Question

categories =  {'player_name': None, 'player_id': None, 'season': None} 

L = ['Player 1', 'player_1', '2020']

如何遍历列表并将其值分配给相应的键？所以它会变成这样：

{'player_name': 'Player 1', 'player_id': 'player_1, 'season': '2020'}

谢谢

Answer 1

如果python> = 3.6，则使用zip() + dict()，如果<3.6，则看起来dict是无序的，所以我不知道。

test.py ：

categories =  {'player_name': None, 'player_id': None, 'season': None}
L = ['Player 1', 'player_1', '2020']
print(dict(zip(categories, L)))

结果：

$ python3 test.py
{'player_name': 'Player 1', 'player_id': 'player_1', 'season': '2020'}

Answer 2

categories = { 'player_name' : None, 'player_id ': None, 'season' : None }
L = ['Player 1', 'player_1', 2020]

j = 0 
for i in cat.keys():
    cat[i]  =  L[j]
    j += 1

这应该可以解决您的问题

Answer 3

如果列表中的项与字典具有键的顺序相同，即，如果player_name是列表中的第一个元素，则字典中的“ player_name”应排在首位

categories = {'player_name': None, 'player_id': None, 'season': None}
L = ['Player 1', 'player_1', '2020']

for key, value in zip(categories.keys(), L):
    categories[key] = value

Answer 4

您可以尝试这样的事情

categories = {'name':None, 'id':None, 'season':None}
L = ['Player 1', 'player_1', '2020']

it = iter(L)
for x in it:
    categories['name'] = x
    categories['id'] = next(it)
    categories['season'] = next(it)

如何为字典键分配列表值？

4 个答案: