How I can use long type (Int64) with Skip in Linq. It support only Int32.
dataContext.Persons.Skip(LongNumber);
You can use a while loop:
// Some init
List<Person> persons = new List<Person>();
List<Person> resultList = persons;
long bigNumber = 3 * (long)int.MaxValue + 12;
while (bigNumber > int.MaxValue)
{
resultList = resultList.Skip(int.MaxValue).ToList();
bigNumber -= int.MaxValue;
}
resultList = resultList.Skip(int.MaxValue).ToList();
// Then what do what you want with this result list
But does your collection have more than int.MaxValue entries?
The following extension method BigSkip allows skipping more than the Int32.MaxValue maximum value of LINQ's Skip method by calling the method multiple times until the long value has been reached. This method has the advantage of not causing iteration over the collection prematurely.
example usage
bigCollection.BigSkip(howMany: int.MaxValue + 1l)
method
using System;
using System.Collections.Generic;
using System.Linq;
static public class LinqExtensions
{
static public IEnumerable<T> BigSkip<T>(this IEnumerable<T> items, long howMany)
=> BigSkip(items, Int32.MaxValue, howMany);
internal static IEnumerable<T> BigSkip<T>(this IEnumerable<T> items, int segmentSize, long howMany)
{
long segmentCount = Math.DivRem(howMany, segmentSize,
out long remainder);
for (long i = 0; i < segmentCount; i += 1)
items = items.Skip(segmentSize);
if (remainder != 0)
items = items.Skip((int)remainder);
return items;
}
}
The method has been split into two: the internal overload is for convenience allowing a smaller segment size to be specified than Int32.MaxValue to make it unit testable on a smaller scale.
bonus
Replace Skip with Take to make a BigTake method; this same extension method pattern can be used to extend the reach of other LINQ methods.
Here is an implementation that includes a fast path, in case the count is in the Int32 range. In that case, any optimizations that are embedded in the native Skip implementation are enabled.
/// <summary>Bypasses a specified number of elements in a sequence and then
/// returns the remaining elements.</summary>
public static IEnumerable<T> LongSkip<T>(this IEnumerable<T> source, long count)
{
if (count >= 0 && count <= Int32.MaxValue)
return source.Skip(checked((int)count));
else
return Iterator(source, count);
static IEnumerable<T> Iterator(IEnumerable<T> source, long count)
{
long current = 0;
foreach (var item in source)
if (++current > count) yield return item;
}
}
"Searching for alternative functionalities for "Skip" and "Take" functionalities"
1 of the link says "Everytime you invoke Skip() it will have to iterate you collection from the beginning in order to skip the number of elements you desire, which gives a loop within a loop (n2 behaviour)"
Conclusion: For large collections, don’t use Skip and Take. Find another way to iterate through your collection and divide it.
In order to access last page data in a huge collection, can you please suggest us a way other than Skip and Take approach?
Looking at the source for Skip, you can see it enumerates over all the items, even over the first n items you want to skip.
It's strange though, because several LINQ-methods have optimizations for collections, like Count and Last.
Skip apparently does not.
If you have an array or IList<T>, you use the indexer to truly skip over them:
for (int i = skipStartIndex; i < list.Count; i++) {
yield return list[i];
}
Internally it is really correct:
private static IEnumerable<TSource> SkipIterator<TSource>(IEnumerable<TSource> source, int count)
{
using (IEnumerator<TSource> enumerator = source.GetEnumerator())
{
while (count > 0 && enumerator.MoveNext())
--count;
if (count <= 0)
{
while (enumerator.MoveNext())
yield return enumerator.Current;
}
}
}
If you want to skip for IEnumerable<T> then it works right. There are no other way except enumeration to get specific element(s). But you can write own extension method on IReadOnlyList<T> or IList<T> (if this interface is implemented in collection used for your elements).
public static class IReadOnlyListExtensions
{
public static IEnumerable<T> Skip<T>(this IReadOnlyList<T> collection, int count)
{
if (collection == null)
return null;
return ICollectionExtensions.YieldSkip(collection, count);
}
private static IEnumerable<T> YieldSkip<T>(IReadOnlyList<T> collection, int count)
{
for (int index = count; index < collection.Count; index++)
{
yield return collection[index];
}
}
}
In addition you can implement it for IEnumerable<T> but check inside for optimization:
if (collection is IReadOnlyList<T>)
{
// do optimized skip
}
Such solution is used a lot of where in Linq source code (but not in Skip unfortunately).
Depends on your implementation, but it would make sense to use indexed arrays for the purpose, instead.
How I can use long type (Int64) with Skip in Linq. It support only Int32.
dataContext.Persons.Skip(LongNumber);
You can use a while loop:
// Some init
List<Person> persons = new List<Person>();
List<Person> resultList = persons;
long bigNumber = 3 * (long)int.MaxValue + 12;
while (bigNumber > int.MaxValue)
{
resultList = resultList.Skip(int.MaxValue).ToList();
bigNumber -= int.MaxValue;
}
resultList = resultList.Skip(int.MaxValue).ToList();
// Then what do what you want with this result list
But does your collection have more than int.MaxValue entries?
The following extension method BigSkip allows skipping more than the Int32.MaxValue maximum value of LINQ's Skip method by calling the method multiple times until the long value has been reached. This method has the advantage of not causing iteration over the collection prematurely.
example usage
bigCollection.BigSkip(howMany: int.MaxValue + 1l)
method
using System;
using System.Collections.Generic;
using System.Linq;
static public class LinqExtensions
{
static public IEnumerable<T> BigSkip<T>(this IEnumerable<T> items, long howMany)
=> BigSkip(items, Int32.MaxValue, howMany);
internal static IEnumerable<T> BigSkip<T>(this IEnumerable<T> items, int segmentSize, long howMany)
{
long segmentCount = Math.DivRem(howMany, segmentSize,
out long remainder);
for (long i = 0; i < segmentCount; i += 1)
items = items.Skip(segmentSize);
if (remainder != 0)
items = items.Skip((int)remainder);
return items;
}
}
The method has been split into two: the internal overload is for convenience allowing a smaller segment size to be specified than Int32.MaxValue to make it unit testable on a smaller scale.
bonus
Replace Skip with Take to make a BigTake method; this same extension method pattern can be used to extend the reach of other LINQ methods.
Here is an implementation that includes a fast path, in case the count is in the Int32 range. In that case, any optimizations that are embedded in the native Skip implementation are enabled.
/// <summary>Bypasses a specified number of elements in a sequence and then
/// returns the remaining elements.</summary>
public static IEnumerable<T> LongSkip<T>(this IEnumerable<T> source, long count)
{
if (count >= 0 && count <= Int32.MaxValue)
return source.Skip(checked((int)count));
else
return Iterator(source, count);
static IEnumerable<T> Iterator(IEnumerable<T> source, long count)
{
long current = 0;
foreach (var item in source)
if (++current > count) yield return item;
}
}
Is there any way to access the IEnumerable<T> collection being build up by yield return in a loop from within the method building the IEnumerable itself?
Silly example:
Random random = new Random();
IEnumerable<int> UniqueRandomIntegers(int n, int max)
{
while ([RETURN_VALUE].Count() < n)
{
int value = random.Next(max);
if (![RETURN_VALUE].Contains(value))
yield return value;
}
}
There is no collection being built up. The sequence that is returned is evaluated lazily, and unless the caller explicitly copies the data to another collection, it will be gone as soon as it's been fetched.
If you want to ensure uniqueness, you'll need to do that yourself. For example:
IEnumerable<int> UniqueRandomIntegers(int n, int max)
{
HashSet<int> returned = new HashSet<int>();
for (int i = 0; i < n; i++)
{
int candidate;
do
{
candidate = random.Next(max);
} while (returned.Contains(candidate));
yield return candidate;
returned.Add(candidate);
}
}
Another alternative for unique random integers is to build a collection of max items and shuffle it, which can still be done just-in-time. This is more efficient in the case where max and n are similar (as you don't need to loop round until you're lucky enough to get a new item) but inefficient in the case where max is very large and n isn't.
EDIT: As noted in comments, you can shorten this slightly by changing the body of the for loop to:
int candidate;
do
{
candidate = random.Next(max);
} while (!returned.Add(candidate))
yield return candidate;
That uses the fact that Add will return false if the item already exists in the set.
Context: C# 3.0, .Net 3.5
Suppose I have a method that generates random numbers (forever):
private static IEnumerable<int> RandomNumberGenerator() {
while (true) yield return GenerateRandomNumber(0, 100);
}
I need to group those numbers in groups of 10, so I would like something like:
foreach (IEnumerable<int> group in RandomNumberGenerator().Slice(10)) {
Assert.That(group.Count() == 10);
}
I have defined Slice method, but I feel there should be one already defined. Here is my Slice method, just for reference:
private static IEnumerable<T[]> Slice<T>(IEnumerable<T> enumerable, int size) {
var result = new List<T>(size);
foreach (var item in enumerable) {
result.Add(item);
if (result.Count == size) {
yield return result.ToArray();
result.Clear();
}
}
}
Question: is there an easier way to accomplish what I'm trying to do? Perhaps Linq?
Note: above example is a simplification, in my program I have an Iterator that scans given matrix in a non-linear fashion.
EDIT: Why Skip+Take is no good.
Effectively what I want is:
var group1 = RandomNumberGenerator().Skip(0).Take(10);
var group2 = RandomNumberGenerator().Skip(10).Take(10);
var group3 = RandomNumberGenerator().Skip(20).Take(10);
var group4 = RandomNumberGenerator().Skip(30).Take(10);
without the overhead of regenerating number (10+20+30+40) times. I need a solution that will generate exactly 40 numbers and break those in 4 groups by 10.
Are Skip and Take of any use to you?
Use a combination of the two in a loop to get what you want.
So,
list.Skip(10).Take(10);
Skips the first 10 records and then takes the next 10.
I have done something similar. But I would like it to be simpler:
//Remove "this" if you don't want it to be a extension method
public static IEnumerable<IList<T>> Chunks<T>(this IEnumerable<T> xs, int size)
{
var curr = new List<T>(size);
foreach (var x in xs)
{
curr.Add(x);
if (curr.Count == size)
{
yield return curr;
curr = new List<T>(size);
}
}
}
I think yours are flawed. You return the same array for all your chunks/slices so only the last chunk/slice you take would have the correct data.
Addition: Array version:
public static IEnumerable<T[]> Chunks<T>(this IEnumerable<T> xs, int size)
{
var curr = new T[size];
int i = 0;
foreach (var x in xs)
{
curr[i % size] = x;
if (++i % size == 0)
{
yield return curr;
curr = new T[size];
}
}
}
Addition: Linq version (not C# 2.0). As pointed out, it will not work on infinite sequences and will be a great deal slower than the alternatives:
public static IEnumerable<T[]> Chunks<T>(this IEnumerable<T> xs, int size)
{
return xs.Select((x, i) => new { x, i })
.GroupBy(xi => xi.i / size, xi => xi.x)
.Select(g => g.ToArray());
}
Using Skip and Take would be a very bad idea. Calling Skip on an indexed collection may be fine, but calling it on any arbitrary IEnumerable<T> is liable to result in enumeration over the number of elements skipped, which means that if you're calling it repeatedly you're enumerating over the sequence an order of magnitude more times than you need to be.
Complain of "premature optimization" all you want; but that is just ridiculous.
I think your Slice method is about as good as it gets. I was going to suggest a different approach that would provide deferred execution and obviate the intermediate array allocation, but that is a dangerous game to play (i.e., if you try something like ToList on such a resulting IEnumerable<T> implementation, without enumerating over the inner collections, you'll end up in an endless loop).
(I've removed what was originally here, as the OP's improvements since posting the question have since rendered my suggestions here redundant.)
Let's see if you even need the complexity of Slice. If your random number generates is stateless, I would assume each call to it would generate unique random numbers, so perhaps this would be sufficient:
var group1 = RandomNumberGenerator().Take(10);
var group2 = RandomNumberGenerator().Take(10);
var group3 = RandomNumberGenerator().Take(10);
var group4 = RandomNumberGenerator().Take(10);
Each call to Take returns a new group of 10 numbers.
Now, if your random number generator re-seeds itself with a specific value each time it's iterated, this won't work. You'll simply get the same 10 values for each group. So instead, you would use:
var generator = RandomNumberGenerator();
var group1 = generator.Take(10);
var group2 = generator.Take(10);
var group3 = generator.Take(10);
var group4 = generator.Take(10);
This maintains an instance of the generator so that you can continue retrieving values without re-seeding the generator.
You could use the Skip and Take methods with any Enumerable object.
For your edit :
How about a function that takes a slice number and a slice size as a parameter?
private static IEnumerable<T> Slice<T>(IEnumerable<T> enumerable, int sliceSize, int sliceNumber) {
return enumerable.Skip(sliceSize * sliceNumber).Take(sliceSize);
}
It seems like we'd prefer for an IEnumerable<T> to have a fixed position counter so that we can do
var group1 = items.Take(10);
var group2 = items.Take(10);
var group3 = items.Take(10);
var group4 = items.Take(10);
and get successive slices rather than getting the first 10 items each time. We can do that with a new implementation of IEnumerable<T> which keeps one instance of its Enumerator and returns it on every call of GetEnumerator:
public class StickyEnumerable<T> : IEnumerable<T>, IDisposable
{
private IEnumerator<T> innerEnumerator;
public StickyEnumerable( IEnumerable<T> items )
{
innerEnumerator = items.GetEnumerator();
}
public IEnumerator<T> GetEnumerator()
{
return innerEnumerator;
}
System.Collections.IEnumerator System.Collections.IEnumerable.GetEnumerator()
{
return innerEnumerator;
}
public void Dispose()
{
if (innerEnumerator != null)
{
innerEnumerator.Dispose();
}
}
}
Given that class, we could implement Slice with
public static IEnumerable<IEnumerable<T>> Slices<T>(this IEnumerable<T> items, int size)
{
using (StickyEnumerable<T> sticky = new StickyEnumerable<T>(items))
{
IEnumerable<T> slice;
do
{
slice = sticky.Take(size).ToList();
yield return slice;
} while (slice.Count() == size);
}
yield break;
}
That works in this case, but StickyEnumerable<T> is generally a dangerous class to have around if the consuming code isn't expecting it. For example,
using (var sticky = new StickyEnumerable<int>(Enumerable.Range(1, 10)))
{
var first = sticky.Take(2);
var second = sticky.Take(2);
foreach (int i in second)
{
Console.WriteLine(i);
}
foreach (int i in first)
{
Console.WriteLine(i);
}
}
prints
1
2
3
4
rather than
3
4
1
2
Take a look at Take(), TakeWhile() and Skip()
I think the use of Slice() would be a bit misleading. I think of that as a means to give me a chuck of an array into a new array and not causing side effects. In this scenario you would actually move the enumerable forward 10.
A possible better approach is to just use the Linq extension Take(). I don't think you would need to use Skip() with a generator.
Edit: Dang, I have been trying to test this behavior with the following code
Note: this is wasn't really correct, I leave it here so others don't fall into the same mistake.
var numbers = RandomNumberGenerator();
var slice = numbers.Take(10);
public static IEnumerable<int> RandomNumberGenerator()
{
yield return random.Next();
}
but the Count() for slice is alway 1. I also tried running it through a foreach loop since I know that the Linq extensions are generally lazily evaluated and it only looped once. I eventually did the code below instead of the Take() and it works:
public static IEnumerable<int> Slice(this IEnumerable<int> enumerable, int size)
{
var list = new List<int>();
foreach (var count in Enumerable.Range(0, size)) list.Add(enumerable.First());
return list;
}
If you notice I am adding the First() to the list each time, but since the enumerable that is being passed in is the generator from RandomNumberGenerator() the result is different every time.
So again with a generator using Skip() is not needed since the result will be different. Looping over an IEnumerable is not always side effect free.
Edit: I'll leave the last edit just so no one falls into the same mistake, but it worked fine for me just doing this:
var numbers = RandomNumberGenerator();
var slice1 = numbers.Take(10);
var slice2 = numbers.Take(10);
The two slices were different.
I had made some mistakes in my original answer but some of the points still stand. Skip() and Take() are not going to work the same with a generator as it would a list. Looping over an IEnumerable is not always side effect free. Anyway here is my take on getting a list of slices.
public static IEnumerable<int> RandomNumberGenerator()
{
while(true) yield return random.Next();
}
public static IEnumerable<IEnumerable<int>> Slice(this IEnumerable<int> enumerable, int size, int count)
{
var slices = new List<List<int>>();
foreach (var iteration in Enumerable.Range(0, count)){
var list = new List<int>();
list.AddRange(enumerable.Take(size));
slices.Add(list);
}
return slices;
}
I got this solution for the same problem:
int[] ints = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10};
IEnumerable<IEnumerable<int>> chunks = Chunk(ints, 2, t => t.Dump());
//won't enumerate, so won't do anything unless you force it:
chunks.ToList();
IEnumerable<T> Chunk<T, R>(IEnumerable<R> src, int n, Func<IEnumerable<R>, T> action){
IEnumerable<R> head;
IEnumerable<R> tail = src;
while (tail.Any())
{
head = tail.Take(n);
tail = tail.Skip(n);
yield return action(head);
}
}
if you just want the chunks returned, not do anything with them, use chunks = Chunk(ints, 2, t => t). What I would really like is to have to have t=>t as default action, but I haven't found out how to do that yet.