Python 3.8.6中如何从自定义Literal类型随机选取选项？

你定义了如下Literal类型别名：

from typing import Literal, get_args

_ACTIONS = Literal["pasteCharAt", "copyCharAt", "deleteCharAt", "replaceCharAt",
                   "selectAll", "deleteAll", "copyAll", "pasteAllAt", "click",
                   "moveLeft", "moveRight"]

尝试直接用random.choice(_ACTIONS)或者random.choice(list(_ACTIONS))时都触发了错误：

• 第一种写法报错：

File "C:\python38\lib\random.py", line 288, in choice
    i = self._randbelow(len(seq))
TypeError: object of type '_GenericAlias' has no len()

• 第二种写法报错：

File "C:\python38\lib\typing.py", line 261, in inner
    return func(*args, **kwds)
  File "C:\python38\lib\typing.py", line 685, in __getitem__
    params = tuple(_type_check(p, msg) for p in params)
  File "C:\python38\lib\typing.py", line 685, in <genexpr>
    params = tuple(_type_check(p, msg) for p in params)
  File "C:\python38\lib\typing.py", line 149, in _type_check
    raise TypeError(f"{msg} Got {arg!r:.100}.")
TypeError: Parameters to generic types must be types. Got 0.

错误原因 #

_ACTIONS是类型别名（属于_GenericAlias类型），不是可迭代的序列对象，所以random.choice无法直接处理它；而list(_ACTIONS)会错误触发类型的__getitem__方法，导致参数校验失败。

解决方案 #

你已经导入了get_args函数，它可以提取出Literal类型定义中的所有选项参数，返回一个元组。直接用这个元组作为random.choice的输入即可：

import random
from typing import Literal, get_args

_ACTIONS = Literal["pasteCharAt", "copyCharAt", "deleteCharAt", "replaceCharAt",
                   "selectAll", "deleteAll", "copyAll", "pasteAllAt", "click",
                   "moveLeft", "moveRight"]

# 获取Literal的所有选项
action_options = get_args(_ACTIONS)
# 随机选择一个值
random_action = random.choice(action_options)
print(f'random.choice(action_options) = {random_action}')

这个方法在Python 3.8+版本中都能正常工作，因为get_args在3.8及以上版本已经支持提取Literal的参数。

内容的提问来源于stack exchange，提问作者Gen Eva