Median sort a list in unordered fashion

What's the best way of sorting a python list in an unordered fashion according to the median of value positions in the list?

Suppose:

a = [1, 3 ,6, 7, 10, 12, 17]

I'm looking for this:

a = [1, 17, 7, 3, 12, 6, 10]

Meaning, the list now looks like [start, end, mid, first_half_mid, second_half_mid, ...]

edit: To clarify further, I'm looking for a way to keep bisecting the list until it covers the whole range!

edit2: Another example to illustrate the problem

input:

[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

desired output:

[1, 10, 6, 3, 9, 2, 5, 8, 4, 7]

Solution

UPDATE: consider left element as a middle if the length of the list is an even numer

def f(a):
    # take left middle element for even-length lists
    mid = len(a)//2 if len(a)%2 else len(a)//2-1
    # take len(a)//2 as a middle element 
    #mid = len(a)//2 
    if len(a) <= 2:
        return a
    elif(len(a) == 3):
        return a[[0,-1,mid]]
    else:
        return np.append(a[[0,-1,mid]], f(np.delete(a, [0,len(a)-1,mid])))

Output:

In [153]: f(a)
Out[153]: array([ 1, 17,  7,  3, 12,  6, 10])

OLD answer:

here is one of many possible solutions:

import numpy as np

def f(a):
    if len(a) <= 2:
        return a
    elif(len(a) == 3):
        return a[[0,-1,len(a)//2]]
    else:
        return np.append(a[[0,-1,len(a)//2]], f(np.delete(a, [0,len(a)-1,len(a)//2])))

a = np.array([1, 3 ,6, 7, 10, 12, 17])

In [114]: a
Out[114]: array([ 1,  3,  6,  7, 10, 12, 17])

In [115]: f(a)
Out[115]: array([ 1, 17,  7,  3, 12, 10,  6])

PS about last two numbers in the result list/array - the question is what is the middle index for the 4-elements list?

What is the middle element for [3,6,10,12] list? In my solution it would be 10 (index: 4//2 == 2)