如何避免 If 语句的墙壁？

Question

我多次听说这样的编码是不理想的：if weapon == "sword": print("Knight") elif weapon == "katana": print("Samurai") elif weapon == "axe": print("Viking")如何以最佳方式编写此类代码？

Answer 1

您可以将这些关联存储在字典中

weapons_roles = {

    "sword": "Knight",

    "katana": "Samurai",

    "axe": "Viking"

}

打印一些东西，只要钥匙不在字典中

print(weapons_roles.get(weapon, "No role"))

仅当武器已知时才打印角色

if weapon in weapons_roles:

    print(weapons_roles[weapon])

Answer 2

试试下面这个：

def example_function(weapon):

    weapon_dict = {'sword': 'Knight', 'katana': 'Samurai', 'axe': 'Viking'}

    return weapon_dict[weapon]