> Split<T>(this IEnumerable<T> list, int nbItems)
{
return (
list
.Select((o, n) => new { o, n })
.GroupBy(g => (int)(g.n / nbItems))
.Select(g => g.Select(x => x.o))
);
}
}
below code returns both given number of chunks also with sorted data
static IEnumerable<IEnumerable<T>> SplitSequentially<T>(int chunkParts, List<T> inputList)
{
List<int> Splits = split(inputList.Count, chunkParts);
var skipNumber = 0;
List<List<T>> list = new List<List<T>>();
foreach (var count in Splits)
{
var internalList = inputList.Skip(skipNumber).Take(count).ToList();
list.Add(internalList);
skipNumber += count;
}
return list;
}
static List<int> split(int x, int n)
{
List<int> list = new List<int>();
if (x % n == 0)
{
for (int i = 0; i < n; i++)
list.Add(x / n);
}
else
{
// upto n-(x % n) the values
// will be x / n
// after that the values
// will be x / n + 1
int zp = n - (x % n);
int pp = x / n;
for (int i = 0; i < n; i++)
{
if (i >= zp)
list.Add((pp + 1));
else
list.Add(pp);
}
}
return list;
}
int[] items = new int[] { 0,1,2,3,4,5,6,7,8,9, 10 };
int itemIndex = 0;
int groupSize = 2;
int nextGroup = groupSize;
var seqItems = from aItem in items
group aItem by
(itemIndex++ < nextGroup)
?
nextGroup / groupSize
:
(nextGroup += groupSize) / groupSize
into itemGroup
select itemGroup.AsEnumerable();
Just came across this thread, and most of the solutions here involve adding items to collections, effectively materialising each page before returning it. This is bad for two reasons - firstly if your pages are large there's a memory overhead to filling the page, secondly there are iterators which invalidate previous records when you advance to the next one (for example if you wrap a DataReader within an enumerator method).
This solution uses two nested enumerator methods to avoid any need to cache items into temporary collections. Since the outer and inner iterators are traversing the same enumerable, they necessarily share the same enumerator, so it's important not to advance the outer one until you're done with processing the current page. That said, if you decide not to iterate all the way through the current page, when you move to the next page this solution will iterate forward to the page boundary automatically.
using System.Collections.Generic;
public static class EnumerableExtensions
{
/// <summary>
/// Partitions an enumerable into individual pages of a specified size, still scanning the source enumerable just once
/// </summary>
/// <typeparam name="T">The element type</typeparam>
/// <param name="enumerable">The source enumerable</param>
/// <param name="pageSize">The number of elements to return in each page</param>
/// <returns></returns>
public static IEnumerable<IEnumerable<T>> Partition<T>(this IEnumerable<T> enumerable, int pageSize)
{
var enumerator = enumerable.GetEnumerator();
while (enumerator.MoveNext())
{
var indexWithinPage = new IntByRef { Value = 0 };
yield return SubPartition(enumerator, pageSize, indexWithinPage);
// Continue iterating through any remaining items in the page, to align with the start of the next page
for (; indexWithinPage.Value < pageSize; indexWithinPage.Value++)
{
if (!enumerator.MoveNext())
{
yield break;
}
}
}
}
private static IEnumerable<T> SubPartition<T>(IEnumerator<T> enumerator, int pageSize, IntByRef index)
{
for (; index.Value < pageSize; index.Value++)
{
yield return enumerator.Current;
if (!enumerator.MoveNext())
{
yield break;
}
}
}
private class IntByRef
{
public int Value { get; set; }
}
}
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