根据选择的＃做出动作

Posted 2021-04-15

可以说我在底部有这个代码。如果我需要改变一些东西，它真的很烦人。编写此代码是否更容易？有阵列或idk的东西？我对Python很陌生，所以任何帮助都会受到赞赏。

    ti = randint(1,10)

    if ti == 1:
        something.action()

    if ti == 2:
        something2.action()

    if ti == 3:
        something3.action()

    if ti == 4:
        something4.action()

    if ti == 5:
        something5.action()

答案

使用字典将键映射到要运行的函数：

>>> def func1():
...     print(1)
... 
>>> def func2():
...     print(2)
... 
>>> mydict = {1: func1, 2: func2}
>>> 
>>> ti = 1
>>> 
>>> mydict.get(ti)()
1
>>> ti = 2
>>> mydict.get(ti)()
2
>>> 

或者使用你的例子：

mydict = {1: something.action, 2: something2.action}
ti = random.randint(1, 2)
mydict.get(ti)()

另一答案

您可以将函数映射到字典：

# the dictionary
# the keys are what you can anticipate your `ti` to equal
# the values are your actions (w/o the () since we don't want to call anything yet)
func_map = { 
    1: something.action,
    2: something2.action,
    3: something3.action
}
ti = randint(1, 10)

# get the function from the map
# we are using `get` to access the dict here,
# in case `ti`'s value is not represented (in which case, `func` will be None)
func = func_map.get(ti)

# now we can actually call the function w/ () (after we make sure it's not None - you could handle this case in the `else` block)
# huzzah!
if func is not None:
    func()

另一答案

您可以使用类实例列表：

import random
class Something:
   def __init__(self, val):
      self.val = val
   def action(self):
     return self.val

s = [Something(i) for i in range(10)]
print(s[random.randint(1,10)-1].action())

另一答案

这是一个switch statement，这是Python本身不支持的东西。

上面提到的字典解决方案的映射函数是实现switch语句的好方法。您还可以使用if / elif，我发现一次性实现更容易，更易读。

if case == 1:
    do something
elif case == 2:
    do something else
elif case == 3:
    do that other thing
else:
    raise an exception