Question

我在从函数中使用另一个函数的输出时遇到问题。我不知道python中函数的语法。当我定义它时，如何在函数中使用另一个函数的输出。 def hero_attribute（hero_selection（））：＃此语法未被接受

#This program will calculate the damge of hero with stats
global hero_str_result
global hero_agi_result
global hero_int_result

def hero_selection():
    print """1. Life Stealer (strength hero)\n
    2. Phantom lancer (agility hero)\n
    3. Phantom Assassin (agility hero)\n
    4. Wrait King (strength hero) \n
    """

    print "Please enter hero selection: "
    hero_num = int(raw_input("> "))
    return hero_num

def hero_attribute(hero_selection()): #This syntax isn't accepted 
    if hero_num == 1: # Life stealer 
        hero_str = 25 
        hero_agi = 18 
        hero_int = 15 
        #Hero growth stats
        str_growth = 2.4 
        agi_growth = 1.9 
        int_growth = 1.75 

    elif hero_num == 2: # Phantom lancer 
        hero_str = 
        hero_agi = ?
        hero_int = ?
        #Hero growth stats
        str_growth = 2.4
        agi_growth = 1.9
        int_growth = 1.75

    elif hero_num == 3: # Phantom Assassin
        hero_str = ?
        hero_agi = ?
        hero_int = ?
        #Hero growth stats
    else: #Wraith King
        hero_str = ?
        hero_agi = ?
        hero_int = ?    
        #hero growth stats
        str_growth = ?
        agi_growth = ?
        int_growth = ?  
    return (hero_str,hero_agi,hero_int,str_growth,agi_growth,int_growth)

def hero_type(hero_num):
    if hero_num == 1:
        hero_type = "str"
    elif hero_num == 2
        hero_type = "agi"
    elif hero_num == 3
        hero_type = "agi"   
    else:
        hero_type = "str"


#the function will ask user what to do with the hero
def hero_build():
    print "What do you want to do with the hero?"
    print """1. Build hero with stat
        2. Build hero with item (not yet)
        3. Build hero with level
        """

    user_choice = int(raw_input("> "))
    if user_choice == 1:
        print "You want to build hero with stats!"
        print "Please enter number of stats that you want to add: "
        hero_stats = int(raw_input=("> "))

        hero_str, hero_agi, hero_int,str_growth,agi_growth,int_growth = hero_attribute() #This function will take the result of hero_str, hero_agi,hero_int
        hero_str_result = hero_str + str_growth * hero_stats
        hero_agi_result = hero_agi + agi_growth * hero_stats
        hero_int_result = hero_int + int_growth * hero_stats    


    return hero_str_result, hero_agi_result, hero_int_result


print "This is the result of your build: ", hero_build()

Answer 1

函数是一段接收参数的代码，并为这些参数指定一个名称。例如：

def square(x):
    return x * x

此函数计算数字的平方;函数体中的这个未知数字将被称为x。

一旦你有了一个功能，你可以调用它，传递你想要的值作为参数...例如

print( square(12) )

将打印144，因为它会调用square传递x=12和12*12 144的函数。

你当然可以传递一个函数调用另一个函数的结果，例如

def three_times(x):
    return 3 * x

print( square( three_times(5) ) )

将显示225，因为函数three_times将被传递5并且它将返回3*5，函数square将被传递15并将返回15*15 }。

在功能定义（def部分）中，参数总是只有名称。您要传递给该功能的内容将写在致电网站。

Answer 2

你想要的是能够传递函数作为参数。然而，这已经从设计中融入到python中：你只需在传递任何其他参数时传递它。

示例：

def apply(f,x):
   return f(x)
def sq(x):
   return x*x
def cb(x):
   return x*x*x
apply(sq,2)
4
apply(cb,2)
8

Apply的定义是第一个参数是一个函数。只有在您真正阅读apply的定义时才知道。在那里你看到f被视为一个函数，而不是被视为“数字”的x。

python中的嵌套函数（从另一个函数获取输出）

2 个答案: