打印给定月份和年份的天数 \[Python\]

Question

我一直在尝试找出一种方法来完成标题中的内容，而不使用任何从 Python 导入的日历/日期时间库。顶部几乎没有用于检查年份是否为闰年的功能，我希望在打印给定二月的天数时能够参考这一点，但我不太确定该怎么做。 （我猜到了类似 output.bla bla 的东西）

到目前为止，我已经想出了这样的事情，这应该清楚我想做什么，但我对 Python 还是有点陌生​​，所以我会喜欢一些提示/帮助来修复我的任务代码.

 # A function to determine if a year is a leap year.
# Do not change this function.
def is_leap_year (year):
    return (year % 4 == 0) and (year % 100 != 0) or (year % 400 == 0)

# You should complete the definition of this function:

 def days_in_month(month, year):

    if month == 'September' or month == 'April' or month == 'June' or month == 'November'
    print 30

    elseif month == 'January' or month == 'March' or month == 'May' or month== 'July' or month == 'August' or month == 'October'\
    or month== 'December'
     print 31

    elseif month == 'February' and output.is_leap_year = True
    print 29

    elseif month == 'February' and output.is_leap_year = False
    print 28

    else print 'Blank'

好的，我已经修复了我的代码，它似乎每个月都输出正确的数据，但二月除外：

 # A function to determine if a year is a leap year.
# Do not change this function.
def is_leap_year (year):
    return (year % 4 == 0) and (year % 100 != 0) or (year % 400 == 0)

# You should complete the definition of this function:

def days_in_month(month, year):

    if month in ['September', 'April', 'June', 'November']:
        print 30

    elif month in ['January', 'March', 'May', 'July', 'August','October','December']:
        print 31

    elif month == 'February' and is_leap_year == True:
        print 29

    elif month == 'February' and is_leap_year == False:
        print 28

有什么提示可以解决 2 月的输出问题吗？

编辑：只需要在引用第一个函数时添加参数年份。这是供将来参考的 100% 工作代码：

 # A function to determine if a year is a leap year.
# Do not change this function.
def is_leap_year(year):
    return (year % 4 == 0) and (year % 100 != 0) or (year % 400 == 0)

# You should complete the definition of this function:

def days_in_month(month, year):

    if month in ['September', 'April', 'June', 'November']:
        print 30

    elif month in ['January', 'March', 'May', 'July', 'August','October','December']:
        print 31

    elif month == 'February' and is_leap_year(year) == True:
        print 29

    elif month == 'February' and is_leap_year(year) == False:
        print 28

    else:
        return None

的

的

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原文由 James Adams 发布，翻译遵循 CC BY-SA 4.0 许可协议

Answer 1

您的代码中存在一些语法错误：

1. def days_in_month(month,year) 之前不应有空格。 Python 使用缩进来分隔代码块。这是您在评论中给出的错误。
2. python中没有 elseif ，应该是 elif
3. output.is_leap_year = True ，应该是 is_leap_year(year) == True 。 False 部分也应该改变。
4. 在 if 声明和 else 之后应该有一个 : ，就像

   if month == 'September' or month == 'April' or month == 'June' or month == 'November':
       print 30
   elif month == 'January' or month == 'March' or month == 'May' or month== 'July' or month == 'August' or month == 'October' or month== 'December':
       print 31
   elif month == 'February' and is_leap_year(year) == True:
       print 29
   elif month == 'February' and is_leap_year(year) == False:
       print 28
   else:
       print 'Blank'

原文由 David.Zheng 发布，翻译遵循 CC BY-SA 3.0 许可协议

Answer 2

一种更像 Python 的方法是在字典中定义映射，然后简单地从字典中检索值。

试试看：

 days_in_month_dict = {"January": 31, "February": 28,
                      "March": 31, "April": 30,
                      "May": 31, "June": 30,
                      "July": 31, "August": 31,
                      "September": 30, "October": 31,
                      "November": 30, "December": 31}

def is_leap_year(year):
    return (year % 4 == 0) and (year % 100 != 0) or (year % 400 == 0)

def days_in_month(year, month):
    if is_leap_year(year) and month == "February":
        return 28

    try:
        #attempt to get value from dictionary
        return days_in_month_dict[month]
    except KeyError:
        #key does not exist, so we caught the error
        return None

原文由 Madison May 发布，翻译遵循 CC BY-SA 3.0 许可协议