python - 根据索引列表更改列表元素

Question

索引列表(称为“indlst”)如何对应于元素 [1][0] 和 [3][1]，例如 [[1,0], [3,1,2]] [2] 给定列表(称为“lst”)，用于在原始列表中更改它们各自的元素？另请注意，索引指的是嵌套到任意深度的元素。例如，给定

indlst = [[1,0], [3,1,2]]
lst = ["a", ["b","c"], "d", ["e", ["f", "g", "h"]]]
required_output = [lst[1][0],lst[3][1][2]]

输出应对应于 ["b","h"]。我知道我可以通过以下代码片段(请参阅 Use list of nested indices to access list element)做到这一点:

for i in indlst:
    temp = lst
    for j in i:
        temp = temp[j]
    print(temp)
b
h

但是，我需要更改原始列表中的这些元素。例如，将每个访问的元素更改为“CHANGED”:

changed_lst = ["a", ["CHANGED","c"], "d", ["e", ["f", "g", "CHANGED"]]]

Answer 1

只需检查每个序列中最内层的索引 - 当您到达它时，更改值，而不是深入序列并仅检索元素:

for indices in indlst:
    base = lst
    for i, index in enumerate(indices):
        if i == len(indices) - 1:
            base[index] = "CHANGED"
            break
        base = base[index]

这种替代形式可能更优雅，尽管它只是一回事:

for indices in indlst:
    base = lst
    for index in indices[:-1]: 
        base = base[index]
    base[indices[-1]] = "CHANGED"