訪問字典中的元組

Question

我有這本字典，它存儲了兩對測驗分數和參與者的ID。該結構是{（quiz1，quiz2）：ID}

scoredict = {('83', '93'): '81937', ('88', '86'): '33576', ('96', '97'): '01084', 
('81', '95'): '48534', ('84', '72'): '11235', ('77', '80'): '01835', ('90', '83'): 
'39488', ('75', '74'): '31049', ('80', '62'): '10188', ('85', '86'): '63011', 
('66', '89'): '58272'} 

我想使該程序通過進入一對測驗分數來查找的ID。例如，如果用戶輸入測驗1和測驗2的83和93，它將返回81937.自從最近48小時以來，我一直在處理這個問題，但是沒有一個代碼能夠工作...

是否有可能找到兩個測驗最接近的可用分數並打印ID？

Answer 1

我已經驗證了您的解決方案可與ipython：

In [1]: scoredict = {('83', '93'): '81937', ('88', '86'): '33576', ('96', '97'): '01084', 
    ...: ('81', '95'): '48534', ('84', '72'): '11235', ('77', '80'): '01835', ('90', '83'): 
    ...: '39488', ('75', '74'): '31049', ('80', '62'): '10188', ('85', '86'): '63011', 
    ...: ('66', '89'): '58272'} 

In [2]: scoredict['83','93'] 
Out[2]: '81937' 

Answer 2

簡單地做：

>>> scoredict[(score1,score2)] 

Answer 3

對於最接近分數，你可以試試這個：

test = (83, 93) 

deviation = float('inf') 
best_match = None 

for score1, score2 in scoredict: 
    error = abs(int(score1) - test[0]) + abs(int(score2) - test[1]) 

    if error < deviation: 
    deviation = error 
    best_match = (score1, score2) 

print scoredict[best_match]