如何在 Python 的 dict 中获取 dict 的路径？

Question

There is dictionary which is generated by adding some data:

some_dict = {'some_value1': {}, 'dict1': {'qwerty': None, 'bar': {}}}

I know that I can add there some data by using update method:

some_dict['dict1']['bar'].update({'some_key': value})

But what if a part of the code doesn't know correct way to update 'bar'? like:

some_dict[???]['bar'].update({'some_key': value})

How can I update it correctly?

Answer 1

So I want to clarify your question. The problem is that you have a dictionary within a dictionary and (ie overall_dict = {..., 'inner_dict': {'foo': 4 }} ) and what you want is for a function to update 'foo' without passing in overall_dict ? Ie the function below

def func1(dict): 
   dict['foo'] = 6

If that is the case, then there isn't too much of a problem. Python uses pointers and therefore if you pass func1(overall_dict['inner_dict']) then it will do what you want. Essentially you can set a variable to point to the inner dictionary. When you update the new variable, then it will also update the overall_dict .

I hope that helps. If you are wondering about other python dictionary questions, pythontutor.com

My tests to make sure python is by pointers is here

This is also a really good textbook for python: link here

Hope that helps.

Answer 2

As far as I know, there is no standard Python function to do what you want. You will need to implement some kind of search algorithm yourself. Broadly speaking you will have the choice between a depth-first search or breadth-first search strategy.

A possible implementation of the former would be something along the lines of:

some_dict = {'some_value1': {}, 'dict1': {'qwerty': None, 'bar': {}}}

def get_nested_dict(d, name):
    stack = [d]
    while stack:
        for k, v in stack.pop().items():
            if k == name:
                return v

            if isinstance(v, dict):
                stack.append(v)


print(some_dict)
get_nested_dict(some_dict, 'bar')['new'] = 1
print(some_dict)

Producing:

{'some_value1': {}, 'dict1': {'qwerty': None, 'bar': {}}}
{'some_value1': {}, 'dict1': {'qwerty': None, 'bar': {'new': 1}}}