Huge difference in timing between sorting a set vs sorting a list in Python

Question

I was wondering whether I should have my data structure as a set or a list. Mostly I will do set operations, but in the end I will need to sort it.

I wondered whether I should first make the set a list, and then use sorted(list(my_set)), or just sort the set immediately sorted(my_set). Arguably, I might consider a general "to list" phase, since having an ordered iterable at that point in time might make sense anyway.

So I decided to test it, expecting a list to be quicker.

Benchmarker:

import time
def sorter(x):
    t1 = time.time()
    for i in range(1000000):
        sorted(x)
    return time.time() - t1

Data:

one = range(1000)
a1 = list(one)
b1 = set(one)
sorter(a1)
# time: 16.5 s 
sorter(b1)
# time: 20.7 s

I then realized it might have to do with the fact that the elements are already in place, and remembered this amazing question & answer.

Then, I tried some random data:

two = numpy.random.randint(1, 1000, 1000)
a2 = list(two)
b2 = set(two)

With the results:

sorter(a2)
# time: 4min 49s
sorter(b2)
# time: 18.9 s

Huge difference, what is going on?

Bonus: It even appears at a timing of one minute, that sorted(set(a_list)) is impressively faster than sorted(a_list).

Indeed in the second case, there can be duplicates which would be filtered, and thus speed up sorting.

Answer 1

I extended your code a bit and hope that this will give you insight into what is happening:

import numpy
import uuid
import random
import time

def sorter(x):
    t1 = time.time()
    for i in range(10000):
        sorted(x)
    return time.time() - t1

def pr(name, x):
    print('sorter {:<12s} {:<11} (length {:>4})'.format(
        name, '{:.8}'.format(sorter(x)), len(x)))

a2sizes = []
b2sizes = []

for x in range(1000):
    two = numpy.random.randint(1, 1000, 1000)
    a2 = list(two)
    b2 = set(two)
    a2sizes.append(len(a2))
    b2sizes.append(len(b2))

print 'average number of elements in a2', sum(a2sizes)/len(a2sizes)
n = sum(b2sizes)/len(b2sizes)
print 'average number of elements in b2', n

this prints out:

average number of elements in a2 1000
average number of elements in b2 632

This is because of the collision in the random number range

print
pr('a2', a2)
# making a list of set gives you already sorted elements
y = list(b2)
pr('y', y)
random.shuffle(y)
pr('shuffled y ', y)
pr('b2', b2)

gives as output:

sorter a2           2.492537    (length 1000)
sorter b2           0.25028086  (length  633)
sorter y            0.19689608  (length  633)
sorter shuffled y   1.4935901   (length  633)

That b2 would be faster because there are less elements is logical, but that this is even faster if you first make a list of the set must have some reason. That it again is slower if you shuffle that list is again logical and the shuffled result is rather close to the result for a2 when compensated for the length of the lists.

So lets try to put something else in the list:

b3 = set()
for x in range(1000):
    b3.add(uuid.uuid4())

print '\nuuid elements', len(b3)

a3 = list(b3)
pr('a3', a3)
random.shuffle(a3)
pr('shuffled a3', a3)
pr('b3', b3)

gives (I would have been rather surprised to have less than 1000 elements):

uuid elements 1000
sorter a3           32.437758   (length 1000)
sorter shuffled a3  32.178433   (length 1000)
sorter b3           32.163802   (length 1000)

So it must have something to do with having numbers in the set:

previous = -1
ordered = True
for popped in b2:
    if popped < previous:
        print 'popped', popped, previous
        ordered = False
    previous = popped

print '\nOrdered', ordered

gives you:

Ordered True

Instead of iterating , a set has a pop() function you can try and use:

pop()

Remove and return an arbitrary element from the set. Raises KeyError if the set is empty.

So lets arbitrarily retrieve elements from the set b2 and see if there is something special:

previous = -1
ordered = True
while(b2):
    popped = b2.pop()
    if popped < previous:
        print 'popped', popped, previous
        ordered = False
    previous = popped

print '\nOrdered', ordered

gives the same result:

Ordered True

So arbitrarily retrieving elements of set of number retrieves those numbers in order, independent off the ordering how these numbers were put in. As the iteration is how the list-making retrieves an element at a time to append to the list, the result of list(b2) is an ordered list and that gets sorted quite fast using the Timsort algorithm used in Python.