LINQ to find the closest number that is greater / less than an input

Question

Suppose I have this number list:

List<int> = new List<int>(){3,5,8,11,12,13,14,21}

Suppose that I want to get the closest number that is less than 11, it would be 8 Suppose that I want to get the closest number that is greater than 13 that would be 14.

The numbers in list can't be duplicated and are always ordered. How can I write Linq for this?

Answer 1

with Linq assuming that the list is ordered I would do it like this:

var l = new List<int>() { 3, 5, 8, 11, 12, 13, 14, 21 };
var lessThan11 = l.TakeWhile(p => p < 11).Last();
var greaterThan13 = l.SkipWhile(p => p <= 13).First();

EDIT:

As I have received negative feedback about this answer and for the sake of people that may see this answer and while it's accepted don't go further, I explored the other comments regarding BinarySearch and decided to add the second option in here (with some minor change).

This is the not sufficient way presented somewhere else:

var l = new List<int>() { 3, 5, 8, 11, 12, 13, 14, 21 };
var indexLessThan11 = ~l.BinarySearch(10) -1;
var value = l[indexLessThan11];

Now the code above doesn't cope with the fact that the value 10 might actually be in the list (in which case one shouldn't invert the index)! so the good way is to do it:

var l = new List<int>() { 3, 5, 8, 11, 12, 13, 14, 21 };
var indexLessThan11 = l.BinarySearch(10);
if (indexLessThan11 < 0) // the value 10 wasn't found
{    
    indexLessThan11 = ~indexLessThan11;
    indexLessThan11 -= 1;
}
var value = l[indexLessThan11];

I simply want to note that:

l.BinarySearch(11) == 3
//and
l.BinarySearch(10) == -4;

Answer 2

Use Array.BinarySearch - no need for LINQ or visiting on average half the elements to find your target.

There are also a variety of SortedXXX classes that may be suitable for what you're doing [that will have such efficient O(log N) searches built-in]

Answer 3

You can do this using a binary search. If your searching for 11, well obviously you'll get the index your after. If you search for 10 and use the bitwise complement of the result, you'll get the closest match.

   List<int> list = new List<int>(){3,5,8,11,12,13,14,21};

   list.Sort();

   int index = list.BinarySearch(10);

   int found =  (~index)-1;

   Console.WriteLine (list[found]); // Outputs 8

The same goes searching in the other direction

int index = list.BinarySearch(15);

Console.WriteLine("Closest match : " + list[+~index]); // Outputs 21

Binary searches are also extremely fast.

Answer 4

closest number below 11:

        int someNumber = 11;
        List<int> list = new List<int> { 3, 5, 8, 11, 12, 13, 14, 21 };

        var intermediate = from i in list
                     where i < someNumber
                     orderby i descending
                     select i;

        var result = intermediate.FirstOrDefault();

closest number above 13:

        int someNumber = 13;
        List<int> list = new List<int> { 3, 5, 8, 11, 12, 13, 14, 21 };

        var intermediate = from i in list
                     where i > someNumber
                     orderby i
                     select i;

        var result = intermediate.FirstOrDefault();

Answer 5

var list = new List<int> {14,2,13,11,5,8,21,12,3};
var tested = 11;

var closestGreater = list.OrderBy(n => n)
                         .FirstOrDefault(n => tested < n); // = 12

var closestLess = list.OrderByDescending(n => n)
                      .FirstOrDefault(n => tested > n); // = 8

if (closestGreater == 0)
    System.Diagnostics.Debug.WriteLine(
        string.Format("No number greater then {0} exists in the list", tested));

if (closestLess == 0)
    System.Diagnostics.Debug.WriteLine(
        string.Format("No number smaler then {0} exists in the list", tested));

Answer 6

Here is my way hope this helps somebody!

List<float> list = new List<float> { 4.0f, 5.0f, 6.0f, 10.0f, 4.5f,  4.0f, 5.0f, 6.0f, 10.0f, 4.5f, 4.0f, 5.0f, 6.0f, 10.0f };
float num = 4.7f;

float closestAbove = list.Aggregate((x , y) => (x < num ? y : y < num ? x : (Math.Abs(x - num)) < Math.Abs(y - num) ? x : y));
float closestBelow = list.Aggregate((x , y) => (x > num ? y : y > num ? x : (Math.Abs(x - num)) < Math.Abs(y - num) ? x : y));

Console.WriteLine(closestAbove);
Console.WriteLine(closestBelow);

This means you dont have to order the list

Credit: addapted from here: How to get the closest number from a List<int> with LINQ?

The Expanded Code

float closestAboveExplained = list.Aggregate((closestAbove , next) => {
    if(next < num){
        return closestAbove;
    }

    if(closestAbove < num){
        return next;
    }

    else{
        if(Math.Abs(closestAbove - num) < Math.Abs(next - num)){
            return closestAbove;
        }
    }
    return next;
});

Answer 7

This is my answer

List<int> myList = new List<int>() { 3, 5, 8, 11, 12, 13, 14, 21 };
    int n = 11;
    int? smallerNumberCloseToInput = (from n1 in myList
                                    where n1 < n
                                    orderby n1 descending
                                    select n1).First();

    int? largerNumberCloseToInput = (from n1 in myList
                                    where n1 > n
                                    orderby n1 ascending
                                    select n1).First();

Answer 8

You can use a query for this such as:

List<int> numbers = new List<int>() { 3, 5, 8, 11, 12, 13, 14, 21 };
List<int> output = (from n in numbers
                            where n > 13 // or whatever
                            orderby n ascending //or descending
                            select n).ToList();