Bubble sorting with one list comprehension

Question

I'd like to see if it would be possible to convert a BubbleSort function, such as:

def BubbleSort(l):
    for i in range(len(l)-1):
        for j in range(len(l)-1-i):
            if (l[j]>l[j+1]):
                l[j],l[j+1]=l[j+1],l[j]
    return l

to a one-liner list comprehension, maybe similar to:

def BubbleSort(l):
    return [something_goes_here for i in range(len(l)-1) for j in range(len(l)-1-i) if (l[j]>l[j+1])]

Sample Input:

print(BubbleSort([1,5,-5,0,10,100]))

Sample Output

[-5, 0, 1, 5, 10, 100]

Answer 1

A solution based on side-effects looks like this:

def bubblesort(l):
    [l.append(l.pop(0) if i == len(l) - 1 or l[0] < l[1] else l.pop(1)) for j in range(0, len(l)) for i in range(0, len(l))]
    return l

This will sort the list l in-place.

The basic idea is to treat l as both the input and output-list. Swaps can then be emulated by either moving the first or the second element of l to the end. The last element must be moved to the end without any comparison to get a new list list. A visual example of one iteration ( [l.append(l.pop(0) if i == len(l) - 1 or l[0] < l[1] else l.pop(1)) for i in range(0, len(l))] ):

1 3 2 6 5 4 |
  3 2 6 5 4 | 1
    3 6 5 4 | 1 2
      6 5 4 | 1 2 3
        6 4 | 1 2 3 5
          6 | 1 2 3 5 4
            | 1 2 3 5 4 6

In this example | denotes the separator between the last element of the original list and the first element that was already appended. Repeating this process len(l) times guarantees that the entire list is sorted.

Note that while this example does perform a bubblesort, it's runtime is O(n^3) , since we need to remove the first or second element from the list in each step, which runs in O(n) .

EDIT:
It becomes more easy to see that this is actually bubblesort from the above algorithm, if we rewrite the sample-iteration as such:

| 1 3 2 6 5 4
1 | 3 2 6 5 4
1 2 | 3 6 5 4
1 2 3 | 6 5 4
1 2 3 5 | 6 4
1 2 3 5 4 | 6
1 2 3 5 4 6 |

Here the separator denotes the end of the list and a circular view of the list is used.

EDIT 2:
Found a more efficient way to solve this that uses slice-assignment:

def bubblesort(l):
    [l.__setitem__(slice(i, i + 2), (l[i:i + 2] if l[i] < l[i + 1] else l[i +  1:i - 1:-1])) for j in range(0, len(l)) for i in range(0, len(l) - 1)]
    return l

Answer 2

Using a list comprehension to hide a for loop is kind of cheating given that the result produced by the comprehension is not the sorted list. But if you're going to do that, you might want to avoid building a list of None elements by performing the swaps in the condition rather than as the output value.

For example:

a = [1, 3, 2, 6, 5, 4]
[_ for n in range(len(a),1,-1) for i in range(n-1) if a[i]>a[i+1] and a.__setitem__(slice(i,i+2),a[i:i+2][::-1])]

Isolating the element swapping part, this would give:

swap = lambda(a,i):a.__setitem__(slice(i,i+2),a[i:i+2][::-1])
[_ for n in range(len(a),1,-1) for i in range(n-1) if a[i]>a[i+1] and swap(a,i)]

Which is no different from:

for n in range(len(a),1,-1):
    for i in range(n-1):
        if a[i]>a[i+1]: 
           swap(a,i)     # same as a[i],a[i+1] = a[i+1],a[i]

The list comprehension is merely a different way to write the for loop and is not actually returning the sorted result.

What would be more in the spirit of a list comprehension is to actually return the sorted result without affecting the original list. You can do this using a temporary list within the comprehension to perform the element swapping and progressively return the position that is guaranteed to be at the right sorted index:

a = [1, 3, 2, 6, 5, 4]
s = [ b.pop(-1) for b in [list(a)] for n in range(len(a),0,-1) if not [_ for i in range(n-1) if b[i]<b[i+1] and b.__setitem__(slice(i,i+2),b[i:i+2][::-1])] ]
print(s) # [1, 2, 3, 4, 5, 6]  

The approach is the same as before except that b is used internally to manage swapping and return the sorted values. Because the guaranteed sorted position is always the last one in b , the swapping condition was reversed so that internally b is sorted in descending order which produces the output in ascending order when taking the last item at each iteration.

Note that all these solutions fail to implement the early exit condition that allows bubble sort to be very efficient on already sorted lists and lists where elements are near their sorted position (ie stop when no swaps in a pass). The number of iterations will always be N*(N+1)/2 no matter the original order of elements giving a time complexity of O(N^2) all the time instead of worst case.

Answer 3

def bubblesort(l):
    [[l.insert(j+1, l.pop(j)) for j in range(i) if l[j] > l[j+1]] for i in range(len(l)-1,0,-1)]
    return l

I use the l.insert(j+1, l.pop(j)) to swap the elements in the list. The rest are just the nested loops.