Question

我正在创建一个用户可以使用的财务建模工具。我希望用户能够通过键入变量名来输入他们想要使用的方法。但是，使用我的代码，raw_input只是给出了一个他们输入的字符串，而不是变量名称本身

loose = 5
accurate = 10
extreme = 15

method = raw_input('Would you like the calculation to be loose, accurate, or extreme in its precision?: ") #the user would either type loose, accurate, or extreme

calculation = x*y*method #x and y is just a generic calculation and method should be equal to either 5, 10, or 15 based on whether the user chose loose, accurate, or extreme

我收到错误，说方法不是int。我能做些什么来使它等于上面的松散，准确或极端值？

Answer 1

将变量更改为字典，然后根据用户的输入使用它来查找相应的整数：

values = {'loose': 5,
          'accurate': 10,
          'extreme': 15}

method = raw_input('Would you like the calculation to be loose, accurate, or extreme in its precision?: ') #the user would either type loose, accurate, or extreme

calculation = x*y*values[method] #x and y is just a generic calculation and method should be equal to either 5, 10, or 15

您还可以使用values.get(method, 0)代替values[method]，如果用户输入的内容不在您的词典中，则会使用0。

Answer 2

~~最好的~~一种方式是：

loose = 5
accurate = 10
extreme = 15

method_selection = raw_input('Would you like the calculation to be loose, accurate, or extreme in its precision?: ')

if method_selection == "loose":
    method = loose
elif method_selection == "accurate":
    method = accurate
else:
    method = extreme

calculation = x*y*method

调整if语句的顺序，为你提供正确的“默认”（即某人放入的方法不符合你的一个）。

Answer 3

直接解决方案是使用eval。

method = eval(raw_input(...))

请注意，建议不要使用eval，因为它可以让用户自由地执行他/她想要的任何内容。

使用raw_input获取用户输入变量名称并使用其值

3 个答案: