在 Python 中编写 DRY 代码

Question

我真的是开发新手，我正在将 Codecademy 的 Battleship 游戏重写为 2 个玩家。我真的很挣扎，但到目前为止，我觉得这是一次很好的锻炼。我想保持良好的 OOP 和 DRY 原则。不过我有个问题。我试图通过用户输入来创建对象，并且我必须创建两个相似的定义来实现我的目标。我只写了一部分程序，我一直在尽可能地测试。要了解我的问题，请查看下面的 get_name1 和 get_name2：

 from random import randint


 class Person(object):

     def __init__(self, name, turn):
         self.name = name
         self.turn = turn


     def get_name1():
         while 1:
             name = input("What is the name of Player 1? ")
             if name.isalpha() == False:
                 print("\nPlease share your name with me.\n")
             else:
                 print("\nNice to meet you %s. It will be fun to play Battleship!\n" % name)
                 return name
                 break

     def get_name2():
         while 1:
             name = input("What is the name of Player 2? ")
             if name.isalpha() == False:
                 print("\nPlease share your name with me.\n")
             else:
                 print("\nNice to meet you %s. It will be fun to play Battleship!\n" % name)
                 return name
                 break



 Player1 = Person(Person.get_name1(), 1)
 Player2 = Person(Person.get_name2(), 2)

 print("Welcome to Battleship, %s!" % Player1.name)
 print("You will take turn %s.\n" % Player1.turn)
 print("Welcome to Battleship, %s!" % Player2.name)
 print("You will take turn %s.\n" % Player2.turn)

有没有一种方法可以将 get_name1 和 get_name2 合并到一个函数中，同时保留 "What is the name of Player 1?" 和 "What is the name of Player 2?" 的唯一输入行并仍然将唯一输入传递给两个不同的 class 对象？

Answer 1

def get_name(name_str):
         while 1:
             name = input("What is the name of %s? " % name_str)
             if name.isalpha() == False:
                 print("\nPlease share your name with me.\n")
             else:
                 print("\nNice to meet you %s. It will be fun to play Battleship!\n" % name)
                 return name
                 break

Player1 = Person(Person.get_name('Player 1'), 1)
Player2 = Person(Person.get_name('Player 2'), 2)

Answer 2

所以这两个函数唯一的区别就是播放器编号？然后将玩家编号作为参数传递给函数，并使用字符串格式进行提示。

def get_name(player_number):
    prompt = 'What is the name of Player {}? '.format(player_number)
    while True:
        name = input(prompt)
        # rest of code goes here..

通常，当您发现两个几乎相同的函数时，您会寻找不同之处，将其设为变量，然后将该变量作为参数传递。那么你只有一个函数，它被它的参数修改了。

Answer 3

最好，用一个classmethod一次性创建一个人名字：

class Person(object):
    def __init__(self, name, turn):
        self.name = name
        self.turn = turn

    @classmethod
    def create(cls, turn):
        while True:
            name = input("What is the name of Player %d? " % turn)
            if name.isalpha():
                break;
            print("\nPlease share your name with me.\n")
        print("\nNice to meet you %s. It will be fun to play Battleship!\n" % name)
        return cls(name, turn)

Player1 = Person.create(1)
Player2 = Person.create(2)

print("Welcome to Battleship, %s!" % Player1.name)
print("You will take turn %s.\n" % Player1.turn)

在 Python 中编写 DRY 代码

Writing DRY Code in Python

python

dry

python-3.x