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如何在不使用datetime模块的情况下计算给定年份的剩余天数?

python

beginner_geek07 · 技术社区 · 2 年前

我一直在努力计算一年中剩余的天数。

我已经定义了一个函数,用于确定一年是否为闰年,以及给定月份的天数。但是,我在定义最后一个函数时遇到了一个难题,没有使用datetime模块。

year = int(input("Enter the year to determine the number of days: "))
month = int(input("Enter the month of the year: "))
day = int(input("Enter the day of the year: "))

def if_leap_year(year):

    if (year % 400 == 0): return 366
    
    elif (year % 100 == 0): return 365
    
    elif (year % 4 == 0): return 366

    else:
        return 365

print(if_leap_year(year))

def days_in_month(month, year):
    if month in {1, 3, 5, 7, 8, 10, 12}:
        return 31
    if month == 2:
        if if_leap_year(year):
            return 29
        return 28
    return 30

print(days_in_month(month, year))


def days_left_in_year(month, day, year):
    month2 = 12
    day2 = 31
    day = day2 -365

我被困在最后一部分。。。任何帮助都将不胜感激。

编辑: 现在我有:

    def days_left_in_year(month, day, year):
        daysInMonth = [31,28,31,30,31,30,31,31,30,31,30,31]
        daysLeft = (if_leap_year(year) if month < 3 else 365) - sum(daysInMonth[:month - 1]) - day
        return daysLeft

我想使用这三个函数来创建类似以下内容的输出:

你什么时候取消了次贷服务?

月份:2

日期:29

年份:2021

抱歉,2021年2月29日似乎不是有效日期。要不要再试一次?

月份:2

日期:29

年份:2020年

我试着使用:

def refund_period():
    year = int(input("Enter the year to determine the number of days: "))
    month = int(input("Enter the month of the year: "))
    day = int(input("Enter the day of the year: "))
    if if_leap_year(year) == 365:
        money_owed = (days_left_in_year(month, day, year) / 365) * 278
    if if_leap_year(year) == 366:
        money_owed = (days_left_in_year(month, day, year) / 366) * 278

但是欠的钱没有打电话,所以现在我又被卡住了(

1 回复 | 直到 2 年前

constantstranger 2 年前

下面是一个经过最小修改的代码版本,可以满足您的要求:

year = int(input("Enter the year to determine the number of days: "))
#month = int(input("Enter the month of the year: "))
#day = int(input("Enter the day of the year: "))

def if_leap_year(year):

    if (year % 400 == 0): return 366
    
    elif (year % 100 == 0): return 365
    
    elif (year % 4 == 0): return 366

    else:
        return 365

#print(if_leap_year(year))

def days_in_month(month, year):
    if month in {1, 3, 5, 7, 8, 10, 12}:
        return 31
    if month == 2:
        if if_leap_year(year):
            return 29
        return 28
    return 30

#print(days_in_month(month, year))

def days_left_in_year(month, day, year):
    daysInMonth = [31,28,31,30,31,30,31,31,30,31,30,31]
    daysLeft = (if_leap_year(year) if month < 3 else 365) - sum(daysInMonth[:month - 1]) - day
    return daysLeft

print(days_left_in_year(1,1,2020))
print(days_left_in_year(3,1,2020))
print(days_left_in_year(1,1,2022))
print(days_left_in_year(3,1,2022))

输出:

更新: 我想你只是想回答问题的第二部分:

def refund_period():
    year = int(input("Enter the year to determine the number of days: "))
    month = int(input("Enter the month of the year: "))
    day = int(input("Enter the day of the year: "))
    if if_leap_year(year) == 365:
        money_owed = (days_left_in_year(month, day, year) / 365) * 278
    if if_leap_year(year) == 366:
        money_owed = (days_left_in_year(month, day, year) / 366) * 278
    return money_owed

print(refund_period())

你的职能 refund_period() 没有归还任何东西,所以我补充说 return money_owed ,然后我加了一行 print(refund_period()) .

下面是一个运行示例:

Enter the year to determine the number of days: 2020
Enter the month of the year: 02
Enter the day of the year: 29
232.42622950819674