1

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 2 月的天数时能够参考，但我不太确定该怎么做。（我猜想有类似输出的东西。bla bla）

到目前为止，我已经想出了类似的东西，这应该清楚我想要做什么，但我对 Python 还是有点陌生​​，所以我希望有一些提示/帮助来修复我的代码以完成任务.

# 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 2 月：

# 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

有什么提示可以修复二月份的输出吗？

编辑：只需要在引用第一个函数时添加参数 year 。以下是 100% 工作代码供将来参考：

# 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

​</p>

​</p>

​</p>

4

5 回答 5

3

更 Pythonic 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
于 2013-08-20T02:18:58.410 回答
2

您的代码中有一些语法错误：

  1. 前面应该没有空格def days_in_month(month,year)。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里面没有elseif，应该是elif
  3. output.is_leap_year = True，应该是is_leap_year(year) == True。False零件也应该改变。

  4. 在if声明之后，else应该有一个:, like

    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'
于 2013-08-20T01:55:35.890 回答
0 
"""
Takes the year and month as input and returns the no. of days
"""
def is_leap_year (year):
    return (year % 4 == 0) and (year % 100 != 0) or (year % 400 == 0)

def days_in_month(month, year):
    if month == 'September' or month == 'April' or month == 'June' or month == 'November':
        result=30
    elif month == 'January' or month == 'March' or month == 'May' or month== 'July' or month == 'August' or month == 'October'or month== 'December':
        result=31

    elif month == 'February' and output.is_leap_year ==True:
        result=29

    elif month == 'February' and output.is_leap_year == False:
        result=28
    return result

print(is_leap_year(2016))
print(days_in_month('September',2016))
于 2018-08-12T09:51:53.733 回答
0 
month = int (input ('month (1-12): '))

if month < 13:
    if month == 2:
        year = int (input ('year: '))
        if year % 4 == 0:
            if year % 100 == 0:
                if year % 400 == 0:
                    print ('29')
                else:
                    print ('28')
            else:
                print ('29')
        else:
            print ('28')

    elif month >= 8:
        if month % 2 == 0:
            print ('31')
        else:
            print ('30')

    elif month % 2 == 0:
        print ('30')
    else:
        print ('31')

else:
    print ('Only 1-12 accepted')
于 2020-05-27T07:02:32.597 回答
0 
month=input("month")
year=int(input("year"))
if year%4==0:
        year=('leap 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 year == "leap year":
        print ("29")

elif month == 'February' and  year != "leap year":
        print ("28")

else:
    print("none")
于 2020-07-28T09:09:00.020 回答