如何壓縮python代碼？

-1

所以我試圖讓這個搜索功能顯示名字輸入時人的星座。我用4個名字做了手動方式，我知道有一種方法可以用我已經（但不使用）的字典壓縮代碼，但我不記得了。如何壓縮python代碼？

Horoscopes = { 
    "A": "Scorpio", 
    "B": "Gemini", 
    "J": "Sagittarius", 
    "P": "Gemini", 
    } 

def horoscope(name): 
    if name == "A" or name == "a": 
     print ("Hello " + name + ", you are a Scorpio!") 
     print("Welcome to the Horoscope Search!") 
     name = input("What is your name? ") 
     horoscope(name) 
    elif name == "B" or name == "b": 
     print ("Hello " + name + ", you are a Gemini!") 
     print("Welcome to the Horoscope Search!") 
     name = input("What is your name? ") 
     horoscope(name) 
    elif name == "J" or name == "j": 
     print ("Hello " + name + ", you are a Sagittarius!") 
     print("Welcome to the Horoscope Search!") 
     name = input("What is your name? ") 
     horoscope(name) 
    elif name == "P" or name == "p": 
     print ("Hello " + name + ", you are a Gemini!") 
     print("Welcome to the Horoscope Search!") 
     name = input("What is your name? ") 
     horoscope(name) 
    else: 
     print ("Sorry " + name + ", you are not registered in our 
     system!") 
     print("Welcome to the Horoscope Search!") 
     name = input("What is your name? ") 
     horoscope(name) 

print("Welcome to the Horoscope Search!") 
name = input("What is your name? ") 
horoscope(name)

來源

2017-07-17 Jessica Tay

你能澄清你的問題，你的目標是什麼，你的問題是不完全清楚在什麼條件你正在尋找成果。 –

嗨，我正在尋找將整個功能塊壓縮成只有一個「如果」和「其他」。因此，如果名稱出現在詞典中，那麼if可能會是這樣的，那麼該名稱將與星座運算的值一起打印 –

請參閱@RacezeQ做了什麼。壓縮的最佳方法是在字典中查找名稱並返回與該名稱關聯的值。 –

你應該定義你的字典用鑰匙從小寫字母開始，這樣你就可以分析所有的答案，以降低字母和這樣比較吧：

Horoscopes = { 
    "a": "Scorpio", 
    "b": "Gemini", 
    "j": "Sagittarius", 
    "p": "Gemini", 
} 

def horoscope(name): 
    if name.lower() in Horoscopes: 
     print("Hello " + name + " you are a " + Horoscopes[name.lower()] + "!") 
    else: 
     print("Sorry " + name + ", you are not registered in our system!")

來源

2017-07-17 15:10:28 RaczeQ

如何編寫「如果」，以便如果名稱未出現在字典中，則代碼將運行其他 –

如果在星座中沒有name.lower（）： print（「Sorry」+ name +「，you 「） else： print（」Hello「+ name +」you is a「+ Horoscopes [name.lower（）] +」！「） – RaczeQ

這東西可實現像這樣：

horoscopes = { 
    "Angelina": "Scorpio", 
    "Bernice": "Gemini", 
    "Jessica": "Sagittarius", 
    "Peniel": "Gemini", 
    } 

print("Welcome to the Horoscope Search!") 
name = input("What is your name? ") 

if name in horoscopes: # If name is a key in horoscopes dict 
    print("Your Horoscope is {}!".format(horoscopes[name]))

請注意這是一個大小寫敏感的檢查星座中給定的名字，即我如果某個名字被輸入爲'angelina'，它將不會與字典鍵'Angelina'相匹配。考慮到這一點，如果字典鍵被稱爲是在下情況下，人們可能會使用字符串方法.lower()：

name = input("What is your name? ").lower()

這樣，無論怎麼輸入名稱，仍然會有匹配。

如果你希望用戶被提示，直到一個有效的名字被輸入，則：

horoscopes = { 
    "angelina": "Scorpio", 
    "bernice": "Gemini", 
    "jessica": "Sagittarius", 
    "peniel": "Gemini", 
    } 

print("Welcome to the Horoscope Search!") 

while True: # Until a name is matched to horoscopes 
    name = input("What is your name? ").lower() 

    if name in horoscopes: # If name is a key in horoscopes dict 
     print("Your Horoscope is {}!".format(horoscopes[name])) 
     break # A valid name has been entered, break from loop 
    else: 
     print("Please enter a valid name!")

來源

2017-07-17 15:28:01 flevinkelming

如何壓縮python代碼？

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