groups_per_user 函數接收一個字典，其中包含組名稱和用戶列表。用戶可以屬于多個組。填寫空白以返回一個字典，其中用戶作為鍵，其組列表作為值def groups_per_user(group_dictionary): user_groups = {} for key,values in group_dictionary.items(): for users in values: user_groups[users] = user_groups.get(users,0)+1 return(user_groups)print(groups_per_user({"local": ["admin", "userA"],"public": ["admin", "userB"],"administrator": ["admin"] }))# output must be {'admin':["local","public","administrator"],'userA':"local" ,'userB':"public"}#output I am getting {'admin': 3, 'userA': 1, 'userB': 1}
2 回答
慕森王
TA貢獻1777條經驗 獲得超3個贊
您可以迭代當前dict并反轉鍵值對：
d = {"local": ["admin", "userA"], "public": ["admin", "userB"], "administrator": ["admin"] }
res = {}
for k,v in d.items():
for e in v:
if e in res:
res[e].append(k)
else:
res[e] = [k]
print(res)
輸出：
{'admin': ['public', 'administrator', 'local'], 'userB': ['public'], 'userA': ['local']}
Qyouu
TA貢獻1786條經驗 獲得超11個贊
請嘗試下面的程序。它只是迭代 group_dictionary 的所有元素并構造 user_dictionary。
def groups_per_user(group_dictionary):
user_groups=dict([(a,[]) for a in sum(group_dictionary.values(),list())])
for u in user_groups:
for g in group_dictionary:
if u in group_dictionary[g]:
user_groups[u].append(g)
return(user_groups)
print(groups_per_user({"local": ["admin", "userA"],"public": ["admin", "userB"],"administrator": ["admin"] }))
輸出
{'admin': ['local', 'public', 'administrator'], 'userA': ['local'], 'userB': ['public']}
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