如何使用具有唯一键的多个列表创建字典

Question

我试图从预定义列表中将键值添加到现有字典中。下面是一个例子：

# I created players dictionary : 
players = {'gyms_visited': [], 
'player_id': [4], 
'player_name': 'cynthia', 
'player_pokemon': {}, 
'time_played': 30.9} 

我想添加的列表a，b我的队员字典，我上面创建：

a = ['player_id', 'player_name','time_played', 'player_pokemon', 'gyms_visited'] 
b = [2, 'teri', 22.2, {}, []] 

此外，有没有一种方法，使Player_id的值（4,2）的钥匙给玩家字典（这是我想要实现）：

{4: {'gyms_visited': [], 
    'player_name': 'cynthia', 
    'player_pokemon': {}, 
    'time_played': 30.9}, 
2: {'player_name':'teri', 
    'time_played':22.2, 
    'gyms_visited': [], 
    'player_pokemon':{}} 

Answer 1

所有你需要的是：

点

# to change already existing dict 
players = {players['player_id']:{key:value for key,value in players.items() if key != 'player_id'}} 
# make dict from lists a and b 
z = {b[a=='player_id']: {key:value for key,value in zip(a,b) if key !='player_id'}} 
# group both 
players={**players,**z} 

的细节在下面：

首先，你需要将已经作出字典更改为所需的输出类型是这样的：

players = {players['player_id']:{key:value for key,value in players.items() if key != 'player_id'}} 
print(players) 

将输出：

{4: {'gyms_visited': [], 
    'player_name': 'cynthia', 
    'player_pokemon': {}, 
    'time_played': 30.9}} 

那么你把你的两个名单a和b，并作出一个额外的字典，他们是这样的：

z = {b[a=='player_id']: {key:value for key,value in zip(a,b) if key !='player_id'}} 
print(z) 

将输出：

{2: {'player_name': 
    'teri', 'time_played': 22.2, 
    'player_pokemon': {}, 
    'gyms_visited': []}} 

和最后，你只要把它们放在同一个字典如下：

players={**players,**z} 
print(players) 

输出：

{4: {'gyms_visited': [], 
    'player_name': 'cynthia', 
    'player_pokemon': {}, 
    'time_played': 30.9}, 
2: {'player_name': 'teri', 
    'time_played': 22.2, 
    'player_pokemon': {}, 
    'gyms_visited': []}} 

请注意如果你想添加更多的元素，你可以在循环中做到这一点。对于为例考虑这些新名单：

a = ['player_id', 'player_name','time_played', 'player_pokemon', 'gyms_visited'] 
b = [6, 'new_Name', 66.6, {}, []] 

做和以前一样：你作为输出

z = {b[a=='player_id']: {key:value for key,value in zip(a,b) if key !='player_id'}} 
players={**players,**z} 

：

{4: {'gyms_visited': [], 
    'player_name': 'cynthia', 
    'player_pokemon': {}, 
    'time_played': 30.9}, 
2: {'player_name': 'teri', 
    'time_played': 22.2, 
    'player_pokemon': {}, 
    'gyms_visited': []}, 
6: {'player_name': 'new_Name', 
    'time_played': 66.6, 
    'player_pokemon': {}, 
    'gyms_visited': []}} 

您可以参考有关merging dicts和**dicts这些链接为了更好的理解。

这里是内置拉链方法的link。

Answer 2

from itertools import izip 
from pprint import pprint 

# Edit your existing dict 
>>> players2 = {players.pop('player_id')[0]: players} 
>>> pprint(players2) 
{4: {'gyms_visited': [], 
    'player_name': 'cynthia', 
    'player_pokemon': {}, 
    'time_played': 30.9}} 

# Add new player info 
>>> players2.update({b[0]: {k: v for k, v in izip(a, b[1:])}}) 
>>> pprint(players2) 
{2: {'gyms_visited': [], 
    'player_name': 'teri', 
    'player_pokemon': {}, 
    'time_played': 22.2}, 
4: {'gyms_visited': [], 
    'player_name': 'cynthia', 
    'player_pokemon': {}, 
    'time_played': 30.9}} 

注意：每当你与一些重复的代码工作，这是值得一试内置itertools package。对于实例，izip通过返回迭代器而不是列表而改进超过zip。