CODEWITHSUNDEEP
c#arrays
Anders Juul
Anders Juul

Reputation: 2604

How to elegantly fill an array/list with values

How do I construct an array of doubles 'in a smart way'? I need it filled with 10-ish values like this, 

var d = new double[] { -0.05, 0.0, 0.05};

but would prefer a more dynamic building like this

    var d = new List<double>();
    for (var dd = -0.05; dd < 0.05; dd += 0.05)
    {
        d.Add(dd);
    }

It looks chunky though and commands too much attention in my code compared to the rather mundane service performed.

Can I write it smarter?

BR, Anders

Upvotes: 0

Views: 115

Answers (4)

Dmitrii Bychenko
Dmitrii Bychenko

Reputation: 186668

Try Enumerable.Range:

  int n = 7;
  double step = 0.05; 

  double[] d = Enumerable
    .Range(-n / 2, n)
    .Select(i => i * step)
    .ToArray();

  Console.Write(string.Join("; ", d));

Outcome

-0.15; -0.1; -0.05; 0; 0.05; 0.1; 0.15

If n = 3 then we'll get

-0.05; 0; 0.05

Upvotes: 2

Martin Verjans
Martin Verjans

Reputation: 4796

All this depends on what you would call "Smart Way". Performance ? Readability ? Maintenance ?

On one hand, if you have only three elements and the values are static, your first line of code is perfectly fine and it is the "Smartest Way". A small improvement would be (if you want a list) :

List<double> d = new List<double> { -0.05, 0, 0.05 };

On another hand, if you have lots of values, a for loop may be easier to write.

Linq is also possible so writing something like this :

List<double> d = Enumerable.Range(-1,3).Select(x => x * 0.05).ToList();

is really good looking because you are using Linq and it rocks, but is it really more readable ? More performant ? Smarter ?

All in all, it all depends on what you want and what you need to do. There is no straight answer to your question.

Upvotes: 2

TheGeneral
TheGeneral

Reputation: 81483

Another approach if you are initalzing a lot of array, you might want to use IEnumerable and yield in a helper class

public static IEnumerable<double> RangeEx(double start, double finish, double step)
{
    for (var dd = start; dd < finish; dd += step) 
        yield return dd;
}

Usage 

var array = RangeEx(-0.05, 0.05, 0.05).ToArray();

Upvotes: 0

Access Denied
Access Denied

Reputation: 9461

You can use linq, not quite sure if it is smart:

double start = -0.05;
double step = 0.05;
var result = Enumerable.Repeat(0, 3).Select((x, i) =>start+step*i).ToArray();

Upvotes: 0

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