python中间值怎么获得_“中间值”算法的Python实现

下面是我的PYTHON实现。为了提高速度，您可能需要使用PYPY。

关于速度的问题：

每列5个数的理论速度是~10N，因此我使用每列15个数，在~5N时使用2X速度，而最佳速度是~4N。但是，对于最先进的解决方案的最佳速度，我可能是错误的。在我自己的测试中，我的程序运行速度略高于使用sort()的程序。当然，你的里程数可能会有所不同。

假设python程序是“median.py”，运行它的示例是“python./median.py 100”。对于速度基准测试，您可能需要注释掉验证代码，并使用PYPY。#!/bin/python

#

# TH @stackoverflow, 2016-01-20, linear time "median of medians" algorithm

#

import sys, random

items_per_column = 15

def find_i_th_smallest( A, i ):

t = len(A)

if(t <= items_per_column):

# if A is a small list with less than items_per_column items, then:

# 1. do sort on A

# 2. return the i-th smallest item of A

#

return sorted(A)[i]

else:

# 1. partition A into columns of items_per_column items each. items_per_column is odd, say 15.

# 2. find the median of every column

# 3. put all medians in a new list, say, B

#

B = [ find_i_th_smallest(k, (len(k) - 1)/2) for k in [A[j:(j + items_per_column)] for j in range(0,len(A),items_per_column)]]

# 4. find M, the median of B

#

M = find_i_th_smallest(B, (len(B) - 1)/2)

# 5. split A into 3 parts by M, { < M }, { == M }, and { > M }

# 6. find which above set has A's i-th smallest, recursively.

#

P1 = [ j for j in A if j < M ]

if(i < len(P1)):

return find_i_th_smallest( P1, i)

P3 = [ j for j in A if j > M ]

L3 = len(P3)

if(i < (t - L3)):

return M

return find_i_th_smallest( P3, i - (t - L3))

# How many numbers should be randomly generated for testing?

#

number_of_numbers = int(sys.argv[1])

# create a list of random positive integers

#

L = [ random.randint(0, number_of_numbers) for i in range(0, number_of_numbers) ]

# Show the original list

#

print L

# This is for validation

#

print sorted(L)[int((len(L) - 1)/2)]

# This is the result of the "median of medians" function.

# Its result should be the same as the validation.

#

print find_i_th_smallest( L, (len(L) - 1) / 2)