设置函数调用python中的raw_input

Question

我该如何完成这项工作？我正在尝试设置一个全局函数，以便稍后可以将其称为原始输入，而不是在需要时调用原始输入。

我认为我的精髓在于我需要的程度，但我很难理解如何格式化，或者甚至可能。

预先感谢您。

def choice_1(): 
     choice_1 == raw_input("> ") 
    def choice_1a():  
     choice_1a = raw_input("> ") 
    def choice_1b():  
     choice_1b = raw_input("> ") 

编辑：我不认为我是在我的问题的目的很清楚。这里是我正在处理的代码的更新，也许这将清除一切。

print "You've arrived at your desk" 


def choice_1(one): 
    choice_1a = raw_input("< ") 
def choice_1a():  
    choice_1a = raw_input("> ") 
def choice_1b():  
    choice_1b = raw_input("> ") 

#Choice_1 
print "What do you want to do?" 
print "We can \n1. Read\n2. Draw\n3. Work on homework" 
print choice_1 
#choice 1 branch 1 
if choice_1 == "1": 
    print "What book should we read today?" 
    print "We can read\n1. Tom Sawyer\n2. Quantum Physics \n3. Ray Bradbury" 
    print choice_1a 
    if choice_1a == "1": 
     print "Great choice!" 
    if choice_1a == "2": 
     print "Heavy stuff there." 
    if choice_1a == "3": 
     print "Entertaining author, that one there!" 
    else: 
     print "Let's go to the library, maybe they'll have that one." 
#choice 1 branch 2 
if choice_1 == "2": 
    print "What would you like to draw?" 
    print "We can draw a\n1. Tiger\n2. Fish\n3. Bear " 
    print choice_1b 
    if choice_1b == "1": 
     print "You drew a Tiger!" 
    if choice_1b == "2": 
     print "You drew a Fish!" 
    if choice_1b == "3": 
     print "You drew a Bear!" 
    else: 
     print "Time for some improvisation." 

#choice 1 branch 3 
if choice_1 == "3": 
    print "" 

这是否清楚了一些混淆？

Answer 1

这会做你想做的。作为@SethMMorton注意到，您曾经在return

def choice(): 
    return raw_input('> ‘) 

顺便说错过了，这不是一个很好的事情，因为，给别人读你的代码，它不会立即清楚什么choice是这样做。

我这样做：

print 'Your name is :' + choice() 

和它的工作如预期。

编辑：你应该让你如果这样的语句：

if choice() == "1"：

Answer 2

def multiplier(x, y): 
    ans = x * y 
    print ans 

这将是一个功能的使用，当准备使用它，你会 这样称呼它

multiplier(5, 10) 
     50