问题

我有以下的输入变量：'PATH'，它是一个包含要创建的键的数组，以及'output_content'，它是一个字典，应该在其中创建这些键，'output content'已经包含键值对，它们不应该丢失。

例如，考虑PATH = ['patterns', 'pattern_1', 'children', 'pattern_1_1']，结果的'output_content'字典应该如下所示：

{
    'patterns': {
        'pattern_1': {
            'children': {
                'pattern_1_1': {}
            }
        }
    },
    'already_existing_key': 'already_existing_value'
}

我尝试了以下方法：

current_dict = output_content
for key in PATH[1:]:
    if key not in current_dict:
        current_dict[key] = {}
    current_dict = current_dict[key]

但很快我意识到这样做不会起作用，因为output_content根据current_dict在变化，所以最终的输出将只是{}。

{
    'patterns': {
        'pattern_1': {
            'children': {
                'pattern_1_1': {}
            }
        }
    },
    'already_existing_key': 'already_existing_value'
}

current_dict = output_content
for key in PATH[1:]:
    if key not in current_dict:
        current_dict[key] = {}
    current_dict = current_dict[key]

答案1

得分: 0

两个关于你的代码的注释：
* 你应该遍历整个列表 `PATH`，而不仅仅是 `PATH[1:]`。我看不到跳过第一个元素的理由。
* 如果条件 `key in current_dict` 被满足，那么你应该确保相关联的值是一个字典。

```python3
def add_path_to_dict(d, path):
    for k in path:
        if k not in d:
            d[k] = {}
        elif not isinstance(d[k], dict):
            return
        d = d[k]

d = {'z': 42}
add_path_to_dict(d, 'abcdefgh')
print(d)
# {'z': 42, 'a': {'b': {'c': {'d': {'e': {'f': {'g': {'h': {}}}}}}}}}

d = {'a': {'b': {'c':{'z': 42}}}}
add_path_to_dict(d, 'abcdefgh')
print(d)
# {'a': {'b': {'c': {'z': 42, 'd': {'e': {'f': {'g': {'h': {}}}}}}}}}

d = {'a': {'b': {'c':42}}}
add_path_to_dict(d, 'abcdefgh')
print(d)
# {'a': {'b': {'c': 42}}}

更进一步的建议：

• 与其使用空的 return，你可以使用 return 'Some error value'，或者你可以 raise ValueError(f'Key {k} already in dict and its value is not a dict.')

<details>
<summary>英文:</summary>

Two remarks with your code:
* You should iterate on the whole list `PATH`, not just `PATH[1:]`. I see no reason to skip the first element.
* If the condition `key in current_dict` is realised, then you should make sure that the associated value is a dict.

```python3
def add_path_to_dict(d, path):
    for k in path:
        if k not in d:
            d[k] = {}
        elif not isinstance(d[k], dict):
            return
        d = d[k]

d = {&#39;z&#39;: 42}
add_path_to_dict(d, &#39;abcdefgh&#39;)
print(d)
# {&#39;z&#39;: 42, &#39;a&#39;: {&#39;b&#39;: {&#39;c&#39;: {&#39;d&#39;: {&#39;e&#39;: {&#39;f&#39;: {&#39;g&#39;: {&#39;h&#39;: {}}}}}}}}}

d = {&#39;a&#39;: {&#39;b&#39;: {&#39;c&#39;:{&#39;z&#39;: 42}}}}
add_path_to_dict(d, &#39;abcdefgh&#39;)
print(d)
# {&#39;a&#39;: {&#39;b&#39;: {&#39;c&#39;: {&#39;z&#39;: 42, &#39;d&#39;: {&#39;e&#39;: {&#39;f&#39;: {&#39;g&#39;: {&#39;h&#39;: {}}}}}}}}}

d = {&#39;a&#39;: {&#39;b&#39;: {&#39;c&#39;:42}}}
add_path_to_dict(d, &#39;abcdefgh&#39;)
print(d)
# {&#39;a&#39;: {&#39;b&#39;: {&#39;c&#39;: 42}}}

Further suggestions:

• Instead of the empty return, you could return 'Some error value' or you could raise ValueError(f'Key {k} already in dict and its value is not a dict.')

答案2

得分: 0

如评论中所指出的，您的代码没有太多问题，只是路径中不必要的 [1:] 切片操作。

如果您想要更加简洁的写法，您可以使用 dict.setdefault 方法，避免需要测试键是否已经存在：

output_content = {'already_existing_key': 'already_existing_value'}
path = ['patterns', 'pattern_1', 'children', 'pattern_1_1']
current_dict = output_content
for key in path:
    current_dict = current_dict.setdefault(key, {})
print(output_content)

演示：https://replit.com/@blhsing/UnusedDiscreteScriptinglanguages

output_content = {'already_existing_key': 'already_existing_value'}
path = ['patterns', 'pattern_1', 'children', 'pattern_1_1']
current_dict = output_content
for key in path:
    current_dict = current_dict.setdefault(key, {})
print(output_content)

答案3

得分: 0

答案中的代码如下：

@Stef 和 @blhsing 的两个答案都修改了原始的 `dict`，这可能不是你想要的行为。如果你需要一个副本，可以尝试以下解决方案：

```py
def add_path(path, dict_):
    copied = dict_.copy()
    dict_ = copied
    
    for index, element in enumerate(path):
        dict_ = dict_.setdefault(element, {})
        
        if not isinstance(dict_, dict):
            break
    
    return copied

<details>
<summary>英文:</summary>

Both answers by @Stef and @blhsing modify the original `dict`, a behaviour which you may not want. Here&#39;s another solution if you need a copy:

```py
def add_path(path, dict_):
	copied = dict_.copy()
	dict_ = copied
	
	for index, element in enumerate(path):
		dict_ = dict_.setdefault(element, {})
		
		if not isinstance(dict_, dict):
			break
	
	return copied