How do I return a list of the 3 lowest values in another list

Question

How do I return a list of the 3 lowest values in another list. For example I want to get the 3 lowest values of this list:

in_list = [1, 2, 3, 4, 5, 6]
inpu         


        

Answer 1

If your lists are long, the most efficient way of doing this is via numpy.partition:

>>> def lowest(a, n): return numpy.partition(a, n-1)[:n]
>>> in_list = [6, 4, 3, 2, 5, 1]
>>> lowest(in_list, 3)
array([1, 2, 3])

This executes in O(N) time, unlike a full sort which would operate in O(NlogN) time. The time savings come from not performing a full sort, but only the minimal amount needed to ensure that the n lowest elements are first. Hence the output is not necessarily sorted.

If you need them to be sorted, you can do that afterwards (numpy.sort(lowest(in_list,3)) => array([1,2,3])). For a large array this will still be faster than sorting the whole thing first.

Edit: Here is a comparison of the speed of numpy.partition, heapq.nsmallest and sorted:

>>> a = numpy.random.permutation(np.arange(1000000))
>>> timeit numpy.partition(a, 2)[:3]
100 loops, best of 3: 3.32 ms per loop
>>> timeit heapq.nsmallest(3,a)
1 loops, best of 3: 220 ms per loop
>>> timeit sorted(a)[:3]
1 loops, best of 3: 1.18 s per loop

So numpy.partition is 66 times faster than heapq.nsmallest for an array with a million elements, and it is 355 times faster than sorted. This doesn't mean that you should never use heapq.nsmallest (which is very flexible), but it demonstrates how important it is to avoid plain lists when speed is important.