Python pop() - List 取出元素

pop() 方法用來移除並回傳指定位置的元素。

基本用法

fruits = ["apple", "banana", "orange"]
removed = fruits.pop()
print(removed)  # orange
print(fruits)   # ['apple', 'banana']

語法

list.pop(index=-1)

• index：要移除的位置索引，預設為 -1（最後一個）
• 回傳：被移除的元素
• 例外：空 list 或索引超出範圍時拋出 IndexError

移除指定位置的元素

fruits = ["apple", "banana", "orange", "grape"]

# 移除最後一個（預設）
last = fruits.pop()
print(last)     # grape
print(fruits)   # ['apple', 'banana', 'orange']

# 移除第一個
first = fruits.pop(0)
print(first)    # apple
print(fruits)   # ['banana', 'orange']

# 移除指定位置
middle = fruits.pop(0)
print(middle)   # banana
print(fruits)   # ['orange']

負數索引

numbers = [1, 2, 3, 4, 5]

print(numbers.pop(-1))  # 5（最後一個）
print(numbers)          # [1, 2, 3, 4]

print(numbers.pop(-2))  # 3（倒數第二個）
print(numbers)          # [1, 2, 4]

索引超出範圍

numbers = [1, 2, 3]

# 會拋出 IndexError
# numbers.pop(10)  # IndexError: pop index out of range

# 空 list 也會報錯
empty = []
# empty.pop()  # IndexError: pop from empty list

# 安全的做法
if numbers:
    item = numbers.pop()

pop() vs remove()

fruits = ["apple", "banana", "orange"]

# pop() - 依索引移除，回傳被移除的元素
removed = fruits.pop(1)
print(removed)  # banana

# remove() - 依值移除，不回傳
fruits = ["apple", "banana", "orange"]
fruits.remove("banana")  # 回傳 None

實際範例

實作堆疊（Stack）

stack = []

# Push
stack.append("first")
stack.append("second")
stack.append("third")

# Pop（後進先出）
print(stack.pop())  # third
print(stack.pop())  # second
print(stack.pop())  # first

實作佇列（Queue）

queue = []

# Enqueue
queue.append("first")
queue.append("second")
queue.append("third")

# Dequeue（先進先出）
print(queue.pop(0))  # first
print(queue.pop(0))  # second
print(queue.pop(0))  # third

處理待辦事項

tasks = ["task1", "task2", "task3"]

while tasks:
    current = tasks.pop(0)
    print(f"Processing: {current}")
    # 處理任務...

撤銷操作

actions = []
undo_stack = []

def do_action(action):
    actions.append(action)
    undo_stack.append(action)
    print(f"Did: {action}")

def undo():
    if undo_stack:
        action = undo_stack.pop()
        print(f"Undid: {action}")
    else:
        print("Nothing to undo")

do_action("Type A")
do_action("Type B")
do_action("Type C")
undo()  # Undid: Type C
undo()  # Undid: Type B

輪流處理

players = ["Alice", "Bob", "Charlie"]

# 輪流取出並放回末尾
for _ in range(6):
    current = players.pop(0)
    print(f"{current}'s turn")
    players.append(current)

分離奇偶數

numbers = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
evens = []

# 倒序處理避免索引問題
for i in range(len(numbers) - 1, -1, -1):
    if numbers[i] % 2 == 0:
        evens.append(numbers.pop(i))

evens.reverse()
print(f"Odds: {numbers}")   # Odds: [1, 3, 5, 7, 9]
print(f"Evens: {evens}")    # Evens: [2, 4, 6, 8, 10]

時間複雜度

• pop()（末尾）：O(1)
• pop(0)（開頭）：O(n)
• pop(i)（中間）：O(n)

如果需要頻繁從開頭移除元素，使用 collections.deque：

from collections import deque

d = deque([1, 2, 3, 4, 5])
d.popleft()  # O(1)
print(list(d))  # [2, 3, 4, 5]