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Home/ Questions/Q 98529
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Asked: May 11, 20262026-05-11T00:12:48+00:00 2026-05-11T00:12:48+00:00

I need to get the lesser n numbers of a list in Python. I

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I need to get the lesser n numbers of a list in Python. I need this to be really fast because it’s in a critical part for performance and it needs to be repeated a lot of times.

n is usually no greater than 10 and the list usually has around 20000 elements. The list is always different each time I call the function. Sorting can’t be made in place.

Initially, I have written this function:

def mins(items, n):     mins = [float('inf')]*n     for item in items:         for i, min in enumerate(mins):             if item < min:                 mins.insert(i, item)                 mins.pop()                 break     return mins 

But this function can’t beat a simple sorted(items)[:n] which sort the entire list. Here is my test:

from random import randint, random import time  test_data = [randint(10, 50) + random() for i in range(20000)]  init = time.time() mins = mins(test_data, 8) print 'mins(items, n):', time.time() - init  init = time.time() mins = sorted(test_data)[:8] print 'sorted(items)[:n]:', time.time() - init 

Results:

mins(items, n): 0.0632939338684 sorted(items)[:n]: 0.0231449604034 

sorted()[:n] is three times faster. I believe this is because:

  1. insert() operation is costly because Python lists are not linked lists.
  2. sorted() is an optimized c function and mine is pure python.

Is there any way to beat sorted()[:n] ? Should I use a C extension, or Pyrex or Psyco or something like that?

Thanks in advance for your answers.

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1 Answer

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  1. 2026-05-11T00:12:49+00:00Added an answer on May 11, 2026 at 12:12 am

    You actually want a sorted sequence of mins.

    mins = items[:n] mins.sort() for i in items[n:]:     if i < mins[-1]:          mins.append(i)         mins.sort()         mins= mins[:n] 

    This runs much faster because you aren’t even looking at mins unless it’s provably got a value larger than the given item. About 1/10th the time of the original algorithm.

    This ran in zero time on my Dell. I had to run it 10 times to get a measurable run time.

    mins(items, n): 0.297000169754 sorted(items)[:n]: 0.109999895096 mins2(items)[:n]: 0.0309998989105 

    Using bisect.insort instead of append and sort may speed this up a hair further.

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