如何压缩python代码？

-1

所以我试图让这个搜索功能显示名字输入时人的星座。我用4个名字做了手动方式，我知道有一种方法可以用我已经（但不使用）的字典压缩代码，但我不记得了。如何压缩python代码？

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)

来源

2017-07-17 Jessica Tay

你能澄清你的问题，你的目标是什么，你的问题是不完全清楚在什么条件你正在寻找成果。 –

嗨，我正在寻找将整个功能块压缩成只有一个“如果”和“其他”。因此，如果名称出现在词典中，那么if可能会是这样的，那么该名称将与星座运算的值一起打印 –

请参阅@RacezeQ做了什么。压缩的最佳方法是在字典中查找名称并返回与该名称关联的值。 –

你应该定义你的字典用钥匙从小写字母开始，这样你就可以分析所有的答案，以降低字母和这样比较吧：

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!")

来源

2017-07-17 15:10:28 RaczeQ

如何编写“如果”，以便如果名称未出现在字典中，则代码将运行其他 –

如果在星座中没有name.lower（）： print（“Sorry”+ name +“，you “） else： print（”Hello“+ name +”you is a“+ Horoscopes [name.lower（）] +”！“） – RaczeQ

这东西可实现像这样：

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]))

请注意这是一个大小写敏感的检查星座中给定的名字，即我如果某个名字被输入为'angelina'，它将不会与字典键'Angelina'相匹配。考虑到这一点，如果字典键被称为是在下情况下，人们可能会使用字符串方法.lower()：

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

这样，无论怎么输入名称，仍然会有匹配。

如果你希望用户被提示，直到一个有效的名字被输入，则：

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!")

来源

2017-07-17 15:28:01 flevinkelming

如何压缩python代码？

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