Manual Sorting of 4 Numbers

Question

I am trying to sort 4 numbers and do it without using any built-in functions. Here is what I have which works in some cases.

#sort numbers

def sort_n(w, x, y, z):
    list = [w, x, y, z]
    while list[3] < list[2]:

        if list[0] > list[1]:
            list[0], list[1] = list[1], list[0]

        if list[1] > list[2]:
            list[1], list[2] = list[2], list[1]

        if list[2] > list[3]:
            list[2], list[3] = list[3], list[2]
    while list[1] < list[2]:

        if list[0] > list[1]:
            list[0], list[1] = list[1], list[0]

        if list[1] > list[2]:
            list[1], list[2] = list[2], list[1]

        if list[2] > list[3]:
            list[2], list[3] = list[3], list[2]

    while list[1] < list[0]:

        if list[0] > list[1]:
            list[0], list[1] = list[1], list[0]

        if list[1] > list[2]:
            list[1], list[2] = list[2], list[1]

        if list[2] > list[3]:
            list[2], list[3] = list[3], list[2]

    print list


sort_n(10, 1, 2, 3)

look up bubble sort ... that is probably the easiest sort to implement ... this is just a mess .... — Joran Beasley, Jul 24 '13 at 23:45
If you simply want to sort without directly using any built-in functions I suggest that you look at how some of those functions are implemented. — keyser, Jul 24 '13 at 23:46
You do realize that the `>` operator is calling the _built-in_ `__gt__` method of the integers, right? :| — voithos, Jul 24 '13 at 23:47
Why did you put those `while` loops there? They don't actually help; in fact, they make your code wrong. — user2357112, Jul 24 '13 at 23:57
hah user2357 is right ... just get rid of the whiles and do your 3 if statements once and it will be sorted — Joran Beasley, Jul 25 '13 at 00:00

score 1 · Answer 1 · answered Jul 24 '13 at 23:49

1

def bubble_sort(a_list):
    changed = True
    while changed:
        changed = False
        for i in range(len(a_list)-1):
            if a_list[i] > a_list[i+1]:
                 changed = True
                 a_list[i],a_list[i+1] = a_list[i+1],a_list[i]
    return a_list

I think anyway ...

answered Jul 24 '13 at 23:49

Joran Beasley

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Yup. `[4,3,10,1]` -> `[1,3,4,10]`. Test case randomly selected by me. Pseudo. – keyser Jul 24 '13 at 23:52
But `range` and `len` are built in functions :) – SethMMorton Jul 24 '13 at 23:56
im pretty sure he meant sort functions – Joran Beasley Jul 24 '13 at 23:58

score 1 · Answer 2 · answered Jul 25 '13 at 03:23

Inplace insertion sort

def isort(a, f):
    for i in range(0, len(a)):
        for j in range(0, i):
            if f(a[i], a[j]):
                a[j], a[i] = a[i], a[j]
    return a

Then you can use it like so

>>> isort([2,9,8,3], lambda x, y: x < y) # Ascending order
[2, 3, 8, 9]
>>> isort([2,9,8,3], lambda x, y: x > y) # Descending order
[9, 8, 3, 2]

However it does use len and range which are builtins...

score 0 · Answer 3 · edited May 23 '17 at 10:25

you may be looking for the optimal sorting network for 4 numbers.

there are more details (in c) at Standard sorting networks for small values of n and more basic introduction here.

unfortunately i can't find a python implementation for the n=4 case. but the question linked above contains:

 - 4-input: 3 networks

[[1 2][3 4][1 3][2 4][2 3]]
[[1 3][2 4][1 2][3 4][2 3]]
[[1 4][2 3][1 2][3 4][2 3]]

and if i understand correctly, you can pick any line and then do the comparisons described there. so taking the first line:

if list[1] > list[2]: list[1], list[2] = list[2], list[1]
if list[3] > list[4]: list[3], list[4] = list[4], list[3]
if list[1] > list[3]: list[1], list[3] = list[3], list[1]
if list[2] > list[4]: list[2], list[4] = list[4], list[2]
if list[2] > list[3]: list[2], list[3] = list[3], list[2]

but i really need to dash and haven't tested that...

Manual Sorting of 4 Numbers

3 Answers3