Python中的枚举中的抽象方法

Question

我的问题是，如何在Python中写这篇文章？这样的事情吗？

它应该如何工作： 我正在从算法中获取数据，该算法将决定输出哪个字母。如果数据中不应适用于一个字符的某些条件，则应检查其他字符的条件。数据和条件当然比此处显示的要复杂。

为什么要枚举： 因为只有这种小主要方法必须写在算法文件中（具有算法）。并将字母的条件封装在另一个文件中并明确结构。

enum Letter {
    
    A () {
        public boolean condition(int[] args) {
            if (args[0] > args[1]) return false;
            if (args[1] > args[2]) return false;
            return true;
        }
    },
    
    B () {
        public boolean condition(int[] args) {
            if (args[0] > args[1]) return false;
            if (args[1] < args[2]) return false;
            return true;
        }
    },
    
    C () {
        public boolean condition(int[] args) {
            if (args[0] < args[1]) return false;
            if (args[1] < args[2]) return false;
            return true;
        }
    };
    
    public abstract boolean condition(int[] args);
    
}

public class Alphabet {
    
    public static void main(String[] args) {
        int[] arr = {1, 2, 3};
        //int[] arr = {1, 2, 1};
        //int[] arr = {3, 2, 1};
        
        for (Letter l : Letter.values()) {
            if (l.condition(arr)) {
                System.out.println(l);
                break;
            }
        }
    }
}

Answer 1

While you could shoehorn an Enum into serving this purpose, it'd be much easier to just use a dict or list of functions that you can iterate over, eg:

letters = [
    ("A", lambda args: args[0] <= args[1] <= args[2]),
    ("B", lambda args: args[0] <= args[1] >= args[2]),
    ("C", lambda args: args[0] >= args[1] >= args[2]),
]

def alphabet(args):
    print(next(letter for letter, condition in letters if condition(args)))

alphabet([1, 2, 3])  # A
alphabet([1, 2, 1])  # B
alphabet([3, 2, 1])  # C

Or with type annotations (these are optional in Python, but they allow you to do static type analysis/testing similar to what you get in a compiled language, and can make tricky code easier to understand):

from typing import Callable, List, Tuple

Condition = Callable[[List[int]], bool]

letters: List[Tuple[str, Condition]] = [
    ("A", lambda args: args[0] <= args[1] <= args[2]),
    ("B", lambda args: args[0] <= args[1] >= args[2]),
    ("C", lambda args: args[0] >= args[1] >= args[2]),
]

def alphabet(args: List[int]) -> None:
    print(next(letter for letter, condition in letters if condition(args)))

Note that in the type-annotated version, we can just define Condition作为“获取INT列表并返回bool的函数”的方便类型别名。我们实际上不需要实际组建一个新类或定义新的抽象接口，因为功能是Python中的一流对象，它们的接口已经完全拟合了此用例。