Python如何统计列表各元素出现次数，不重复输出元素及对应计数

代码错误原因 #

你当前代码没有输出是两个核心错误导致的：

• 打印语句放在了continue关键字之后，continue执行后会直接跳过本次循环剩余代码进入下一轮，你写的打印逻辑永远不会被触发
• 遍历finished_sandwiches列表的同时执行remove操作修改列表本身，会导致遍历索引错位，跳过部分元素；同时全局的temp计数变量没有在统计新品类时重置，计数逻辑完全错误

实现方案 #

方案1：使用标准库collections.Counter（推荐，代码最简洁） #

这是Python专门用来做频次统计的内置工具，完全符合你的需求：

from collections import Counter

sandwich_orders = ['chicken','egg','ham','beef','egg','fish','beef','beef']
finished_sandwiches = []

while sandwich_orders:
    in_between = sandwich_orders.pop()
    print("I made your " + in_between.title() + " sandwich.")
    finished_sandwiches.append(in_between)

print("The following sandwiches were made: ")
count_result = Counter(finished_sandwiches)
for sandwich_name, count in count_result.items():
    print(f"{sandwich_name.title()}({count})")

方案2：纯基础语法实现（无需导入额外库，适合巩固基础） #

如果你还没学到Counter，用基础的字典做统计也可以实现：

sandwich_orders = ['chicken','egg','ham','beef','egg','fish','beef','beef']
finished_sandwiches = []

while sandwich_orders:
    in_between = sandwich_orders.pop()
    print("I made your " + in_between.title() + " sandwich.")
    finished_sandwiches.append(in_between)

print("The following sandwiches were made: ")
count_dict = {}
# 先统计每个品类的数量
for sandwich in finished_sandwiches:
    if sandwich not in count_dict:
        count_dict[sandwich] = 0
    count_dict[sandwich] += 1
# 统一输出结果
for sandwich_name, count in count_dict.items():
    print(sandwich_name.title() + "(" + str(count) + ")")

内容的提问来源于stack exchange，提问作者davecodes_029