我編寫了一個(gè)函數(shù)，其中包含一個(gè)縮寫的星期幾到該天的全名的字典。當(dāng)我輸入縮寫時(shí)，我得到了正確的輸出日，但是為了嘗試另一個(gè)縮寫，我必須重新輸入該函數(shù)。我有：def weekday() day = input('Enter day abbreviation ' ) days = {'Mo':'Monday','Tu':'Tuesday', 'we':'Wednesday', 'Th':'Thursday', 'Fr':'Friday', 'Sa':'Saturday','Su':Sunday'} while day in days: print(days.get(day))我的問題是，它一遍又一遍地打印全日名稱，而我希望它打印全日名稱，然后再次打印“輸入日縮寫”。它看起來應(yīng)該像這樣：>>>weekday():Enter day abbreviation: TuTuesdayEnter day abbreviation: SuSundayEnter day abbreviation:...相反，我得到：>>>weekday():Enter day abbreviation: TuTuesdayTuesdayTuesdayTuesdayTuesday... # it continues without stopping我知道這是一個(gè)非常簡(jiǎn)單的解決方案，但我無法弄清楚。
3 回答
蕪湖不蕪
TA貢獻(xiàn)1796條經(jīng)驗(yàn) 獲得超7個(gè)贊
您永遠(yuǎn)不會(huì)重讀“ day”，因此“ day of day”始終是正確的，并且會(huì)無休止地執(zhí)行。
def weekday()
day = input('Enter day abbreviation ' )
days = {'Mo':'Monday','Tu':'Tuesday',
'we':'Wednesday', 'Th':'Thursday',
'Fr':'Friday', 'Sa':'Saturday','Su':'Sunday'}
while day in days:
print(days.get(day))
day = input('Enter day abbreviation ' )
絕地?zé)o雙
TA貢獻(xiàn)1946條經(jīng)驗(yàn) 獲得超4個(gè)贊
days = {'Mo':'Monday','Tu':'Tuesday',
'we':'Wednesday', 'Th':'Thursday',
'Fr':'Friday', 'Sa':'Saturday','Su':'Sunday'}
while True:
day = input('Enter day abbreviation ' )
if day in days:
print (days[day])
else:
break
輸出：
$ python3 so.py
Enter day abbreviation Mo
Monday
Enter day abbreviation Tu
Tuesday
Enter day abbreviation we
Wednesday
Enter day abbreviation foo
另一種使用方式dict.get：
days = {'Mo':'Monday','Tu':'Tuesday',
'we':'Wednesday', 'Th':'Thursday',
'Fr':'Friday', 'Sa':'Saturday','Su':'Sunday'}
obj = object() #returns a unique object
day = input('Enter day abbreviation ' )
while days.get(day,obj) != obj:
print (days[day])
day = input('Enter day abbreviation ' )
青春有我
TA貢獻(xiàn)1784條經(jīng)驗(yàn) 獲得超8個(gè)贊
您想在每次迭代中再次獲取輸入：
while True:
day = input('Enter day abbreviation ' )
acquired_day = days.get(day)
if acquired_day is None: break
print(acquired_day)
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