Question

我有一个嵌套的字典：

my_dict = {
1: {'player_id': 1,
'player_name': 'Bryan_Demapan',
'time_played': 0.0,
'player_pokemon': {},
'gyms_visited': []},
2: {'player_id': 2,
'player_name': 'Tom Syneal',
'gyms_visited': [],
'player_pokemon': {},
'time_played': 0.0}}

假设我有一个列表

new_list = ['A', 'B', 'C']

我该如何做一个for循环语句，将新列表插入到所有带有“ gyms_visited”键的空列表中？

新词典应如下所示

my_dict = {
1: {'player_id': 1,
'player_name': 'Bryan_Demapan',
'time_played': 0.0,
'player_pokemon': {},
'gyms_visited': ['A', 'B', 'C']},
2: {'player_id': 2,
'player_name': 'Tom Syneal',
'gyms_visited': ['A', 'B', 'C'],
'player_pokemon': {},
'time_played': 0.0}}

Answer 1

只需遍历my_dict的值并将new_list分配给键'gyms_visited'

In [529]: for k, v in my_dict.items():
     ...:     v['gyms_visited'] = new_list.copy()
     ...:

In [530]: my_dict
Out[530]:
{1: {'gyms_visited': ['A', 'B', 'C'],
  'player_id': 1,
  'player_name': 'Bryan_Demapan',
  'player_pokemon': {},
  'time_played': 0.0},
 2: {'gyms_visited': ['A', 'B', 'C'],
  'player_id': 2,
  'player_name': 'Tom Syneal',
  'player_pokemon': {},
  'time_played': 0.0}}

如果仅分配new_list，则一旦更改new_list中的任何值，就会遇到麻烦

In [529]: for k, v in my_dict.items():
     ...:     v['gyms_visited'] = new_list
     ...:

In [531]: new_list[1] = 100

In [532]: my_dict
Out[532]:
{1: {'gyms_visited': ['A', 100, 'C'],
  'player_id': 1,
  'player_name': 'Bryan_Demapan',
  'player_pokemon': {},
  'time_played': 0.0},
 2: {'gyms_visited': ['A', 100, 'C'],
  'player_id': 2,
  'player_name': 'Tom Syneal',
  'player_pokemon': {},
  'time_played': 0.0}}

Answer 2

尝试一下：

print({idx:{k:(new_list if k=='gyms_visited' else v) for k,v in i.items()} for idx,i in enumerate(my_dict.values(),1)})

嵌套词典理解将起作用。

输出：

{1: {'player_id': 1, 'player_name': 'Bryan_Demapan', 'time_played': 0.0, 'player_pokemon': {}, 'gyms_visited': ['A', 'B', 'C']}, 2: {'player_id': 2, 'player_name': 'Tom Syneal', 'gyms_visited': ['A', 'B', 'C'], 'player_pokemon': {}, 'time_played': 0.0}}

我实际上喜欢@aydow的解决方案，除了它，您还可以像这样：

for k,v in data.items():
   v['gyms_visited']=new_list[:]

或者：

import copy
for k,v in data.items():
   v['gyms_visited']=copy.deepcopy(new_list)

Answer 3

您可以使用字典理解：

new_list = ['A', 'B', 'C']
data = {1: {'player_id': 1, 'player_name': 'Bryan_Demapan', 'time_played': 0.0, 'player_pokemon': {}, 'gyms_visited': []}, 2: {'player_id': 2, 'player_name': 'Tom Syneal', 'gyms_visited': [], 'player_pokemon': {}, 'time_played': 0.0}}
new_data = {a:{c:[i for i in new_list if i not in d] if c == 'gyms_visited' else d \
    for c, d in b.items()} for a, b in data.items()}

输出：

{
  "1": {
    "player_id": 1,
    "player_name": "Bryan_Demapan",
    "time_played": 0.0,
    "player_pokemon": {},
    "gyms_visited": ["A", "B", "C"]
 },
 "2": {
    "player_id": 2,
    "player_name": "Tom Syneal",
    "gyms_visited": ["A", "B", "C"],
    "player_pokemon": {},
    "time_played": 0.0
   }
}

Answer 4

您也可以仅extend()列表：

for key in my_dict:
    my_dict[key]['gyms_visited'].extend(new_list)

并且正如您在此处看到的那样，所有列表都具有不同的id()并且没有引用相同的对象：

print(id(new_list))
# 2704861952904

for key in my_dict:
    print(id(my_dict[key]['gyms_visited']))
# 2704833143368
# 2704833143432

就性能而言，list.extend()为O（N），其中N = 3是要扩展的列表的长度。这很可能是一系列list.append()调用，每个调用都是O（1）。

使用[:]或.copy()（即O（N））进行复制将是相同的。

使用for循环在嵌套字典内的空列表中插入列表值

4 个答案: