串列、元組
與字典處理

Lists, Tuples & Dictionaries

用撲克牌建出完整的牌組和遊戲！

四種容器，各有用途

Python 提供不同的「容器」來存放多筆資料，就像生活中不同的收納工具：

清單 → list：一副牌

hand = ["A", "7", "K"]

# 一副牌的 13 個牌面
ranks = ["A", "2", "3", "4", "5", "6", "7",
         "8", "9", "10", "J", "Q", "K"]

print(ranks[0])       # A（索引從 0 開始）
print(ranks[-1])      # K（-1 是最後一個）
print(ranks[0:3])     # ['A', '2', '3']（切片）
print(len(ranks))     # 13（長度）

試試看：用索引取出指定的牌

填入正確的索引，讓每個 print 印出註解說的那張牌。填完按「執行看看」！

ranks = ["A", "2", "3", "4", "5", "6", "7", "8", "9", "10", "J", "Q", "K"]

print(ranks[])      # 印出第一張：A
print(ranks[])     # 印出最後一張：K
print(ranks[])      # 印出第五張：5

試試看：用切片取出多張牌

切片 [a:b] 會取出索引 a 到 b-1 的元素。試著填入正確的切片！

nums = [10, 20, 30, 40, 50]

print(nums[])     # [20, 30]
print(nums[])      # [10, 20]  從頭取 2 個
print(nums[])      # [40, 50]  從索引 3 到底

手牌操作：新增、移除、排序

hand.append("K")
hand.pop(0)
len(hand)

hand = ["7", "A", "3"]

hand.append("K")         # 加到末尾 → ['7','A','3','K']
hand.insert(0, "2")      # 插入到最前面
hand.remove("3")         # 移除指定的牌
popped = hand.pop()      # 移除並回傳最後一張
hand.sort()              # 排序
hand.reverse()           # 反轉順序
print("A" in hand)       # True/False — 手上有沒有 A？

試試看：用方法操作手牌

選對方法，觀察每一步手牌怎麼變化。故意填錯也可以執行看看差別！

hand = ["7", "A", "3"]

hand.("K")
print("加入 K：", hand)

hand.()
print("排序後：", hand)

last = hand.()
print("抽出：", last)
print("剩下：", hand)

list 語法速查表

用 for 走訪手牌

hand = ["A♠", "7♥", "K♦", "3♣", "10♠"]

# 方法 1：直接走訪每張牌
for card in hand:
    print(card)

# 方法 2：用 enumerate 同時取得索引和值
for i, card in enumerate(hand):
    print(f"第 {i+1} 張：{card}")

# 方法 3：用 range + 索引
for i in range(len(hand)):
    print(f"hand[{i}] = {hand[i]}")

試試看：用 for 走訪 list

完成迴圈，讓它印出每張牌和編號。

hand = ["A♠", "7♥", "K♦", "3♣", "10♠"]

 i, card  (hand):
    print(f"第 {i+1} 張：{card}")

List Comprehension 串列推導式

一行就能產生或過濾 list：

# 傳統寫法：產生一副牌
suits = ["♠", "♥", "♦", "♣"]
ranks = ["A","2","3","4","5","6","7","8","9","10","J","Q","K"]
deck = []
for s in suits:
    for r in ranks:
        deck.append(r + s)

# List Comprehension：一行搞定！
deck = [r + s for s in suits for r in ranks]
print(len(deck))    # 52
print(deck[:5])     # ['A♠', '2♠', '3♠', '4♠', '5♠']

# 過濾：只要黑桃
spades = [card for card in deck if "♠" in card]
print(spades)       # ['A♠', '2♠', ..., 'K♠']

洗牌與發牌

import random

def create_deck():
    """建立一副 52 張的撲克牌"""
    suits = ["♠", "♥", "♦", "♣"]
    ranks = ["A","2","3","4","5","6","7","8","9","10","J","Q","K"]
    return [r + s for s in suits for r in ranks]

def deal(deck, n=5):
    """從牌堆抽 n 張牌（會從牌堆移除）"""
    hand = []
    for i in range(n):
        card = deck.pop()    # 從牌堆頂抽一張
        hand.append(card)
    return hand

# --- 使用 ---
deck = create_deck()
random.shuffle(deck)        # 洗牌（就地打亂）
print(f"牌堆共 {len(deck)} 張")

my_hand = deal(deck, 5)
print(f"你的手牌：{my_hand}")
print(f"牌堆剩 {len(deck)} 張")

Tuple — 不可修改的組合

一張牌可以用 (花色, 牌面) 的 tuple 表示：

# tuple 用小括號
card = ("♠", "A")
print(card[0])     # ♠
print(card[1])     # A

# tuple 不能修改！
# card[0] = "♥"   # ❌ TypeError!

# 用 tuple 建牌組
deck = [("♠","A"), ("♥","K"), ("♦","7"), ("♣","3")]
for suit, rank in deck:
    print(f"{rank}{suit}")

試試看：tuple 解包

tuple 可以一次把值拆給多個變數，叫做「解包」。完成下面的程式！

deck = [("♠", "A"), ("♥", "K"), ("♦", "7"), ("♣", "3")]

for ,  in deck:
    print(f"{}{}")

Dict — 牌面對應點數

用「鍵」查找「值」，就像查字典：

# 每張牌面對應幾點
card_values = {
    "A": 14, "K": 13, "Q": 12, "J": 11,
    "10": 10, "9": 9, "8": 8, "7": 7,
    "6": 6, "5": 5, "4": 4, "3": 3, "2": 2
}

print(card_values["A"])      # 14
print(card_values["7"])      # 7
print(card_values.get("Joker", 0))  # 0（找不到時的預設值）

字典常用操作：玩家記分板

scores = {"小明": 0, "小華": 0}

# 新增 / 修改
scores["小美"] = 0          # 新增玩家
scores["小明"] = 10         # 小明加分

# 刪除
del scores["小美"]

# 遍歷
for player, score in scores.items():
    print(f"{player}：{score} 分")

# 常用方法
print(scores.keys())     # dict_keys(['小明', '小華'])
print(scores.values())   # dict_values([10, 0])
print("小明" in scores)  # True — 檢查玩家是否存在

試試看：dict 的三種遍歷方式

dict 有 keys()、values()、items() 三種方法，輸出完全不同。填對後執行看看！

scores = {"小明": 100, "小華": 85, "小美": 90}

print("--- 只印名字 ---")
for name in scores.():
    print(name)

print("--- 只印分數 ---")
for val in scores.():
    print(val)

print("--- 名字 + 分數 ---")
for name, score in scores.():
    print(f"{name}：{score} 分")

Set — 去重與集合運算

# set 自動去除重複
hand = ["A", "7", "A", "K", "7"]
unique = set(hand)
print(unique)       # {'A', '7', 'K'}

# 用途 1：檢查手牌有幾種不同的牌
print(f"有 {len(set(hand))} 種不同的牌")

# 用途 2：找出兩個玩家的共同牌
p1 = {"A", "K", "Q", "7"}
p2 = {"K", "3", "7", "J"}
print(p1 & p2)     # {'K', '7'}  交集 — 共同的牌
print(p1 | p2)     # 聯集 — 所有的牌
print(p1 - p2)     # {'A', 'Q'} 差集 — p1 有但 p2 沒有

試試看：set 集合運算

& 交集、| 聯集、- 差集。填入正確的運算符號，執行看看結果！

p1 = {"A", "K", "Q", "7"}
p2 = {"K", "3", "7", "J"}

common = p1  p2
print("共同的牌：", common)

all_cards = p1  p2
print("全部的牌：", all_cards)

only_p1 = p1  p2
print("只有 P1 有：", only_p1)

四種資料結構比較

試試看：選對容器

根據需求選出最適合的容器類型。

# 手牌可以加牌、出牌 → 用可修改的有序容器
hand = "A", "7", "K"

# 一張牌的花色和牌面固定不變 → 用不可修改的容器
card = "♠", "A"

# 牌面對應點數，用名字查值 → 用鍵值對
values = "A": 14, "K": 13

print(hand, card, values)

巢狀結構：玩家資訊

list 裡面放 dict，組合出複雜的資料：

players = [
    {"name": "小明", "hand": ["A♠", "K♥", "7♦"], "score": 0},
    {"name": "小華", "hand": ["Q♠", "J♣", "10♥"], "score": 0},
]

# 取得小明的手牌
print(players[0]["hand"])     # ['A♠', 'K♥', '7♦']

# 印出所有玩家的資訊
for p in players:
    print(f"{p['name']}：{p['hand']}，{p['score']} 分")

# 找出分數最高的玩家
best = max(players, key=lambda p: p["score"])
print(f"目前領先：{best['name']}")

試試看：從巢狀結構取值

list 裡面放 dict，要用兩層取值。先 [索引] 再 [鍵]。

players = [
    {"name": "小明", "hand": ["A♠", "K♥"], "score": 10},
    {"name": "小華", "hand": ["Q♠", "J♣"], "score": 5},
]

# 印出小明的名字
print(players[][])

# 印出小華的手牌
print(players[][])

動手做：撲克牌比大小（完整版）

import random

def create_deck():
    suits = ["♠", "♥", "♦", "♣"]
    ranks = ["A","2","3","4","5","6","7","8","9","10","J","Q","K"]
    return [r + s for s in suits for r in ranks]

values = {"A":14,"K":13,"Q":12,"J":11,
          "10":10,"9":9,"8":8,"7":7,"6":6,"5":5,"4":4,"3":3,"2":2}

def get_value(card):
    return values[card[:-1]]   # 去掉花色取牌面

deck = create_deck()
random.shuffle(deck)
p_score, c_score = 0, 0

print("撲克牌比大小！共 5 回合\n")
for r in range(1, 6):
    p, c = deck.pop(), deck.pop()
    pv, cv = get_value(p), get_value(c)
    print(f"第 {r} 回合")
    print(f"你：{p}({pv}點)  vs  電腦：{c}({cv}點)")
    if pv > cv:
        p_score += 1
        print("→ 你贏！\n")
    elif pv < cv:
        c_score += 1
        print("→ 電腦贏！\n")
    else:
        print("→ 平手！\n")

print(f"最終比分 — 你 {p_score} : {c_score} 電腦")

第五章完成！

學會了 list、tuple、dict、set 四種資料結構，
還用它們組出了一個完整的撲克牌遊戲！