递归访问字典和修改

Question

我有以下字典：

my_dict = {'key1': {'key2': {'foo': 'bar'} } }

我想将一个条目附加到key1-> key2-> key3上，其值为’blah’，产生：

my_dict = {'key1': {'key2': {'foo': 'bar', 'key3': 'blah'} } }

我正在寻找一种与键的数量无关的通用解决方案，即即使不存在从key3向下的键，key1-> key2-> key3-> key4->
key5也应该起作用。这样我得到：

my_dict = {'key1': {'key2': {'foo': 'bar', 'key3': {'key4': {'key5': 'blah'} } } } }

提前致谢。

Answer 1

您可以使用该reduce()函数遍历一系列嵌套字典：

def get_nested(d, path):
    return reduce(dict.__getitem__, path, d)

演示：

>>> def get_nested(d, path):
...     return reduce(dict.__getitem__, path, d)
... 
>>> my_dict = {'key1': {'key2': {'foo': 'bar', 'key3': {'key4': {'key5': 'blah'}}}}}
>>> get_nested(my_dict, ('key1', 'key2', 'key3', 'key4', 'key5'))
'blah'

当密钥不存在时，此版本引发异常：

>>> get_nested(my_dict, ('key1', 'nonesuch'))
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<stdin>", line 2, in get_nested
KeyError: 'nonesuch'

但您可以替换dict.__getitem__为lambda d, k: d.setdefault(k, {})，使其创建空字典：

def get_nested_default(d, path):
    return reduce(lambda d, k: d.setdefault(k, {}), path, d)

演示：

>>> def get_nested_default(d, path):
...     return reduce(lambda d, k: d.setdefault(k, {}), path, d)
... 
>>> get_nested_default(my_dict, ('key1', 'nonesuch'))
{}
>>> my_dict
{'key1': {'key2': {'key3': {'key4': {'key5': 'blah'}}, 'foo': 'bar'}, 'nonesuch': {}}}

要在给定路径上 设置 值，请遍历除最后一个键以外的所有键，然后在常规词典分配中使用最后一个键：

def set_nested(d, path, value):
    get_nested_default(d, path[:-1])[path[-1]] = value

这使用该get_nested_default()函数根据需要添加空字典：

>>> def set_nested(d, path, value):
...     get_nested_default(d, path[:-1])[path[-1]] = value
... 
>>> my_dict = {'key1': {'key2': {'foo': 'bar'}}}
>>> set_nested(my_dict, ('key1', 'key2', 'key3', 'key4', 'key5'), 'blah')
>>> my_dict
{'key1': {'key2': {'key3': {'key4': {'key5': 'blah'}}, 'foo': 'bar'}}}