Question

我最初只使用

给出的单个随机数据

pivots = random.randrange(l,r)

这里l和r将是定义我的范围的整数

我希望通过选择三个随机枢轴的中位数来大大增加我的枢轴可能成为一个好枢轴的可能性，从而改善运行时间。下面是我使用的代码，它导致我的运行时间增加了20％-30％。

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

如何更快地实现上述目标？

编辑：下面添加了整个代码

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

编辑2：这是更正后的代码。选择3个枢轴的算法存在错误

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

您可以通过以下方式选择枢轴：

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]

示例：

Answer 2

虽然偶尔可以通过随机选择来表现，但仍然值得研究median-of-medians算法的枢轴选择（以及一般的秩选择），该算法在O（n）时间内运行。它与你目前正在做的事情相差不远，但它背后有更强的保证，它选择了一个好的＆＃34;枢轴而不是仅取三个随机数的中位数。

如何在python中改进快速排序枢轴选择？

2 个答案: