Array of sequence

Question

Let's say I have a sequence of integers, from 0 (always starting at 0) to 3. Now, I have an array of integers, which will hold those sequence looped one after the other, starting from a certain point. For example:

An array of 10 elements, the sequence is 0 to 3, and start at 2, should yield 2, 3, 0, 1, 2, 3, 0, 1, 2, 3.

An array of 5 elements, the sequence 0 to 5, and start at 5, should yield 5, 0, 1, 2, 3.

An array of 5 elements, the sequence 0 to 10, and start at 3, should yield 3, 4, 5, 6, 7.

I'm suffering brain freeze! What's the best way to create this array if you know the array size, max number in sequence, and starting value?

My best attempt was:

private static int[] CreateIndexers(int index, int size, int players) 
{
  var indexers = new int[size];
  for (int i = 0; i < size; i++)
  { 
    var division = i / players; 
    var newInt = division + i >= players ? ((division + i) - players) : division + i;
    indexers[i] = newInt; 
  } 

  return indexers; 
}

Can you post some code? Explain what you tried and where you are stuck. As it stands, this reads like you want us to write your code for you. — Oded
– Oded, Commented Feb 21, 2013 at 21:15
Echoing the "What have you tried?" route - this question is far too "write it for me" as written. — JerKimball
– JerKimball, Commented Feb 21, 2013 at 21:19

Servy · Accepted Answer · 2013-02-21 21:26:13Z

9

public static IEnumerable<int> Foo(int count, int start, int max)
{
    return Enumerable.Range(0, count)
        .Select(n => (n + start) % (max + 1));
}

edited Feb 21, 2013 at 21:26

answered Feb 21, 2013 at 21:20

Servy

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3 Comments

Sergey Berezovskiy Over a year ago

+1 Clever, very nice. It took me half a minute to anderstand how it works :)

vlad Over a year ago

+1 similar to what I have, but you can simplify by starting at start

Servy Over a year ago

@vlad Yeah, I suppose you can. It doesn't seem like the kind of operation worth spending too much time over though.

Lee · Accepted Answer · 2013-02-21 21:19:02Z

6

public int[] Cycle(int max, int start, int count)
{
    int cycles = count / max + 1;
    return Enumerable.Repeat(Enumerable.Range(0, max+1), cycles)
        .SelectMany(seq => seq)
        .Skip(start)
        .Take(count)
        .ToArray();
}

answered Feb 21, 2013 at 21:19

Lee

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2 Comments

BlackBear Over a year ago

Awesome, I'd have never thought at this!

Servy Over a year ago

You may as well just repeat for int.MaxValue, since the Take ensures you never got to the end anyway.

vlad · Accepted Answer · 2013-02-21 21:26:56Z

3

Use LINQ:

public static IEnumerable<int> Foo(int length, int start, int end)
{
    return Enumerable.Range(start, length).Select(n => n % (end + 1));
}

answered Feb 21, 2013 at 21:26

vlad

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Comments

Daniel Mann · Accepted Answer · 2013-02-21 21:27:05Z

0

Here's a less "LINQy" version of a solution:

    public static IEnumerable<int> GetSequence(int start, int end, int count)
    {
        var fullSequence = new List<int>();
        var baseRange = Enumerable.Range(0, end + 1);
        fullSequence.AddRange(baseRange.Skip(start));

        while (fullSequence.Count < count)
        {
            fullSequence.AddRange(baseRange);
        }

        return fullSequence.Take(count);
    }

edited Feb 21, 2013 at 21:27

answered Feb 21, 2013 at 21:15

Daniel Mann

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2 Comments

Sergey Berezovskiy Over a year ago

You can write 'you can use Linq'

Servy Over a year ago

For a less LINQy solution, it still has a lot of LINQ in it...(not that that's a bad thing, just saying).

Amy B · Accepted Answer · 2013-02-21 21:39:50Z

int numberOfElements = 10;
int sequenceStartElement = 0;
int sequenceCount = 4;
int firstElement = 2;

IEnumerable<int> sequence = Enumerable.Range(sequenceStartElement, sequenceCount)
int[] array = sequence
     //you could figure out a lesser number to Repeat... but it's deferred, doesn't matter.
  .Repeat(numberOfElements) 
  .SkipWhile(x => x != firstElement)
  .Take(numberOfElements)
  .ToArray();

What's the best way to create this array if you know the array size, max number in sequence, and starting value?

I see there are three listed inputs, instead of the four previously considered. Here's how to do it with your approach.

private static int[] CreateIndexers(int firstElement, int numberOfElements, int sequenceMax) 
{
  int sequenceCount = sequenceMax + 1
  var indexers = new int[numberOfElements];
  for (int i = 0; i < numberOfElements; i++)
  { 
    indexers[i] = (i + firstElement) % sequenceCount;
  } 
  return indexers; 
}

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