如何关联拆分后的弹珠ID与时间列表，找出最快弹珠及计算时间统计值

核心思路是将拆分后的弹珠编号与时间重新绑定，有两种简洁的实现方式：

方法一：利用列表索引关联（基于现有代码修改） #

借助两个列表的索引对应关系，找到最短时间对应的弹珠ID：

def minute_converter(x):
    total_sec = int(x) if isinstance(x, float) else int(x)
    sec = total_sec % 60
    minutes = total_sec // 60
    print(f"{minutes} mins {sec} secs.")


def average(x):
    avg = [int(n) for n in x]
    return sum(avg) / len(avg)


print('Marble Racer Program')

data = ['331;;591', '010;;902', '809;;030', '756;;201', '128;;382']
marble_number = [x.split(';;', 1)[0] for x in data]
marble_times = [x.split(';;', 1)[1] for x in data]
print(marble_number)
print(marble_times)

# 找到最短/最长时间对应的弹珠ID
min_time = min(marble_times)
best_marble_id = marble_number[marble_times.index(min_time)]

max_time = max(marble_times)
worst_marble_id = marble_number[marble_times.index(max_time)]

# 输出结果
print(f'Marble with best time is {best_marble_id}: ', end='')
minute_converter(min_time)
print(f'Average time: ', end='')
minute_converter(average(marble_times))
print(f'Marble with worst time is {worst_marble_id}: ', end='')
minute_converter(max_time)
print(f'We have seen a total of {len(marble_number)} marbles.')

输出：

Marble Racer Program
['331', '010', '809', '756', '128']
['591', '902', '030', '201', '382']
Marble with best time is 809: 0 mins 30 secs.
Average time: 7 mins 1 secs.
Marble with worst time is 010: 15 mins 2 secs.
We have seen a total of 5 marbles.

方法二：整合为元组列表（更推荐） #

直接将弹珠ID与转换为整数的时间配对，避免字符串比较的潜在错误（比如时间'99'和'100'，字符串比较会误判'100'更小）：

def minute_converter(total_sec):
    sec = int(total_sec) % 60
    minutes = int(total_sec) // 60
    print(f"{minutes} mins {sec} secs.")


def average(times):
    return sum(times) / len(times)


print('Marble Racer Program')

data = ['331;;591', '010;;902', '809;;030', '756;;201', '128;;382']
# 转换为(弹珠ID, 时间秒数)的元组列表
marble_data = [(item.split(';;')[0], int(item.split(';;')[1])) for item in data]
print([item[0] for item in marble_data])
print([str(item[1]) for item in marble_data])

# 找出最快/最慢弹珠
best_marble = min(marble_data, key=lambda x: x[1])
worst_marble = max(marble_data, key=lambda x: x[1])
avg_time = average([item[1] for item in marble_data])

# 输出结果
print(f'Marble with best time is {best_marble[0]}: ', end='')
minute_converter(best_marble[1])
print(f'Average time: ', end='')
minute_converter(avg_time)
print(f'Marble with worst time is {worst_marble[0]}: ', end='')
minute_converter(worst_marble[1])
print(f'We have seen a total of {len(marble_data)} marbles.')

这种方式的优势是数据结构更清晰，直接绑定ID与时间，无需维护两个独立列表，同时避免了字符串时间比较的风险。

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