How do I find the closest array element to an arbitrary (non-member) number?

Question

Seemingly similar questions: \"Finding closest number in an array\" (in Java) and \"find nearest match to array of doubles\" (actually a geography problem).

Answer 1

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

    if (results.Any(ab => ab < target)) {
        lessThanTarget = results.Where<int>(i => i < target).Max();
    }

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

Answer 2

One way to do this using LINQ is like this:

public int GetClosestIndex( List<double> doublelist, double targetvalue )
{
  return doublelist.IndexOf(doublelist.OrderBy(d => Math.Abs(d - targetvalue)).ElementAt(0));
}

It might have some performance issues, but If the list is not that long, it should not pose a problem. Also, if two elements are equally distant from the target value, it will return the first index of those.

Answer 3

Something like this:

double[] values = new double[]
{
    1.8,
    2.4,
    2.7,
    3.1,
    4.5
};

double difference = double.PositiveInfinity;
int index = -1;

double input = 2.5;

for (int i = 0; i < values.Length; i++)
{
    double currentDifference = Math.Abs(values[i] - input);

    if (currentDifference < difference)
    {
        difference = currentDifference;
        index = i;
    }

    // Stop searching when we've encountered a value larger
    // than the inpt because the values array is sorted.
    if (values[i] > input)
        break;
}

Console.WriteLine("Found index: {0} value {1}", index, values[index]);

Answer 4

Array.BinarySearch, which returns:

The index of the specified value in the specified array, if value is found. If value is not found and value is less than one or more elements in array, a negative number which is the bitwise complement of the index of the first element that is larger than value. If value is not found and value is greater than any of the elements in array, a negative number which is the bitwise complement of (the index of the last element plus 1).

Now that won't get you 100% of the way there, since you'll know the number is either less than or greater than the match, but it really only leaves you with two indices to check.

Answer 5

Perhaps not the fastest solution, but certainly pleasant eye-candy:

double search;
double[] array;

var nearest = (
    from value in array
    orderby Math.Abs(value - search)
    select value).First();

var index = array.IndexOf(nearest);

Note that this will absolutely be slower than a binary search algorithm, because it need to process each element in the array and sorting means building a hash table of those items.