24

What is the fastest way to sort an array of whole integers bigger than 0 and less than 100000 in Python? But not using the built in functions like sort.

Im looking at the possibility to combine 2 sport functions depending on input size.

4
  • 10
    Why not use built in functions? Commented Oct 4, 2010 at 13:17
  • What's the largest size the array might be? Commented Oct 4, 2010 at 13:17
  • 3
    @Anders: Don't reinvent the wheel. The built in sort() should suffice for your case. Commented Oct 4, 2010 at 13:19
  • 10
    I'm curious: why do you need to implement your own sorting routine? It smells like a homework assignment to me :-) Commented Oct 4, 2010 at 13:31

10 Answers 10

29

If you are interested in asymptotic time, then counting sort or radix sort provide good performance.

However, if you are interested in wall clock time you will need to compare performance between different algorithms using your particular data sets, as different algorithms perform differently with different datasets. In that case, its always worth trying quicksort:

def qsort(inlist):
    if inlist == []: 
        return []
    else:
        pivot = inlist[0]
        lesser = qsort([x for x in inlist[1:] if x < pivot])
        greater = qsort([x for x in inlist[1:] if x >= pivot])
        return lesser + [pivot] + greater

Source: http://rosettacode.org/wiki/Sorting_algorithms/Quicksort#Python

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3 Comments

Good advice, except choice of variable list, which can cause nice errors. I post other, faster version.
Running the list comprehension twice over the same set of variables is probably also less than optimal.
@Tony Veijalainen “There are only two hard things in Computer Science: cache invalidation and naming things” -- I've changed the variable name
8

Since you know the range of numbers, you can use Counting Sort which will be linear in time.

1 Comment

(I did not downvote). Note that this is not a good algorithm if the array of integers is significantly smaller than 100000, as it will waste memory (and thus time) to construct the 100000 element list.
5

Early versions of Python used a hybrid of samplesort (a variant of quicksort with large sample size) and binary insertion sort as the built-in sorting algorithm. This proved to be somewhat unstable. S0, from python 2.3 onward uses adaptive mergesort algorithm.

Order of mergesort (average) = O(nlogn). Order of mergesort (worst) = O(nlogn). But Order of quick sort (worst) = n*2

if you uses list=[ .............. ]

list.sort() uses mergesort algorithm.

For comparison between sorting algorithm you can read wiki

For detail comparison comp

2 Comments

it's timsort which is more adaptive than mergesort
Timsort is an adaptive, stable, natural mergesort.
4

Radix sort theoretically runs in linear time (sort time grows roughly in direct proportion to array size ), but in practice Quicksort is probably more suited, unless you're sorting absolutely massive arrays.

If you want to make quicksort a bit faster, you can use insertion sort] when the array size becomes small.

It would probably be helpful to understand the concepts of algorithmic complexity and Big-O notation too.

3 Comments

When you say the array size becomes small do you mean less than 64 in size?
I'd say more about less than 10, but there's no right answer; the best idea is to experiment with different values and see which ends up faster.
radix sort is O(kn) where k is the number of digits in your integer. But the number of digits in a base $b$ integer R is $O(log_b R)$ so radix sort's "linear time" is basically extremely misleading. It's asymptotically not any different than a comparison sort. If it WAS, we could convert comparable items to integers, and then radix sort to beat the $O(n \log n)$ lower bound for sorting.
3

I might be a little late to the show, but there's an interesting article that compares different sorts at https://www.linkedin.com/pulse/sorting-efficiently-python-lakshmi-prakash

One of the main takeaways is that while the default sort does great we can do a little better with a compiled version of quicksort. This requires the Numba package.

enter image description here

Here's a link to the Github repo: https://github.com/lprakash/Sorting-Algorithms/blob/master/sorts.ipynb

1 Comment

So default sort is kicking.
1

We can use count sort using a dictionary to minimize the additional space usage, and keep the running time low as well. The count sort is much slower for small sizes of the input array because of the python vs C implementation overhead. The count sort starts to overtake the regular sort when the size of the array (COUNT) is about 1 million.

If you really want huge speedups for smaller size inputs, implement the count sort in C and call it from Python.

(Fixed a bug which Aaron (+1) helped catch ...) The python only implementation below compares the 2 approaches...

import random
import time

COUNT = 3000000

array = [random.randint(1,100000) for i in range(COUNT)]
random.shuffle(array)

array1 = array[:]

start = time.time()
array1.sort()
end = time.time()
time1 = (end-start)
print 'Time to sort = ', time1*1000, 'ms'

array2 = array[:]

start = time.time()
ardict = {}
for a in array2:
    try:
        ardict[a] += 1
    except:
        ardict[a] = 1

indx = 0
for a in sorted(ardict.keys()):
    b = ardict[a]
    array2[indx:indx+b] = [a for i in xrange(b)]
    indx += b

end = time.time()
time2 = (end-start)
print 'Time to count sort = ', time2*1000, 'ms'

print 'Ratio =', time2/time1

6 Comments

+1 Ratio = 1.16710428623 on my machine. clever use of a dict. It's worth noting though that changing the dict construction phase from try: ardict[a] += 1; except: ardict[a] = 1 to if a in ardict: ardict[a] += 1; else: ardict[a] = 1 drops the ratio to Ratio = 0.696179723863 Sometimes (often) it is better to look before you leap. I knew to do this because try is only cheaper than if if the exception rarely occurs. An actual exception is still very expensive.
unfortunately this algorithm is wrong. Try array = [1,10, 100, 1000, 10000, 100000, 1000000]. The dangers of skating on undocumented implementation details strike again.
Thanks aaron - fixed the bug of not sorting the dict keys. That should slow it down a bit. However, it will preserve its almost O(n) nature if number of distinct elements compared to the array size is low. I would love to see a 3D plot of distinct elements, array length as x and y dimensions and ratio of running time and the 3rd dimension. Maybe I will do it in a day or 2.
@Aaron: On my comp, the (try, except) works better. I coded it the (if then else) way first and switched to the (try, except) to speed up things. Also, the version with the bug removed runs faster than the previous one - because of the underlying C sort being used in sorting keys. Thats free beer right there. :-)
For me the except solution was also faster, little faster generation was by using tuple with generator if list is not necessary to produce (as it is mutable): array2=tuple(a for a in sorted(ardict) for i in xrange(ardict[a]))
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0

The built in functions are best, but since you can't use them have a look at this:

http://en.wikipedia.org/wiki/Quicksort

Comments

0
def sort(l):
    p = 0
    while(p<len(l)-1):
        if(l[p]>l[p+1]):
            l[p],l[p+1] = l[p+1],l[p]
            if(not(p==0)):
                p = p-1
        else:
            p += 1
    return l

this is a algorithm that I created but is really fast. just do sort(l) l being the list that you want to sort.

1 Comment

That's a bubble sort
0

@fmark Some benchmarking of a python merge-sort implementation I wrote against python quicksorts from http://rosettacode.org/wiki/Sorting_algorithms/Quicksort#Python and from top answer.

  1. Size of the list and size of numbers in list irrelevant

merge sort wins, however it uses builtin int() to floor

import numpy as np
x = list(np.random.rand(100))


# TEST 1, merge_sort 
def merge(l, p, q, r):
    n1 = q - p + 1
    n2 = r - q
    left = l[p : p + n1]
    right = l[q + 1 : q + 1 + n2]

    i = 0
    j = 0
    k = p
    while k < r + 1:
        if i == n1:
            l[k] = right[j]
            j += 1
        elif j == n2:
            l[k] = left[i]
            i += 1
        elif  left[i] <= right[j]:
            l[k] = left[i]
            i += 1
        else:
            l[k] = right[j]
            j += 1
        k += 1

def _merge_sort(l, p, r):
    if p < r:
        q = int((p + r)/2)
        _merge_sort(l, p, q)
        _merge_sort(l, q+1, r)
        merge(l, p, q, r)

def merge_sort(l):
    _merge_sort(l, 0, len(l)-1)

# TEST 2
def quicksort(array):
    _quicksort(array, 0, len(array) - 1)

def _quicksort(array, start, stop):
    if stop - start > 0:
        pivot, left, right = array[start], start, stop
        while left <= right:
            while array[left] < pivot:
                left += 1
            while array[right] > pivot:
                right -= 1
            if left <= right:
                array[left], array[right] = array[right], array[left]
                left += 1
                right -= 1
        _quicksort(array, start, right)
        _quicksort(array, left, stop)

# TEST 3
def qsort(inlist):
    if inlist == []: 
        return []
    else:
        pivot = inlist[0]
        lesser = qsort([x for x in inlist[1:] if x < pivot])
        greater = qsort([x for x in inlist[1:] if x >= pivot])
        return lesser + [pivot] + greater

def test1():
    merge_sort(x)

def test2():
    quicksort(x)

def test3():
    qsort(x)

if __name__ == '__main__':
    import timeit
    print('merge_sort:', timeit.timeit("test1()", setup="from __main__ import test1, x;", number=10000))
    print('quicksort:', timeit.timeit("test2()", setup="from __main__ import test2, x;", number=10000))
    print('qsort:', timeit.timeit("test3()", setup="from __main__ import test3, x;", number=10000))

Comments

0

Bucket sort with bucket size = 1. Memory is O(m) where m = the range of values being sorted. Running time is O(n) where n = the number of items being sorted. When the integer type used to record counts is bounded, this approach will fail if any value appears more than MAXINT times.

def sort(items):
  seen = [0] * 100000
  for item in items:
    seen[item] += 1
  index = 0
  for value, count in enumerate(seen):
    for _ in range(count):
      items[index] = value
      index += 1

Comments

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