我想以编程方式访问字典。我知道如何使用递归函数执行此操作，但有更简单的方法吗？
example = {'a': {'b': 'c'},
'1': {'2': {'3': {'4': '5'}}}}
keys = ('a', 'b')
example[keys] = 'new'
# Now it should be
# example = {'a': {'b': 'new'},
# '1': {'2': {'3': {'4': '5'}}}}
keys = ('1', '2', '3', '4')
example[keys] = 'foo'
# Now it should be
# example = {'a': {'b': 'new'},
# '1': {'2': {'3': {'4': 'foo'}}}}
keys = ('1', '2')
example[keys] = 'bar'
# Now it should be
# example = {'a': {'b': 'new'},
# '1': {'2': 'bar'}}
答案 0 :(得分：4)
您似乎想要做的是定义您自己的支持这种索引的字典类。我们可以通过使用以下事实来获得相当简洁的语法：当您执行d[1, 2, 3]时，Python实际上将元组(1, 2, 3)传递给__getitem__。
class NestedDict:
def __init__(self, *args, **kwargs):
self.dict = dict(*args, **kwargs)
def __getitem__(self, keys):
# Allows getting top-level branch when a single key was provided
if not isinstance(keys, tuple):
keys = (keys,)
branch = self.dict
for key in keys:
branch = branch[key]
# If we return a branch, and not a leaf value, we wrap it into a NestedDict
return NestedDict(branch) if isinstance(branch, dict) else branch
def __setitem__(self, keys, value):
# Allows setting top-level item when a single key was provided
if not isinstance(keys, tuple):
keys = (keys,)
branch = self.dict
for key in keys[:-1]:
if not key in branch:
branch[key] = {}
branch = branch[key]
branch[keys[-1]] = value
以下是使用示例
# Getting an item
my_dict = NestedDict({'a': {'b': 1}})
my_dict['a', 'b'] # 1
# Setting an item
my_dict = NestedDict()
my_dict[1, 2, 3] = 4
my_dict.dict # {1: {2: {3: 4}}}
# You can even get a branch
my_dict[1] # NestedDict({2: {3: 4}})
my_dict[1][2, 3] # 4
然后，您还可以定义NestedDict，__iter__和__len__，从而使__contains__实施更加丰富。
此外，这可以很容易地集成到您的代码中，因为任何预先存在的字典都可以通过NestedDict(your_dict)转换为嵌套字典。
答案 1 :(得分：1)
此解决方案创建另一个具有相同键的字典，然后更新现有字典：
cd code/clang-variables
docker run -it -v $PWD:/home clang
root@5196c095092d:/home# ./clang-variables test.cpp -- -std=c++14
使用字符串，列表或元组来访问密钥
答案 2 :(得分：0)
在适应this answer（基本上是化妆品）后，我得到了：
from functools import reduce
import operator
def nested_get(dictionary, *keys):
return reduce(operator.getitem, keys, dictionary)
def nested_set(dictionary, value, *keys):
nested_get(dictionary, *keys[:-1])[keys[-1]] = value
example = {
'a': {'b': 'c'},
'1': {
'2': {
'3': {'4': '5'}
}
}
}
nested_set(example, "foo", "1", "2", "3")
print(example)
keys = ["1", "2"]
nested_set(example, "yay", *keys)
print(example)
输出：
{'a': {'b': 'c'}, '1': {'2': {'3': 'foo'}}}
{'a': {'b': 'c'}, '1': {'2': 'yay'}}
相同的想法（使用变量通过引用传递的事实），但这次调整one of my answers：
def nested_set(element, value, *keys):
if type(element) is not dict:
raise AttributeError('nested_set() expects dict as first argument.')
if len(keys) < 2:
raise AttributeError('nested_set() expects at least three arguments, not enough given.')
_keys = keys[:-1]
_element = element
for key in _keys:
_element = _element[key]
_element[keys[-1]] = value
example = {"foo": { "bar": { "baz": "ok" } } }
nested_set(example, "yay", "foo", "bar")
print(example)
输出
{'foo': {'bar': 'yay'}}
这个不需要任何花哨的进口，所以我倾向于更喜欢它.. 选择你的味道
答案 3 :(得分：0)
您可以使用较小的递归函数和字典理解：
import functools
def all_examples(f):
def wrapper():
def update_dict(d, path, target):
return {a:target if path[-1] == a else update_dict(b, path, target) if isinstance(b, dict) else b for a, b in d.items()}
current_d = {'a': {'b': 'c'},'1': {'2': {'3': {'4': '5'}}}}
final_ds = []
for i in f():
current_d = update_dict(current_d, *i)
final_ds.append(current_d)
return final_ds
return wrapper
@all_examples
def input_data():
return [[('a', 'b'), 'new'], [('1', '2', '3', '4'), 'foo'], [('1', '2'), 'bar']]
for i in input_data():
print(i)
输出：
{'a': {'b': 'new'}, '1': {'2': {'3': {'4': '5'}}}}
{'a': {'b': 'new'}, '1': {'2': {'3': {'4': 'foo'}}}}
{'a': {'b': 'new'}, '1': {'2': 'bar'}}
答案 4 :(得分：0)
遍历您字典的方法。您可以通过继承dict。
遍历算法是@MartijnPeters的礼貌（在那里）。
import operator
class ndict(dict):
def get_traverse(self, mapList):
return reduce(operator.getitem, mapList, self)
def set_traverse(self, mapList, value):
self.get_traverse(mapList[:-1])[mapList[-1]] = value
d = ndict({'a': {'b': 'c'}, '1': {'2': {'3': {'4': '5'}}}})
d.get_traverse(['a', 'b']) # 'c'
d.set_traverse(['a', 'b'], 4) # {'a': {'b': 4}, '1': {'2': {'3': {'4': '5'}}}}
答案 5 :(得分：0)
我想添加对dict-digger的引用，这是一个我有时想要使用的开源模块，它提供了开箱即用的功能。 （我与该项目无关）
答案 6 :(得分：0)
我结合了jpp和OlivierMelançon的答案来创建一个NestedDict类。我更喜欢像通常访问字典一样访问字典，然后使用列表作为参数
import operator
from functools import reduce
class NestedDict(dict):
def __getitem__(self, item):
if isinstance(item, list):
return self.get_traverse(item)
return super(NestedDict, self).__getitem__(item)
def __setitem__(self, key, value):
if isinstance(key, list):
for _key in key[:-1]:
if not _key in self:
self[_key] = {}
self = self[_key]
self[key[-1]] = value
return self
return super(NestedDict, self).__setitem__(key, value)
def get_traverse(self, _list):
return reduce(operator.getitem, _list, self)
nested_dict = NestedDict({'aggs': {'aggs': {'field_I_want': 'value_I_want'}, 'None': None}})
path = ['aggs', 'aggs', 'field_I_want']
nested_dict[path] # 'value_I_want'
nested_dict[path] = 'changed'
nested_dict[path] # 'changed'
对于任何有兴趣的人。我使用自动查找路径的功能增强了该类。 （在doctest中使用）
import operator
from functools import reduce
from copy import deepcopy
class NestedDict(dict):
"""
Dictionary that can use a list to get a value
:example:
>>> nested_dict = NestedDict({'aggs': {'aggs': {'field_I_want': 'value_I_want'}, 'None': None}})
>>> path = ['aggs', 'aggs', 'field_I_want']
>>> nested_dict[path]
'value_I_want'
>>> nested_dict[path] = 'changed'
>>> nested_dict[path]
'changed'
"""
def __getitem__(self, item):
if isinstance(item, list):
return self.get_traverse(item)
return super(NestedDict, self).__getitem__(item)
def __setitem__(self, key, value):
if isinstance(key, list):
for _key in key[:-1]:
if _key not in self:
self[_key] = {}
self = self[_key]
self[key[-1]] = value
return self
return super(NestedDict, self).__setitem__(key, value)
def get_traverse(self, _list):
return reduce(operator.getitem, _list, self)
_paths = []
_path = []
def find(self, key, _dict=None, _main_loop=True):
""" Find a list of paths to a key
:param key: str, the key you want to find
:param _dict: used for recursive searching
:return: list with paths
:example:
>>> nested_dict = NestedDict({'aggs': {'aggs': {'field_I_want': 'value_I_want'}, 'None': None}})
>>> paths = nested_dict.find('field_I_want')
>>> paths
[['aggs', 'aggs', 'field_I_want']]
>>> nested_dict[paths[0]] = 'changed'
>>> nested_dict[paths[0]]
'changed'
"""
if _main_loop:
self._path, self._paths = [], []
_dict = self
for _key in _dict.keys():
self._path.append(_key)
if _key == key:
self._paths.append(deepcopy(self._path))
if isinstance(_dict[_key], dict):
self.find(key, _dict[_key], _main_loop=False)
self._path.pop()
if _main_loop:
return self._paths