Question

我一直试图找到一种方法来完成标题中的内容，而不使用Python中任何导入的日历/日期时间库。顶部几乎没有功能来检查年份是否是闰年，我希望能够在打印给定二月份的天数时参考，但是我不太清楚如何这样做。（我已经猜到了输出.bla bla）

到目前为止，我已经想出了类似的东西，这应该说清楚我想要做什么，但我仍然对Python有点新意，所以我想要一些提示/帮助来修复我的代码为了这项任务。

# 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，似乎每个月输出正确的数据，但是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

要解决2月份输出的任何提示吗？

编辑：只需要在引用第一个函数时添加参数年份。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

Answer 1

代码中出现一些语法错误：

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。
python中没有elseif，它应该是elif
output.is_leap_year = True，应为is_leap_year(year) == True。 False部分也应该更改。

在if语句和else之后应该有:，例如

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'

Answer 2

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。

尝试一下：

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

Answer 3

Rectangle {
    id: calendarPopUp
    visible: false;
    border.width: 0
    color: "#ffffff"
    anchors.left: parent.left
    anchors.top: parent.top
    width: parent.width
    z: 10;
    VMComboBox{
        id: yearPicker
        width: parent.width
        //vmModel: ["1999", "2000"]
        vmModel: {
            var ans = [];
            var cyear = 2018;
            var fyear = cyear - 120;
            for (var i = cyear; i >= fyear; i--){
                ans.push(i)
            }
            calendar.minimumDate = new Date(fyear, 0, 1);
            calendar.maximumDate = new Date(cyear, 0, 1);
            return ans;
        }
        font.family: viewHome.robotoR.name
        anchors.top: parent.top
        anchors.topMargin: 2
        anchors.horizontalCenter: parent.horizontalCenter
        onCurrentIndexChanged: {
            var year = yearPicker.currentText;
            console.log("Changing visible year to: " + year);
            calendar.visibleYear = parseInt(year);
        }
    }
    Calendar {
        id: calendar
        width: parent.width
        anchors.top: yearPicker.bottom
        anchors.horizontalCenter: parent.horizontalCenter
    }
}

Answer 4

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')

Answer 5

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")

打印给定月份和年份的天数[Python]

5 个答案: