我有以下代碼：
from collections import defaultdict
db1 = {'Adam': {'Cleaning': 4, 'Tutoring': 2, 'Baking': 1},
'Betty': {'Gardening': 2, 'Tutoring': 1, 'Cleaning': 3},
'Charles': {'Plumbing': 2, 'Cleaning': 5},
'Diane': {'Laundry': 2, 'Cleaning': 4, 'Gardening': 3}}
def by_skill(db1 : {str:{str:int}}) -> [int,[str,[str]]]:
order_skills = defaultdict(lambda:defaultdict(list))
for k,v in db1.items():
for key,value in v.items():
order_skills[value][key].append(k)
dict(order_skills)
order_skills_sorted = sorted( sorted(order_skills.items()), reverse=True )
return order_skills_sorted
if __name__ == '__main__':
print(by_skill(db1))
隨著輸出：
[(5, defaultdict(<class 'list'>, {'Cleaning': ['Charles']})), (4, defaultdict(<class 'list'>, {'Cleaning': ['Adam', 'Diane']})), (3, defaultdict(<class 'list'>, {'Cleaning': ['Betty'], 'Gardening': ['Diane']})), (2, defaultdict(<class 'list'>, {'Tutoring': ['Adam'], 'Gardening': ['Betty'], 'Plumbing': ['Charles'], 'Laundry': ['Diane']})), (1, defaultdict(<class 'list'>, {'Baking': ['Adam'], 'Tutoring': ['Betty']}))]
但我需要輸出（格式為可讀性，而不是要求）：（按字母順序）
[(5, [('Cleaning', ['Charles'])]),
(4, [('Cleaning', ['Adam', 'Diane'])]),
(3, [('Cleaning', ['Betty']), ('Gardening', ['Diane'])]),
(2, [('Gardening', ['Betty']), ('Laundry', ['Diane']),
('Plumbing', ['Charles']), ('Tutoring', ['Adam'])]),
(1, [('Baking', ['Adam']), ('Tutoring', ['Betty'])])]
我是否必須第三次呼叫 sorted 才能實作這一目標？
uj5u.com熱心網友回復：
由于評分范圍在 1 到 5 之間，您可以構建一個以技能為鍵的姓名串列字典的以評分為鍵的字典，然后遍歷評分以在線性時間內按技能提取名稱：
by_skill = {}
for name, skills in db1.items():
for skill, rating in skills.items():
by_skill.setdefault(rating, {}).setdefault(skill, []).append(name)
print([
(
rating,
sorted(
(skill, sorted(names)) for skill, names in by_skill.get(rating, {}).items()
)
)
for rating in range(5, 0, -1)
])
這輸出：
[(5, [('Cleaning', ['Charles'])]), (4, [('Cleaning', ['Adam', 'Diane'])]), (3, [('Cleaning', ['Betty']), ('Gardening', ['Diane'])]), (2, [('Gardening', ['Betty']), ('Laundry', ['Diane']), ('Plumbing', ['Charles']), ('Tutoring', ['Adam'])]), (1, [('Baking', ['Adam']), ('Tutoring', ['Betty'])])]
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