如何随机切片

Question

我正在尝试进行战舰，我正在将船只在一个列表中"板"中制作，船的长度将由切片确定。我正在尝试通过随机移动船只，但我很快就会意识到随机。选择只能一次移动一个元素，但船只有2-5个元素。我正在尝试查看是否可以将列表中的切片随机化为切片上的单元。random.shuffle似乎不起作用。

import random
board = "o"
board = [board] * 49
ship_two_space = board[0:2]
ship_three_space = board[0:3]
ship_three_space_second = board[0:3]
ship_four_space = board[0:4]
ship_five_space = board[0:5]

Answer 1

船位位置不能重叠 - 使用集合记录占用空间

occupied = set()

定义规格

N = 49
board = ['o'] * N
ship_specs = [('two_space', 2), ('three_space1', 3), ('three_space2', 3),
              ('four_space', 4), ('five_space', 5)]

开始制作船 - 使用Collections.Meaduple到 define 船

Ship = collections.namedtuple('Ship', ('name','indices', 'slice'))
def get_position(length):
    '''Determine the unique position indices of a ship'''
    start = random.randint(0, N-length)
    indices = range(start, start + length)
    if not occupied.intersection(indices):
        occupied.update(indices)
        return indices, slice(start, start + length)
    else:
        return get_position(length)
ships = []
for name, length in ship_specs:
    indices, _slice = get_position(length)
    ships.append(Ship(name, indices, _slice))

print(board)
for ship in ships:
    board[ship.slice] = str(len(ship.indices))*len(ship.indices)
print(board)
print([(ship.name, ship.slice) for ship in ships])
>>> 
['o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o']
['o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', '2', '2', '4', '4', '4', '4', '3', '3', '3', 'o', '5', '5', '5', '5', '5', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', '3', '3', '3', 'o', 'o', 'o']
[('two_space', slice(10, 12, None)), ('three_space1', slice(43, 46, None)), ('three_space2', slice(16, 19, None)), ('four_space', slice(12, 16, None)), ('five_space', slice(20, 25, None))]
>>>