Question

如果我知道另一个元素，如何从Python 3中获取一对元素（2个元素的列表）？例如，在回合制策略游戏中，我想交换玩家：

players = [playerA, playerB]
active_player = playerA

players_tmp = list(players) 
players_tmp.remove(active_player)
active_player = players_tmp[0]

这段代码很完美，但很难看。

当然，在这种特殊情况下，我总是可以存储活动播放器的索引（0或1），然后执行：

self._active_player = 1 - self._active_player

然后编写@property方法。但它并不那么有趣。

有没有漂亮的方法呢？

Answer 1

对于找到一对中的其他元素的任务，我只是写一个漂亮而清晰的函数：

def other_of(pair, item):
    a, b = pair
    return a if item == b else b
…
active_player = other_of(players, active_player)

如果它适用于您的情况，itertools.cycle可能是循环选项的更好方式：

import itertools

player_rotation = itertools.cycle(players)
…
active_player = next(player_rotation)

Answer 2

您可以利用True可用作索引1的事实

>>> l = ['a','b']
>>> l['a' is l[0]]
'b'
>>> l['b' is l[0]]
'a'

因此您可以将您的功能定义为

def find(l,i):
    return l[i is l[0]]

演示

>>> find(["Player1","Player2"],"Player1")
'Player2'
>>> find(["Player1","Player2"],"Player2")
'Player1'

Answer 3

当您想要更换玩家并获取第一个元素时，只需交换，然后您就不必关心谁是活跃的，因为每次都会得到一个不同的玩家：

players = ["playerA", "playerB"]

players[0],players[1] = players[1],players[0]

active = players[0]

或者更好，因为minitech只是致电list.reverse

  players = ["playerA", "playerB"]
  players.reverse()
  active = players[0]
  print(active)
  playerB

Answer 4

使用set;

players = {playerA, playerB}
active_player = {playerA}

active_player = players - active_player

或;

players = {playerA, playerB}
active_player = playerA

active_player = (players - {active_player}).pop()

Answer 5

这是切换标记的常用方法：

if active_player == playerA:
    active_player = playerB
else:
    active_player = playerA

Answer 6

更好的解决方案是拥有一台发电机。

players = ['sdf','asdf','sasd']
active = 'sdf'
temp_players = (player for player in players if player != active)
active = temp_players.next()

通过这种方式，您可以获得所有玩家的列表，并且每次都可以获得一名玩家。如果你一次在一支球队中有多名球员，这也会更容易。

Answer 7

我建议：

players = ['playerA', 'playerB']
def active_player(): return players[0]
print(active_player()) 
players.reverse()
print(active_player())

给出：

playerA
playerB

Answer 8

# conditional expressions
active_player = players[0] if players[1] == active_player else players[1]

# negative index
active_player = players[players.index(active_player) - 1]

从对中获取另一个元素

8 个答案: