如何在python中重新排列列表

Question

I want to rearrange my list in python but there are some rules about it. These next few pictures describe how beavers jump into a hole, while others go accross and the ones in the hole go out.

Original array

Beavers fill in the hole

Others go accross

Everyone else come out

Now there's where it gets complicated. What if i have more than 1 hole . I've managed to figure out everything myself up to this point. Now all i'm asking is how would any of you approach this and if there's any tricks i could use for sorting the list. This is my code so far:

bobri = "54321"
luknje = [4, 2, 4]
bobri_original = bobri
bobri_original_list = list(bobri_original)

bobri_list = list(bobri)

i = 1
gli = 0
for st in range(len(luknje)):
    i = 1
    globina = luknje[st]
    # Prvega zapisem in zbrisem iz glavnega seznama
    last_bober = bobri_list[-1]
    bobri_list.remove(str(last_bober))
    bobri_luknja = [last_bober]

    if globina == 1:
        for n in range(len(bobri_list)):
            bobri_luknja.append(bobri_list[n])
        bobri_list = bobri_luknja
    else:
        while i <= globina - 1:
            last_bober = bobri_list[-1]
            bobri_list.remove(str(last_bober))
            bobri_luknja.append(last_bober)

            if i == globina - 1:
                for n in range(len(bobri_list)):
                    bobri_luknja.append(bobri_list[n])
                bobri_list = bobri_luknja
            i += 1
            gli += 1

Answer 1

Zdravo (:

def cross_all_holes(initial_list, hole_depths=[]):
    # Crossing single hole function.
    def cross_the_hole(current_list, hole_depth):
        hole_is_filled = False
        list_len = len(current_list)
        result_list = []
        hole = []
        # While not result list contain all the items
        while len(result_list) < list_len:
            # Filling the hole
            if not hole_is_filled:
                element = current_list.pop()
                hole.append(element)
                # Checking if it's filled
                hole_is_filled = len(hole) == hole_depth
            else:
                # Crossing the filled hole
                if len(current_list):
                    element = current_list.pop()
                # Emptying hole
                elif len(hole):
                    element = hole.pop()
                result_list = [element] + result_list
        return result_list

    # Repeat as much as needed.
    current_list = initial_list
    for hole_depth in hole_depths:
        current_list = cross_the_hole(current_list, hole_depth)
    return current_list

Thanks for the task, it was funny (:

In [20]: cross_all_holes([1,2,3,4,5,6,7,8,9,10], [1,3])
Out[20]: [9, 8, 7, 10, 1, 2, 3, 4, 5, 6]

Answer 2

Never thought I would write a function with inputs as 'beavers' and 'holes', but thanks for this moment. The beavers is simply the list of the number of each beaver, and holes is the list with each hole, and the amount of beavers getting into each:

def rearrange(beavers, holes):
    l = len(beavers)
    for n in range(len(holes)):
        back = beavers[-holes[n]:][::-1]
        front = beavers[:l - holes[n]]
        beavers = back + front
    return(beavers)

rearrange([7,6,5,4,3,2,1], [3])
Out: [1, 2, 3, 7, 6, 5, 4]

rearrange([7,6,5,4,3,2,1], [3,4])
Out: [4, 5, 6, 7, 1, 2, 3]