使用for循環在嵌套字典中的空列表中插入列表值

Question

我有一個嵌套字典：

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}}

假設我有一個清單

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

我如何制作一個for循環語句，將我的新列表插入到所有帶有'gyms_visited'鍵的空列表中？

新詞典應該是這樣的

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

只是迭代的值my_dict並分配new_list的關鍵'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}}

如果您只是簡單地指定new_list那么只要更改new_list任何值，就會遇到麻煩

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

試試這個：

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)})

嵌套字典理解會起作用。

輸出：

{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}}

我其實喜歡@ aydow的解決方案，除此之外，你可以這樣做：

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

要么：

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()}

輸出：

{
  "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

你也可以只extend()列表：

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

正如您在此處所看到的，所有列表都具有不同的id()並且不引用相同的對象：

print(id(new_list))
# 2704861952904

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

在性能方面， list.extend()是O（N），其中N = 3是要擴展的列表的長度。 這很可能是一系列list.append()調用，每個調用都是O（1）。

對於使用[:]或.copy()進行復制，這是相同的，即O（N）。