Help with a python problem. Only need to complete

running_medians(iterable): 

 # YOUR CODE HERE 

--------------------------------------------------------------------------

2. Computing the Running Median

The median of a series of numbers is simply the middle term if ordered by magnitude, or, if there is no middle term, the average of the two middle terms. E.g., the median of the series [3, 1, 9, 25, 12] is 9, and the median of the series [8, 4, 11, 18] is 9.5.

If we are in the process of accumulating numerical data, it is useful to be able to compute the running median where, as each new data point is encountered, an updated median is computed. This should be done, of course, as efficiently as possible.

The following function demonstrates a naive way of computing the running medians based on the series passed in as an iterable.

In [10]:

def running_medians_naive(iterable): 

 values = [] 

 medians = [] 

 for i, x in enumerate(iterable): 

 values.append(x) 

 values.sort() 

 if i%2 == 0: 

 medians.append(values[i//2]) 

 else: 

 medians.append((values[i//2] + values[i//2+1]) / 2) 

 return medians 

In [*]:

running_medians_naive([3, 1, 9, 25, 12]) 

Out[11]:

[3, 2.0, 3, 6.0, 9] 

In [*]:

running_medians_naive([8, 4, 11, 18]) 

Note that the function keeps track of all the values encountered during the iteration and uses them to compute the running medians, which are returned at the end as a list. The final running median, naturally, is simply the median of the entire series.

Unfortunately, because the function sorts the list of values during every iteration it is incredibly inefficient. Your job is to implement a version that computes each running median in O(log N) time using, of course, the heap data structure!

Hints

You will need to use two heaps for your solution: one min-heap, and one max-heap.

The min-heap should be used to keep track of all values greater than the most recent running median, and the max-heap for all values less than the most recent running median this way, the median will lie between the minimum value on the min-heap and the maximum value on the max-heap (both of which can be efficiently extracted)

In addition, the difference between the number of values stored in the min-heap and max-heap must never exceed 1 (to ensure the median is being computed). This can be taken care of by intelligently pop-ping/add-ing elements between the two heaps.

In [8]:

def running_medians(iterable): 

 # YOUR CODE HERE 

 values = [] 

 medians = [] 

 for i, x in enumerate(iterable): 

 values.append(x) 

 values.sort() 

 if i%2 == 0: 

 medians.append(values[i//2]) 

 else: 

 medians.append((values[i//2] + values[i//2+1]) / 2) 

 return medians 

In [9]:

# (2 points) 

from unittest import TestCase 

tc = TestCase() 

tc.assertEqual([3, 2.0, 3, 6.0, 9], running_medians([3, 1, 9, 25, 12])) 

In [ ]:

# (2 points) 

import random 

from unittest import TestCase 

tc = TestCase() 

vals = [random.randrange(10000) for _ in range(1000)] 

tc.assertEqual(running_medians_naive(vals), running_medians(vals)) 

In [ ]:

# (4 points) MUST COMPLETE IN UNDER 10 seconds! 

import random 

from unittest import TestCase 

tc = TestCase() 

vals = [random.randrange(100000) for _ in range(100001)] 

m_mid = sorted(vals[:50001])[50001//2] 

m_final = sorted(vals)[len(vals)//2] 

running = running_medians(vals) 

tc.assertEqual(m_mid, running[50000]) 

tc.assertEqual(m_final, running[-1])