如何从子列表中找到中位数的中位数？

Question

Working on a function which would take a list, keep breaking it down into sublists of 3 items each by finding the median of each sublist, then outputting the final median. For example, given the list [99, 42, 17, 7, 1, 9, 12, 77, 15] , the function would break it down to [[99, 42, 17], [7, 1, 9], [12, 77, 15]] then find the medians, [42,7,15] , which would then give the final output of 15 . Using the same logic, the list [55, 99, 131, 42, 88, 11, 17, 16, 104, 2, 8, 7, 0, 1, 69, 8, 93, 9, 12, 11, 16, 1, 77, 90, 15, 4, 123] should also return 15 , but I keep getting 16 instead. Really can't figure out what I am doing wrong.

Here's what I have so far:

def test(lst):
    import numpy as np
    if 1162261467 % len(lst) == 0:
      lst = np.array(lst)
      lst1 = np.split(lst,3)
      lst2 = np.array(lst1).tolist()
      l = [i for x in lst2 for i in x]
      m = np.median(l)
    return m
print (test([55, 99, 131, 42, 88, 11, 17, 16, 104, 2, 8, 7, 0, 1, 69, 8, 93, 9, 12, 11, 16, 1, 77, 90, 15, 4, 123]
))# Outputs 16

Answer 1

This should work:

import numpy as np
def test(lst):

    if 1162261467 % len(lst) == 0:
      lst = np.array(lst)
      lst1 = np.split(lst,len(lst)/3) # <-- you had bug here, here is the correct one
      lst2 = np.array(lst1).tolist()

      l = [np.median(x) for x in lst2] # <-- Another bug is here, here is the corrrect one.
      m = np.median(l)
    return m
print (test([55, 99, 131, 42, 88, 11, 17, 16, 104, 2, 8, 7, 0, 1, 69, 8, 93, 9, 12, 11, 16, 1, 77, 90, 15, 4, 123]
))

Answer 2

Since the starting array is a power of three you can always count on the further arrays being a multiple of three. This makes it quite concise with numpy:

import numpy as np

l = np.array([55, 99, 131, 42, 88, 11, 17, 16, 104, 2, 8, 7, 0, 1, 69, 8, 93, 9, 12, 11, 16, 1, 77, 90, 15, 4, 123])

while len(l) > 1:
    l = np.median(l.reshape(-1, 3), axis=1)
    print(l)

print("median:", int(l[0]))

Prints:

[99. 42. 17.  7.  1.  9. 12. 77. 15.]
[42.  7. 15.]
[15.]
median: 15

Answer 3

This seems to do it. I am kind of mystified by the 1162261467 in your original. Ah, its 3^19. Also you were just reconstructing the original list in your list comprehension.

import numpy as np
def test(lst):
    if len(lst) == 3:
        lst = np.array(lst)
        m = np.median(lst)
        return m
    else:
        lst = np.array(lst)
        lst1 = np.split(lst,len(lst)/3)
        lst2 = np.array(lst1).tolist()
        l = [test(l) for l  in lst2]
        return test(l)