如何设置给定列表位置的嵌套python dict的"第n"元素？

Question

如果我有一个dict例如

d = { 'a': {
            "x": [],
            "y": {
                "z": {
                    "1": 'loser'
                        }
                 }
            }
    }

print(d['a']['y']['z']['1']) #=> loser

但是,我不知道这个词典中有多少嵌套项目.相反,我有一个键列表,如:

['a', 'y', 'z', '1']

什么是优雅的设置方式d['a']['y']['z']['1'] = 'winner'？

这是我尝试过的:

l = ['a', 'y', 'z', '1']

def change_the_value(d, l, value):
    if len(l) == 1:
        d[l[0]] = value
    if len(l) == 2:
        d[l[0]][l[1]] = value
    if len(l) == 3:
        d[l[0]][l[1]][l[2]] = value
    if len(l) == 4:
        d[l[0]][l[1]][l[2]][l[3]] = value
    # ... ad infinitum
    return d

change_the_value(d, l, 'winner')
print(d) # => {'a': {'x': [], 'y': {'z': {'1': 'winner'}}}}

Answer 1

你可以使用一个简单的for循环:

_path = ['a', 'y', 'z', '1']
d = {'a': {'x': [], 'y': {'z': {'1': 'loser'}}}}
_start = d
for i in _path[:-1]:
   _start = _start[i]

_start[_path[-1]] = 'winner'
print(d)

输出:

{'a': {'x': [], 'y': {'z': {'1': 'winner'}}}}

您也可以使用递归(如果您不介意为其创建新结构d):

def update(_d, _path):
  return {a:'winner' if a == _path[-1] and len(_path) == 1 else 
    update(b, _path[1:]) if a == _path[0] else b for a, b in _d.items()}

print(update(d, ['a', 'y', 'z', '1']))

输出:

{'a': {'x': [], 'y': {'z': {'1': 'winner'}}}}

Answer 2

如果您可以确定密钥列表有效,则可以使用functools.reduce.

>>> from functools import reduce 
>>>
>>> keys = ['a', 'y', 'z', '1']                                                                                     
>>> d = { 'a': { 
...:             "x": [], 
...:             "y": { 
...:                 "z": { 
...:                     "1": 'loser' 
...:                         } 
...:                  } 
...:             } 
...:     }                                                                                                             
>>>                                                                                                                    
>>> the_dict = reduce(dict.get, keys[:-1], d)                                                                          
>>> the_dict[keys[-1]] = 'winner'                                                                                      
>>> d                                                                                                                  
{'a': {'x': [], 'y': {'z': {'1': 'winner'}}}}