I'm writing some code in Python that takes a set of [x,y] coordinates (in a list of tuples) and also an x-value. I need to find the two closest x coordinates to my x-value.

To find the closest x, I created a list of all the x-coordinates, used that list to create a lambda function that found the closest x to my x-value, and then found the index of that x.

Finding the 2nd closest value was a bit trickier. I created a new list that excluded the closest x and followed the same process.

Code is below. Any suggestions on how to improve this or make it more efficient?

def closest_2_points(list_of_tuples, x_value):

    x_points = []

    for i in list_of_tuples:

    # find closest point
    closest_xpoint_1 = min(x_points, key=lambda z:abs(z-x_value))
    closest_xpoint_1_idx = x_points.index(closest_xpoint_1)

    # find 2nd closest point
    x_points_2 = [x for x in x_points if x != closest_xpoint_1]    
    closest_xpoint_2 = min(x_points_2, key=lambda z:abs(z-x_value))
    closest_xpoint_2_idx = x_points.index(closest_xpoint_2)

    closest_points = [(list_of_tuples[closest_xpoint_1_idx][0], list_of_tuples[closest_xpoint_1_idx][1]), (list_of_tuples[closest_xpoint_2_idx][0], list_of_tuples[closest_xpoint_2_idx][1]) ]

    return closest_points

When running the function, it should look something like this:

closest_2_points([(4,6),(2,5),(0,4),(-2,3)], 6)

And return something like this:

[(4, 6), (2, 5)]

2 Answers 2


If there are duplicate points in your list, and the closest point turns out to be one of these points, this line of code will remove all of the duplicates:

x_points_2 = [x for x in x_points if x != closest_xpoint_1]

So, instead of returning the 2 closest points, it may return just one of the closest, and some other nearby point.

Finding the n-smallest, n-closest, etc is usually best done using Python's heapq.nsmallest(n, iterable, key=None) function.

Return a list with the n smallest elements from the dataset defined by iterable. key, if provided, specifies a function of one argument that is used to extract a comparison key from each element in iterable (for example, key=str.lower). Equivalent to: sorted(iterable, key=key)[:n].

[This function] perform best for smaller values of n. For larger values, it is more efficient to use the sorted() function. Also, when n==1, it is more efficient to use the built-in min() and max() functions. If repeated usage of these functions is required, consider turning the iterable into an actual heap.

Example: a closest n points function, with n defaulting to 2:

import heapq

def closest_points(list_of_tuples, x_value, n=2):
    return heapq.nsmallest(n, list_of_tuples, lambda pnt:abs(pnt[0] - x_value))


>>> closest_points([(4, 6), (2, 5), (0, 4), (-2, 3)], 6)
[(4, 6), (2, 5)]
  • \$\begingroup\$ Thanks for the feedback. In this particular instance, removing duplicates would not be an issue (in fact I'd want to remove them), but I neglected to mention that particular assumption in the original post. Thanks for your answer; I was not aware of the heapq library. \$\endgroup\$ Commented Apr 24, 2020 at 18:53

Welcome to codereview Ragnar! This was a fun one

Using list comprehensions, you can create a list that contains a tuple with 3 values:

  1. Distance to x
  2. Index of the tuple
  3. x value

So then you can order the list, and take as many items from the tuple as desired

x = 6
points = [(4,6),(2,5),(0,4),(-2,3)]
closest_points = sorted([(x-xvalue[0], index, xvalue[0]) 
                        for index, xvalue in enumerate(points)])

Then you can slice the list and take values from there

> [(2, 0, 4), (4, 1, 2)]

In this example, you would take:

  • 0 as index of first element, 4 as the x value
  • Then 1 as index of second element, 2 as value
  • 3
    \$\begingroup\$ Please explain how this is better than the OPs. Can you also explain why \$O(n\log n)\$ time is better than \$O(n)\$. \$\endgroup\$
    – Peilonrayz
    Commented Apr 24, 2020 at 12:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.