Compare value with array and get closest value to it

Question

I'm a rookie in C# and I'm trying to learn that language.

Can you guys give me a tip how I can compare an array with a value picking the lowest from it?

like:

Double[] w = { 1000, 2000, 3000, 4000, 5000 };

double min = double.MaxValue;
double max = double.MinValue;

foreach (double value in w)
{
    if (value < min)
        min = value;
    if (value > max)
        max = value;
}

Console.WriteLine(" min:", min); 

gives me the lowest value of w, how can I compare now?

If I have:

int p = 1001 + 2000;  // 3001

how can I compare now with the list of the array and find out that the (3000) value is the nearest value to my "Searchvalue"?

Answer 1

You can do this with some simple mathematics and there are different approaches.

LINQ

Double searchValue = ...;

Double nearest = w.Select(p => new { Value = p, Difference = Math.Abs(p - searchValue) })
                  .OrderBy(p => p.Difference)
                  .First().Value;

Manually

Double[] w = { 1000, 2000, 3000, 4000, 5000 };

Double searchValue = 3001;
Double currentNearest = w[0];
Double currentDifference = Math.Abs(currentNearest - searchValue);

for (int i = 1; i < w.Length; i++)
{
    Double diff = Math.Abs(w[i] - searchValue);
    if (diff < currentDifference)
    {
        currentDifference = diff;
        currentNearest = w[i];
    }
}

Answer 2

Double[] w = { 1000, 2000, 3000, 4000, 5000 };
var minimumValueFromArray = w.Min();

produces

1000, as expected, cause we execute Enumerable.Min.

The same is for Enumerable.Max, to figure out the max value:

Double[] w = { 1000, 2000, 3000, 4000, 5000 };
var maximumValueFromArray = w.Max();

Considering that you're comparing with the double.MinValue and double.MaxValue, I would assume that you want just pick the smallest and biggest value from array.

If this is not what you're searching for, please clarify.

Answer 3

based on your code, you can achieve this in a very easy way

    Double[] w = { 1000, 2000, 3000, 4000, 5000 }; // it has to be sorted

    double search = 3001;
    double lowerClosest = 0;
    double upperClosest = 0;


        for (int i = 1; i < w.Length; i++)
        {
            if (w[i] > search)
            {
                upperClosest = w[i];
                break; // interrupts your foreach
            }

        }
        for (int i = w.Length-1; i >=0; i--)
        {
            if (w[i] <= search)
            {
                lowerClosest = w[i];
                break; // interrupts your foreach
            }

        }

    Console.WriteLine(" lowerClosest:{0}", lowerClosest);
    Console.WriteLine(" upperClosest:{0}", upperClosest);
    if (upperClosest - search > search - lowerClosest)
        Console.WriteLine(" Closest:{0}", lowerClosest);
    else
        Console.WriteLine(" Closest:{0}", upperClosest);

    Console.ReadLine();

depending on the position of your search value this will be less than O(n)

Answer 4

                Performance wise custom code will be more use full. 

                List<int> results;
                int targetNumber = 0;
                int nearestValue=0;
                if (results.Any(ab => ab == targetNumber ))
                {
                    nearestValue= results.FirstOrDefault<int>(i => i == targetNumber );
                }
                else
                {
                    int greaterThanTarget = 0;
                    int lessThanTarget = 0;
                    if (results.Any(ab => ab > targetNumber ))
                    {
                        greaterThanTarget = results.Where<int>(i => i > targetNumber ).Min();
                    }
                    if (results.Any(ab => ab < targetNumber ))
                    {
                        lessThanTarget = results.Where<int>(i => i < targetNumber ).Max();
                    }

                    if (lessThanTarget == 0 )
                    {
                        nearestValue= greaterThanTarget;
                    }
                    else if (greaterThanTarget == 0)
                    {
                        nearestValue= lessThanTarget;
                    }
                    else if (targetNumber - lessThanTarget < greaterThanTarget - targetNumber )
                    {
                        nearestValue= lessThanTarget;
                    }
                    else
                    {
                            nearestValue= greaterThanTarget;
                    }
                }