提问人：MrPoulet 提问时间：2/17/2020 最后编辑：MrPoulet 更新时间：2/17/2020 访问量：60
为什么我的字典在可变和不可变之间交替？
Why are my dicts alternating between mutable and immutable?
问：
请考虑以下代码
from pprint import pprint
test_dict = {}
new = {}
new['slot'] = {}
for k in range(5):
test_dict[k] = {}
test_dict[k].update(new)
if k == 3:
test_dict[k]['slot']['this should only be in 3'] = []
pprint(test_dict)
print('Round 2, without SLOT')
test_dict = {}
new = {}
for k in range(5):
test_dict[k] = {}
test_dict[k].update(new)
if k == 3:
test_dict[k]['this should only be in 3'] = []
pprint(test_dict)
使用此输出
> python -i .\test2.py
{0: {'slot': {'this should only be in 3': []}},
1: {'slot': {'this should only be in 3': []}},
2: {'slot': {'this should only be in 3': []}},
3: {'slot': {'this should only be in 3': []}},
4: {'slot': {'this should only be in 3': []}}}
Round 2, without SLOT
{0: {}, 1: {}, 2: {}, 3: {'this should only be in 3': []}, 4: {}}
帮助我理解为什么在第一种情况下，“应该只是......”列表出现在每种情况下，而不是第二种情况下。字典是不可变的，但我不明白为什么我会得到不同的结果。
谢谢
答：
0赞
chepner
2/17/2020
#1
该方法不会为 中的每个条目创建副本。updatenew['slot']test_dict
后
test_dict = {}
new = {}
new['slot'] = {}
for k in range(5):
test_dict[k] = {}
test_dict[k].update(new)
if k == 3:
test_dict[k]['slot']['this should only be in 3'] = []
test_dict[k]['slot']是对每个 . 的相同引用。您可以通过以下方式确认这一点：dictk = 0, 1, ..., 4id
>>> for k in range(5): id(test_dict[k]['slot'])
...
4422633104
4422633104
4422633104
4422633104
4422633104
>>> id(new['slot'])
4422633104
评论
0赞
MrPoulet
2/17/2020
所以你是说如果我做update（new），我正在创建一个新的“新”字典。但是如果我写 update（new['slot']），我每次都会插入相同的 'slot' 副本。
0赞
chepner
2/17/2020
不，本质上只是一个包装器。唯一可能创建的新内容是 中的新键。test_dict[k].update(new)for i in new: test_dict[k][i] = new[i]test_dict[k]
1赞
user3657941
2/17/2020
#2
您将存储在以下每个键中的相同实例：dictnew['slot']test_dict
from pprint import pprint
test_dict = {}
new = {}
new['slot'] = {}
new['slot']['id'] = id(new['slot'])
for k in range(5):
test_dict[k] = {}
test_dict[k].update(new)
if k == 3:
test_dict[k]['slot']['this should only be in 3'] = []
pprint(test_dict)
输出
{0: {'slot': {'id': 4433735760, 'this should only be in 3': []}},
1: {'slot': {'id': 4433735760, 'this should only be in 3': []}},
2: {'slot': {'id': 4433735760, 'this should only be in 3': []}},
3: {'slot': {'id': 4433735760, 'this should only be in 3': []}},
4: {'slot': {'id': 4433735760, 'this should only be in 3': []}}}
一个可能的解决方法是在每次需要时创建一个新的：dict
from pprint import pprint
test_dict = {}
for k in range(5):
test_dict[k] = {}
new = {'slot': dict()}
new['slot']['id'] = id(new['slot'])
test_dict[k].update(new)
if k == 3:
test_dict[k]['slot']['this should only be in 3'] = []
pprint(test_dict)
输出
{0: {'slot': {'id': 4399711968}},
1: {'slot': {'id': 4399712528}},
2: {'slot': {'id': 4399713088}},
3: {'slot': {'id': 4399713648, 'this should only be in 3': []}},
4: {'slot': {'id': 4399730768}}}
评论
new['slot'].update(new)new['slot']