Finding k closest numbers to a given number

Say I have a list [1,2,3,4,5,6,7]. I want to find the 3 closest numbers to, say, 6.5. Then the returned value would be [5,6,7].

Finding one closest number is not that tricky in python, which can be done using

min(myList, key=lambda x:abs(x-myNumber))

But I am trying not to put a loop around this to find k closest numbers. Is there a pythonic way to achieve the above task?

Answer

The short answer

The heapq.nsmallest() function will do this neatly and efficiently:

>>> from heapq import nsmallest
>>> s = [1,2,3,4,5,6,7]
>>> nsmallest(3, s, key=lambda x: abs(x - 6.5))
[6, 7, 5]

Essentially this says, “Give me the three input values that have the smallest absolute difference from the number 6.5“.

Optimizing for repeated lookups

In the comments, @Phylliida, asked how to optimize for repeated lookups with differing start points. One approach would be to pre-sort the data and then use bisect to locate the center of a small search segment:

from bisect import bisect

def k_nearest(k, center, sorted_data):
    'Return *k* members of *sorted_data* nearest to *center*'
    i = bisect(sorted_data, center)
    segment = sorted_data[max(i-k, 0) : i+k]
    return nsmallest(k, segment, key=lambda x: abs(x - center))

For example:

>>> s.sort()
>>> k_nearest(3, 6.5, s)
[6, 7, 5]
>>> k_nearest(3, 0.5, s)
[1, 2, 3]
>>> k_nearest(3, 4.5, s)    
[4, 5, 3]
>>> k_nearest(3, 5.0, s)
[5, 4, 6]