处理重复日期输入时触发EOFError问题求助

我的代码在zyBooks的常规输入下都能正常运行，但当输入4个完全相同的日期时（如下），会抛出EOFError：

2022-01-01
2022-01-01
2022-01-01
2022-01-01

错误栈信息如下：

Traceback (most recent call last):
  File "main.py", line 18, in <module>
    date = read_date()
  File "main.py", line 7, in read_date
    date_string = input()
EOFError: EOF when reading a line

原代码 #

from datetime import date, timedelta, time

# 1 Complete read_date()
from datetime import datetime 

def read_date():
    date_string = input()
    return datetime.strptime(date_string, '%Y-%m-%d')
    
"""Read a string representing a date in the format 2121-04-12, create a
date object from the input string, and return the date object"""

# 2. Use read_date() to read four (unique) date objects, putting the date objects in a list
date_list = []
for i in range(4):
    date = read_date()
    while date in date_list:
        date = read_date()
    date_list.append(date)

# 3. Use sorted() to sort the dates, earliest first
sorted_dates = sorted(date_list)

# 4. Output the sorted_dates in order, earliest first, in the format mm/dd/yy
for date in sorted_dates:
    print(date.strftime('%m/%d/%Y'))

# 5. Output the number of days between the last two dates in the sorted list
#    as a positive number
lastdate = sorted_dates[-1]
secondlastdate = sorted_dates[-2]
daysdifference = (lastdate - secondlastdate).days
print(daysdifference)

# 6. Output the date that is 3 weeks from the most recent date in the list
three_weeks = lastdate + timedelta(weeks = 3)
print(three_weeks.strftime('%B %d, %Y'))

# 7. Output the full name of the day of the week of the earliest day
earliest_day = sorted_dates[0]
print(earliest_day.strftime('%A'))

问题原因与修复方案 #

原因

代码逻辑要求读取4个唯一的日期，但当输入只有4个相同日期时，第一次读取的日期已经在date_list里，进入while循环后会继续调用read_date()尝试获取新输入，但此时已经没有更多输入了，所以触发EOFError。

修复方案

方案1：添加EOF异常处理

修改read_date()函数，捕获EOFError并给出提示后退出，避免程序崩溃：

def read_date():
    try:
        date_string = input()
        return datetime.strptime(date_string, '%Y-%m-%d')
    except EOFError:
        print("错误：输入不足，无法获取足够的唯一日期")
        exit(1)

方案2：调整输入逻辑，提前处理所有输入

如果是批处理输入场景，先一次性读取所有输入，再筛选唯一日期，从根源避免循环等待输入的问题：

from datetime import date, timedelta, datetime

def read_dates_from_input():
    import sys
    # 读取所有非空输入行
    lines = [line.strip() for line in sys.stdin if line.strip()]
    dates = []
    for line in lines:
        try:
            dates.append(datetime.strptime(line, '%Y-%m-%d'))
        except ValueError:
            print(f"无效日期格式：{line}")
            exit(1)
    return dates

# 获取所有输入日期并提取唯一值
all_dates = read_dates_from_input()
unique_dates = list(set(all_dates))

# 检查唯一日期数量是否满足要求
if len(unique_dates) < 4:
    print("错误：输入的唯一日期数量不足4个")
    exit(1)

# 取前4个唯一日期（若超过4个则截断）
date_list = unique_dates[:4]

# 后续逻辑保持不变
sorted_dates = sorted(date_list)

for date in sorted_dates:
    print(date.strftime('%m/%d/%Y'))

lastdate = sorted_dates[-1]
secondlastdate = sorted_dates[-2]
daysdifference = (lastdate - secondlastdate).days
print(daysdifference)

three_weeks = lastdate + timedelta(weeks = 3)
print(three_weeks.strftime('%B %d, %Y'))

earliest_day = sorted_dates[0]
print(earliest_day.strftime('%A'))

内容的提问来源于stack exchange，提问作者TillPickle