Question

我试图找出如何打印以下列表，而不是从第一个项目开始。需要说明的是：如果列表为[0,1,2,3,4,5,6,7,8]，我想打印类似4,5,6,7,8,0,1,2,3

的内容

以下是代码：

you_can_move_on = False

List = [0,1,2,3,4,5,6,7,8]

next_player = 3

while not you_can_move_on:
    next_player = self.get_next_player_index(next_player)
    you_can_move_on = self.check_if_I_can_move_on
    print(next_player)


def get_next_player_index(self, i):
    if i == len(self.players):
        return 0
    else:
        return i+1

def check_if_I_can_move_on(self):
    return False

Answer 1

我认为应该是

print(l[3:] + l[:3])

Answer 2

如果您想轮换玩家，请使用deque来循环播放：

from collections import deque

d = deque([0,1,2,3,4,5,6,7,8])

d.rotate(len(d)//2 + 1)

print(d)
deque([4, 5, 6, 7, 8, 0, 1, 2, 3])

如果你想跟踪每个玩家，你可以继续旋转：

from collections import deque

d = deque([0,1,2,3,4,5,6,7,8])

print(d)
next_p = d[0]
print(next_p)
d.rotate()
print(d)
next_p = d[0]
print(next_p)

输出：

deque([0, 1, 2, 3, 4, 5, 6, 7, 8])
0
deque([8, 0, 1, 2, 3, 4, 5, 6, 7])
8

你可以在get_next_player函数中移动逻辑，只需让self.players成为deque，这将在所有玩家每次革命开始时使用不同的玩家后自动重启：

def get_next_player(self):
    if self.players[0] is None:
        self.players.popleft()
        self.players.rotate(-1)
        self.players.appendleft(None)
     else:
        n_p = self.players[0]
        self.players.rotate()
    return n_p

该功能负责旋转，你可以看到下面的逻辑：

# sim round
In [25]: print([get_next_player() for _ in range(10)])
[1, 0, 8, 7, 6, 5, 4, 3, 2, 1]

In [26]: players 
# now next player starts
Out[26]: deque([None, 1, 2, 3, 4, 5, 6, 7, 8, 0])
# sim round
In [27]: print([get_next_player() for _ in range(10)])
[2, 1, 0, 8, 7, 6, 5, 4, 3, 2]

In [28]: players
Out[28]: deque([None, 2, 3, 4, 5, 6, 7, 8, 0, 1])
# next player starts
In [29]: print([get_next_player() for _ in range(10)])
[3, 2, 1, 0, 8, 7, 6, 5, 4, 3]

In [30]: players
Out[30]: deque([None, 3, 4, 5, 6, 7, 8, 0, 1, 2])

如何在不从第一项开始的情况下打印整个列表

2 个答案: