翻转列表结构列表中相同项目的特定位置

Question

我有以下数据结构：

pool = [[[0,0,0,0,0,0,0,0],"ze","Zero"],
[[0,0,3,0,3,0,0,0],"bd","BasicDilemma"],
[[0,0,3,2,3,0,0,2],"lk","LowLock"],
[[0,1,3,2,0,3,1,2],"DlCo",""],
[[0,1,3,2,0,3,2,1],"DlPc",""],
[[0,1,3,2,1,3,0,2],"DlAs",""],
[[0,1,3,2,1,3,2,0],"DlHa",""],
[[0,1,3,2,2,3,0,1],"DlSh",""],
[[0,1,3,2,2,3,1,0],"DlNc",""]]

def ListFlip (pool):
    for game in range (0, len(pool)):
        game[0][2], game[0][3] = game[0][3], game[0][2]
        game[0][6], game[0][7] = game[0][7], game[0][6]
    return (pool)

我需要翻转此列表列表中每个项目的特定索引位置，仅翻转数值。

结构是：

[0,1,2,3,4,5,6,7] -> [0,1,3,2,4,5,7,6]

所以对于所有的项目，我需要翻转[2] and [3]和[6] and [7]的位置

例如：

[[0,1,3,2,0,3,1,2],"DlCo",""] -> [[0,1,2,3,0,3,2,1],"DlCo",""]

我认为这是这样做的方式，但它不起作用。有谁知道我做错了什么？

谢谢！

Answer 1

这一行：

for game in range (0, len(pool)):

应该是：

for game in pool:

由于第一个只获取池中每个游戏的索引，所以索引game[0][2]在这里是无效的。

您的代码现在可以正常工作了：

pool = [[[0,0,0,0,0,0,0,0],"ze","Zero"],
        [[0,0,3,0,3,0,0,0],"bd","BasicDilemma"],
        [[0,0,3,2,3,0,0,2],"lk","LowLock"],
        [[0,1,3,2,0,3,1,2],"DlCo",""],
        [[0,1,3,2,0,3,2,1],"DlPc",""],
        [[0,1,3,2,1,3,0,2],"DlAs",""],
        [[0,1,3,2,1,3,2,0],"DlHa",""],
        [[0,1,3,2,2,3,0,1],"DlSh",""],
        [[0,1,3,2,2,3,1,0],"DlNc",""]]

def ListFlip(pool):
    for game in pool:
        game[0][2], game[0][3] = game[0][3], game[0][2]
        game[0][6], game[0][7] = game[0][7], game[0][6]

    return pool

print(ListFlip(pool))

哪些输出：

[[[0, 0, 0, 0, 0, 0, 0, 0], 'ze', 'Zero'], 
 [[0, 0, 0, 3, 3, 0, 0, 0], 'bd', 'BasicDilemma'], 
 [[0, 0, 2, 3, 3, 0, 2, 0], 'lk', 'LowLock'], 
 [[0, 1, 2, 3, 0, 3, 2, 1], 'DlCo', ''], 
 [[0, 1, 2, 3, 0, 3, 1, 2], 'DlPc', ''], 
 [[0, 1, 2, 3, 1, 3, 2, 0], 'DlAs', ''], 
 [[0, 1, 2, 3, 1, 3, 0, 2], 'DlHa', ''], 
 [[0, 1, 2, 3, 2, 3, 1, 0], 'DlSh', ''], 
 [[0, 1, 2, 3, 2, 3, 0, 1], 'DlNc', '']]