递归访问字典和修改 [英] recursive access to dictionary and modification

问题描述

我有以下字典：

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

我想在key1-> key2-> key3，值为'blah'，产生：

and I would like to append an entry to key1->key2->key3 with value 'blah' yielding:

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

我正在寻找一个通用的解决方案与键的数量无关，即key1-> key2-> key3-> key4-> key5应该也可以正常工作，即使从key3向下的键不存在也是如此。所以我得到：

I am looking for a generic solution that is independent of the number of keys, i.e. key1->key2->key3->key4->key5 should work as well, even though keys from key3 on downwards do not exist. So that I get:

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

提前感谢

推荐答案

您可以使用 reduce（）函数遍历一系列嵌套字典：

You can use the reduce() function to traverse a series of nested dictionaries:

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'

当一个键不存在时，此版本引发异常：

This version throws an exception when a key doesn't exist:

>>> 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'

，但您可以用 lambda d，k：d.setdefault（k，{}） dict .__ getitem __ $ c>可以创建空字典：

but you could replace dict.__getitem__ with lambda d, k: d.setdefault(k, {}) to have it create empty dictionaries instead:

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': {}}}

在给定的路径上设置一个值，遍历所有键，但最后一个，然后在常规字典分配中使用最终键：

To set a value at a given path, traverse over all keys but the last one, then use the final key in a regular dictionary assignment:

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

这将使用 get_nested_default（）函数根据需要添加空字典：

This uses the get_nested_default() function to add empty dictionaries as needed:

>>> 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'}}}

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