根据索引列表更改列表元素

Question

如何将索引列表（称为＆＃34; indlst＆＃34;），如[[1,0]，[3,1,2]]对应元素[1] [0]和[ 3] [1] [2]给定列表（称为＆＃34; lst＆＃34;），用于改变原始列表中的各自元素？另请注意，索引是指嵌套到任意深度的元素。例如，给定

// Your secret key can be anything you want
$secretKey = 'oldlaravel.dev';

Route::group(['domain' => 'oldlaravel.dev'], function() use ($secretKey) {
    Route::get('/', function() use ($secretKey) {

        $hashedToken = Hash::make($secretKey);
        return Redirect::to("http://newlaravel.dev?token=$hashedToken", 301);

    });
});

Route::group(['domain' => 'newlaravel.dev'], function() use ($secretKey) {
    Route::get('/', function() use ($secretKey) {

        if (Request::get('token')) {
            if (Hash::check($secretKey, Request::get('token'))) {
                echo 'You got here by redirect from my old domain!';
            } else {
                echo 'Who the hell are you?!'; //came here with an invalid token parameter
            }
        } else {
            echo 'You got here directly!';
        }

    });
});

输出应对应于[＆＃34; b＆＃34;，＆＃34; h＆＃34;]。我知道我可以通过以下代码段获得这一点（参见Use list of nested indices to access list element）：

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

但是，我需要在原始列表中更改这些元素。例如，将每个访问过的元素更改为＆＃34; CHANGED＆＃34;：

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

Answer 1

我的想法

我的解决方案是从最深层的元素开始，按照以下方式返回：

• 将最深的元素插入第二个最深元素
• 将修改后的第二个最深元素插入第三个最深元素
• 等

执行该操作的一些代码

def set_from_indices(value, indices, array):
    for i in reversed(range(len(indices))): # Pulls back the deepest element available each loop
        current = array.copy() # Without .copy() any changes to current would also change array.
        for index, indice in enumerate(indices):
            if index == i: # If we are on the deepest element available this loop
                current[indice] = value.copy() if isinstance(value, list) else value # Without the .copy() current[indice] just becomes '...'
                value = current.copy() # Make the 
                break
            current = current[indice]
    return value

indlst = [[0,1], [0,0,1], [0,0,0,1]]
lst = [[[[4,3],2],1],0]
for indices in indlst:
    lst = set_from_indices("CHANGED", indices, lst)