仅当变量==1时执行Python定义

我编写了几个定义，我希望用户选择要运行的定义，但是定义总是打印我如何打开和删除定义？ 如果用户选择2，则不应打印。我错过了什么

Senerio = 0
while Senerio not in (1, 2, 3, 4):
    Senerio = int(input("choose a commission/Aggregation senerio number \n"
                        "1. put in car \n"                     
                        "2. put in plane \n"
                        "3. put on boat \n"
                        "4. put on fire "))

if Senerio == 1:
Car = 1
Plane = 0
boat = 0
fire = 0

def incar():
    print('Im putting this in a car')
incar()

if car == 1:
    incar

这应该可以解决问题

Senerio = 0
while Senerio not in (1, 2, 3, 4):
    Senerio = int(input("choose a commission/Aggregation senerio number \n"
                        "1. put in car \n"                     
                        "2. put in plane \n"
                        "3. put on boat \n"
                        "4. put on fire "))

if Senerio == 1:
    car = 1
    plane = 0
    boat = 0
    fire = 0

def incar():
    print('Im putting this in a car')

if car == 1:
    incar()

---

我发现，将if语句包装在定义周围可以满足我的需求 像这样：

if car == 1:
    def incar():
        print('Im putting this in a car')
    incar()
    incar

---

发布的此代码不会打印任何内容。它只是抛出了一个语法错误。我明白了，我在这里键入的内容中漏掉了：但是如果Senerio=1它仍然印着“我正在把这个放进车里”