如何將列表值分配給字典鍵？

Question

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

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

如何遍歷列表並將其值分配給相應的鍵？ 所以它會變成這樣：

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

謝謝

Answer 1

如果 python >= 3.6，則使用zip() + dict() ，如果 < 3.6，看起來 dict 是無序的，所以我不知道。

測試.py ：

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

結果：

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

Answer 2

cat = { '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

這應該可以解決您的問題

Answer 3

如果列表中的項目與字典的鍵順序相同，即如果 player_name 是列表中的第一個元素，則字典中的 'player_name' 應該排在第一位

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

你可以試試這樣的

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)