在列表的层次结构后用索引（长度更改）放置列表以更改值

Question

我具有以下列表层次结构：

boundaries = [[0, 1, 2, 3, 4], [[5, 6, 7, 8, 9], [[0, 4, 8, 7], [4, 3, 9, 8]]],
              [[2, 1, 6, 5]], [[1, 0, 7, 6]]]

带有索引的两个列表：

index1 = [1, 0, 2]      # is the same as [1][0][2]  
index2 = [1, 1, 0, 1]   # is the same as [1][1][0][1]

但是，当我尝试执行以下描述时，它不起作用。

boundaries[index1] = 45  
boundaries[index2] = 45 

这是不可能的，因为index1和index2是列表。

我该如何解决？

Answer 1

这可能是一个错误的设计...。但是，您可以像这样反复迭代到子列表中：

boundaries = [[0, 1, 2, 3, 4], [[5, 6, 7, 8, 9], [[0, 4, 8, 7], [4, 3, 9, 8]]],
              [[2, 1, 6, 5]], [[1, 0, 7, 6]]] 

index1 = [1, 0, 2]      # is the same as [1][0][2]  
index2 = [1, 1, 0, 1]   # is the same as [1][1][0][1]


def iterateively_descent(idx, l, val):
    # iteratively descent until the last sublist
    for k in idx[:-1]:
        l = l[k] 
    # set the value of the last sublist
    l[idx[-1]] = val

print(boundaries)

iterateively_descent(index1,boundaries,45)
print(boundaries)

iterateively_descent(index2,boundaries,145)
print(boundaries)

输出：
[[0, 1, 2, 3, 4], [[5, 6, 7, 8, 9], [[0, 4, 8, 7], [4, 3, 9, 8]]],
 [[2, 1, 6, 5]], [[1, 0, 7, 6]]]

[[0, 1, 2, 3, 4], [[5, 6, 45, 8, 9], [[0, 4, 8, 7], [4, 3, 9, 8]]],
 [[2, 1, 6, 5]], [[1, 0, 7, 6]]]

[[0, 1, 2, 3, 4], [[5, 6, 45, 8, 9], [[0, 145, 8, 7], [4, 3, 9, 8]]], 
 [[2, 1, 6, 5]], [[1, 0, 7, 6]]]

Answer 2

我的最初想法与Patrick所建议的相同，虽然可以解决问题，但似乎有些奇怪。您是否考虑过使用numpy数组？如果这是一个选择，那么它可能会使您的生活更轻松。然后，您可以执行以下操作：

import numpy as np

boundaries = np.array([[[1, 2, 3], [4, 5, 6], [7, 8, 9]]])
index = (0, 1, 1)
print("Before:")
print(boundaries)
boundaries[index] = 42
print("After: ")
print(boundaries)

输出将是：
Before:
[[[1 2 3]
  [4 5 6]
  [7 8 9]]]
After:
[[[ 1  2  3]
  [ 4 42  6]
  [ 7  8  9]]]

这里的主要区别是您需要安装numpy（不确定是否可以选择），并且索引必须为元组（我不认为numpy允许使用列表进行索引）。