Python 定义变量和执行变量

此代码有一些问题。当我输入2尝试执行

whaleq（）

时，它反而执行

mathschallenge（）

，我只希望在用户输入1时执行

有人能帮我看看吗？下面的代码与python idle中的代码完全相同

print('The first thing you should think about doing is asking the locals if they have seen     anything!')

locallist = ['Ann-Marie, the butcher (1)',
             'Bella, the flourist (2)',
             'Caitlyn, the painter (3)',
             'Daniel, the scientist (4)',
             'Lauren, the Zookeeper (5)']
print(locallist)

localpick = int(input('Type the number of the local you want to talk to '))

def localx():
    if localpick == 1:
        mathschallenge()
    if localpick == 2:
        whaleq()

def mathschallenge():
    print('maths challenge yolo')
    quest = input('What is 10+10?')
    if quest == '20':
        print('correct')
    else:
        print('incorrect')

def whaleq():
    mammal = input('What is the largest mammal?')
    if mammal == 'whale':
        print('correct')
    else:
        print('incorrect')

input('Press the enter key to exit')

---

代码对我来说很好，除了重新组织一点之外，我感动了：

print('The first thing you should think about doing is asking the locals if they have seen     anything!')

locallist = ['Ann-Marie, the butcher (1)', 'Bella, the flourist (2)', 'Caitlyn, the 
painter         (3)', 'Daniel, the scientist (4)', 'Lauren, the Zookeeper (5)']
print(locallist)

localpick = int(input('Type the number of the local you want to talk to '))

从代码的顶部到底部，并在上面添加

localx（）

：

input('Press the enter key to exit')

整个代码如下所示：

def localx():
    if localpick == 1:
        mathschallenge()
    if localpick == 2:
        whaleq()

def mathschallenge():
    print('maths challenge yolo')
    quest = input('What is 10+10?')
    if quest == '20':
        print('correct')
    else:
        print('incorrect')

def whaleq():
    mammal = input('What is the largest mammal?')
    if mammal == 'whale':
        print('correct')
    else:
        print('incorrect')


print('The first thing you should think about doing is asking the locals if they have seen     anything!')

locallist = ['Ann-Marie, the butcher (1)', 'Bella, the flourist (2)', 'Caitlyn, the painter     (3)', 'Daniel, the scientist (4)', 'Lauren, the Zookeeper (5)']
print(locallist)

localpick = int(input('Type the number of the local you want to talk to '))
localx()
input('Press the enter key to exit')

我希望这有帮助， ~z~博比

---

注意：这是在python 3.4.2中测试的，在从用户获取'localpick'值之后，您是否正在调用'localx'函数。我希望这不是您的确切代码。它没有正确缩进。将localx更改为localpick，并且没有做任何不同的操作。@RvdK我必须粘贴代码，然后在这里用4个空格缩进。因此，localpick可能是保存用户输入的变量。localx是一个函数，一旦你得到了你的用户选项，它就需要被调用，而这个选项是缺失的。所以，在从用户处获取输入后，您需要调用函数localx。感谢@bobbeh和alfasin的帮助