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simulacion-permeabilidad/fftma_module/bin/bin.py

40 lines
1.1 KiB
Python

import numpy as np
A = [1,2,3,4,5,4.000000001,1000,8,18,9,991473,74168,6387,3678,8796,1343]
def median_of_medians(A, i):
#divide A into sublists of len 5
sublists = [A[j:j+5] for j in range(0, len(A), 5)]
medians = [sorted(sublist)[len(sublist)/2] for sublist in sublists]
if len(medians) <= 5:
pivot = sorted(medians)[len(medians)/2]
else:
#the pivot is the median of the medians
pivot = median_of_medians(medians, len(medians)/2)
#partitioning step
low = [j for j in A if j < pivot]
high = [j for j in A if j > pivot]
k = len(low)
if i < k:
return median_of_medians(low,i)
elif i > k:
return median_of_medians(high,i-k-1)
else: #pivot = k
return pivot
#Here are some example lists you can use to see how the algorithm works
#A = [1,2,3,4,5,1000,8,9,99]
#B = [1,2,3,4,5,6]
#print median_of_medians(A, 0) #should be 1
#print median_of_medians(A,7) #should be 99
#print median_of_medians(B,4) #should be 5
b=np.sort(A)
c=np.array([median_of_medians(A, i) for i in range(len(A))])
print b
print c
print bool(np.sum(b==c))