如何改善python中快速排序数据透视的选择？

Question

I was originally using only a single random pivot given by

pivots = random.randrange(l,r)

Here l and r will be integers that define my range

I wanted to improve the run time by greatly increasing the likely hood that my pivot would be a good pivot by selecting the median of three random pivots. Below is the code I used and it caused my run time to increase by 20%-30%.

rr = random.randrange
pivots = [ rr(l,r) for i in range(3) ]
pivots.sort()

How do I implement the above to be much faster?

Edit: Entire code added below

import random

def quicksort(array, l=0, r=-1):
    # array is list to sort, array is going to be passed by reference, this is new to me, try not to suck
    # l is the left bound of the array to be acte on
    # r is the right bound of the array to act on

    if r == -1:
        r = len(array)

    # base case
    if r-l <= 1:
        return

    # pick the median of 3 possible pivots
    #pivots = [ random.randrange(l,r) for i in range(3) ]
    rr = random.randrange
    pivots = [ rr(l,r) for i in range(3) ]
    pivots.sort()

    i = l+1 # Barrier between below and above piviot, first higher element
    array[l], array[pivots[1]] = array[pivots[1]], array[l]

    for j in range(l+1,r):
        if array[j] < array[l]:
            array[i], array[j] = array[j], array[i]
            i = i+1

    array[l], array[i-1] = array[i-1], array[l]

    quicksort(array, l, i-1)
    quicksort(array, i, r)

    return array

Edit 2: This is the corrected code due. There was an error in the algorithm for picking the 3 pivots

import random

def quicksort(array, l=0, r=-1):
    # array is list to sort, array is going to be passed by reference, this is new to me, try not to suck
    # l is the left bound of the array to be acte on
    # r is the right bound of the array to act on

    if r == -1:
        r = len(array)

    # base case
    if r-l <= 1:
        return

    # pick the median of 3 possible pivots
    mid = int((l+r)*0.5)
    pivot = 0
    #pivots = [ l, mid, r-1]
    if array[l] > array[mid]:
        if array[r-1]> array[l]:
            pivot = l
        elif array[mid] > array[r-1]:
            pivot = mid
    else:
        if array[r-1] > array[mid]:
            pivot = mid
        else:
            pivot = r-1

    i = l+1 # Barrier between below and above piviot, first higher element
    array[l], array[pivot] = array[pivot], array[l]

    for j in range(l+1,r):
        if array[j] < array[l]:
            array[i], array[j] = array[j], array[i]
            i = i+1

    array[l], array[i-1] = array[i-1], array[l]

    quicksort(array, l, i-1)
    quicksort(array, i, r)

    return array

Answer 1

You could choose the pivot in this way:

alen = len(array)
pivots = [[array[0],0], [array[alen//2],alen//2], [array[alen-1],alen-1]]]
pivots.sort(key=lambda tup: tup[0]) #it orders for the first element of the tupla
pivot = pivots[1][1]

Example:

Answer 2

Though it can be outperformed by random choice on occasion, it's still worth looking into the median-of-medians algorithm for pivot selection (and rank selection in general), which runs in O(n) time. It's not too far off of what you are currently doing, but there is a stronger assurance behind it that it picks a "good" pivot as opposed to just taking the median of three random numbers.