Python文本RPG游戏库存格式化显示：每行显示4个物品

问题描述 #

我是Python新手，正在制作一款文本RPG游戏的库存系统。现有代码中用for循环显示玩家库存时，每个物品会单独换行，希望修改为每行显示4个物品，且不想通过逐个索引打印物品来实现。

现有代码 #

class PlayerAttributes:
        inventory = []
        def __init__(self, name, inventory):
            self.name = name
            self.inventory = inventory # LIST
    class Item:
        def __init__(self, item_name, damage):
            self.item_name = item_name
            self.damage = damage

class Weapons(Item):
    weapon_1 = Item("Me Sword", 100)


Player_1 = PlayerAttributes("Bob", [])

def get_name():
    Player_1.name = input("Enter name here: ").capitalize()
    commands()

def stats():
    print("Name = " + str(Player_1.name), "\n",
          "Inventory: ")
    for x in Player_1.inventory:
        print(str(x.item_name))

def commands():
    prompt = None
    prompt_choices = {"stats", "quit", "give"}
    while prompt not in prompt_choices:
        prompt = input("Enter Command: ").lower()
    if prompt == "stats":
        stats()
        commands()
    elif prompt == "quit":
        quit()
    elif prompt == "give":
        Player_1.inventory.append(Weapons.weapon_1)
        commands()

get_name()

当前输出（执行3次give后输入stats） #

Enter name here: Test
Enter Command: give
Enter Command: give
Enter Command: give
Enter Command: stats
Name = Test 
 Inventory: 
Me Sword
Me Sword
Me Sword

期望输出（执行8次give后输入stats） #

Enter name here: Test
Enter Command: give
Enter Command: give
Enter Command: give
Enter Command: give
Enter Command: give
Enter Command: give
Enter Command: give
Enter Command: give
Enter Command: stats
Name = Test 
 Inventory: Me Sword, Me Sword, Me Sword, Me Sword
            Me Sword, Me Sword, Me Sword, Me Sword

解决方案 #

方法1：切片分组遍历 #

把库存列表按每4个一组拆分，每组用, 连接后打印，第二行及以后添加缩进对齐。修改后的stats函数如下：

def stats():
    print(f"Name = {Player_1.name}\n Inventory: ", end="")
    # 提取所有物品名称到列表
    item_names = [x.item_name for x in Player_1.inventory]
    # 按每4个元素为一组遍历
    for i in range(0, len(item_names), 4):
        group = item_names[i:i+4]
        # 第一行直接打印，后续行加缩进对齐
        if i == 0:
            print(", ".join(group))
        else:
            print("            " + ", ".join(group))

方法2：计数器控制换行 #

遍历物品时用计数器判断是否到达每组的末尾，达到时换行并添加缩进。修改后的stats函数如下：

def stats():
    print(f"Name = {Player_1.name}\n Inventory: ", end="")
    # 从1开始计数，方便判断每组第4个元素
    for idx, item in enumerate(Player_1.inventory, 1):
        # 非组尾元素后面加逗号空格，组尾元素不加
        print(item.item_name, end=", " if idx % 4 != 0 else "")
        # 每4个元素换行并添加缩进，最后一组末尾不额外换行
        if idx % 4 == 0 and idx != len(Player_1.inventory):
            print("\n            ", end="")
    print()  # 最后补一个换行收尾

方法3：迭代器批量取元素 #

借助itertools.islice每次从迭代器中取4个元素，适合处理较大的库存列表。需要先导入itertools，再修改stats函数：

import itertools

def stats():
    print(f"Name = {Player_1.name}\n Inventory: ", end="")
    inventory_iter = iter(Player_1.inventory)
    first_group = True
    while True:
        # 每次从迭代器中取4个元素
        group = list(itertools.islice(inventory_iter, 4))
        if not group:
            break
        group_names = [item.item_name for item in group]
        # 第一行直接打印，后续行加缩进
        if first_group:
            print(", ".join(group_names))
            first_group = False
        else:
            print("            " + ", ".join(group_names))

内容的提问来源于stack exchange，提问作者HardlySalty