Question

我在python中有一个深度嵌套的字典，占用了大量的空间。有没有办法缩写像这样的东西

master_dictionary['sub_categories'][sub_cat_name]['attributes'][attribute_name]['special_type']['nested_children'][child_cat_name][color] = blue

以此为例

nested_child_info[color] = blue

还有它编辑字典吗？我希望这是有道理的。

Answer 1

nested_child_info = master_dictionary['sub_categories'][sub_cat_name]['attributes'][attribute_name]['special_type']['nested_children'][child_cat_name]
nested_child_info[color] = blue

nested_child_info是引用，因此更改其内容将会更改master_dictionary的内容。

Answer 2

是的，你可以。

>>> dict1 = {'foo':{'bar':{'baz':0}}}
>>> dict2 = dict1['foo']['bar']
>>> dict2['baz'] = 1
>>> dict1
{'foo': {'bar': {'baz': 1}}} # dict1 has been modified

Answer 3

与@fferri相似。您始终必须在 long 列表中指定项目。使用reduce和getitem获取引用到最终字典：

from functools import reduce
from operator import getitem

d = {1:{2:{3:{4:5}}}}

foo = 2
items = [1,foo,3]
result = d
info = reduce(getitem, items, d)


>>> info[4]
5
>>> d
{1: {2: {3: {4: 5}}}}
>>> info[4] = 99
>>> d
{1: {2: {3: {4: 99}}}}

我还在玩一个类，但它似乎没有很多优点 - 除了你可以自定义一个键错误异常，以便错误消息告诉你哪个键缺少哪个深度。

class Drilldown:
    def __init__(self, d, path):
        #self.final = reduce(getitem, path, d)
        self.final = d
        for i, item in enumerate(path, 1):
            try:
                self.final = self.final[item]
            except KeyError as e:
                msg = ''.join('[{}]' for _ in range(i))
                msg = msg.format(*path[:i])
                msg = 'The last key in the path "{}" does not exist'.format(msg)
                e.args = [msg]
                raise
    def __call__(self, item):
        return self.final[item]
    def __setitem__(self, item, value):
        self.final[item] = value
    def __getitem__(self, item):
        return self.final[item]
    def __str__(self):
        return str(self.final)
    def __repr__(self):
        return repr(self.final)

>>> z = 19
>>> items = [1,2,z]
>>> q = Drilldown(d,items)
Traceback (most recent call last):
  File "<pyshell#68>", line 1, in <module>
    q = Drilldown(d,items)
  File "C:\pyProjects33\tmp.py", line 32, in __init__
    self.final = self.final[item]
KeyError: 'The last key in the path "[1][2][19]" does not exist'

>>> 
>>> #normal usage
>>> items = [1,2,3]
>>> q = Drilldown(d,items)
>>> d
{1: {2: {3: {4: 5}}}}
>>> q
{4: 5}
>>> q(4)
5
>>> q[4]
5
>>> q[4] += 20
>>> q
{4: 25}
>>> d
{1: {2: {3: {4: 25}}}}
>>> q['foo'] = '99'
>>> q
{4: 25, 'foo': '99'}
>>> d
{1: {2: {3: {4: 25, 'foo': '99'}}}}
>>>

Answer 4

你可以这样做吗？

thing = {1: {2: {3: {4: {'hello': 'world'}}}}}
a = thing[1]
b = a[2]
c = b[3]
d = c[4]
print(d) # {'hello': 'world'}

Answer 5

由于字典是可变的，实际上这恰好就像你期望的那样：

>>> test = {'outer': 'thing', 'inner': {'thing': 'im the inner thing'}}
>>> inner_thing = test['inner']
>>> inner_thing
{'thing': 'im the inner thing'}
>>> inner_thing['thing'] = 'im something new'
>>> inner_thing
{'thing': 'im something new'}
>>> test
{'outer': 'thing', 'inner': {'thing': 'im something new'}}

这是因为可变对象在python中通过引用传递，而不是作为副本传递（关于此的大量文章，所以不详细说明）。

但是，值得注意的是，您可能实际上并不想更改原始字典，因为它可能会对使用此变量的其他代码产生不良影响（这取决于您的代码库）。在这种情况下，我通常会复制我需要变异的数据以避免副作用：

>>> from copy import deepcopy
>>> test = {'outer': 'thing', 'inner': {'thing': 'im the inner thing'}}
>>> new_test = deepcopy(test)
>>> inner_thing = test['inner']
>>> inner_thing['thing'] = 'im something new'
>>> test
{'outer': 'thing', 'inner': {'thing': 'im something new'}}
>>> new_test
{'outer': 'thing', 'inner': {'thing': 'im the inner thing'}}

Answer 6

如果你有一些＆＃34;固定＆＃34;键总是可以创建一个函数：

考虑这个例子：

d = dict(a=dict(sub_categories=dict(b=1)))

def changevalue(value, lvl1, lvl2):
    d[lvl1]['sub_categories'][lvl2] = value

changevalue(2,'a','b')

print(d)

#{'a': {'sub_categories': {'b': 2}}}

在你的情况下，你想要解除：

[sub_cat_name]，[attribute_name]，[child_cat_name]，[color] ...也许

在Python中处理深层嵌套字典的便捷方法

6 个答案: