Question

嘿伙计们，我对这个python问题有另一个问题。

years 1982 – 2048中包含复活节的计算公式如下：

让

a = year %19 
b = year %4 
c = year % 7
d = (19a + 24)%30
e= (2b + 4c + 6d + 5) % 7

复活节的日期是March 22 + d + e（可能是四月份）。编写一个输入一年的程序，验证它是否在适当的范围内，然后打印出当年复活节的日期。另外，编写一个函数easterii（）来执行此操作。此函数不接受任何参数并返回值。使用return语句。

这是我目前的代码。当我输入日期时，它总是输出"There was a problem, try again."

感谢任何意见，谢谢。

def easterCal(year):
    a = year % 19
    b = year % 4
    c = year % 7
    d = ((19 * a) + 24) % 30
    e = ((2 * b) + (4 * c) + (6 * d) + 5) % 7
    date = 22+d+e
    return date

def easterii():
    try:
        year = eval(input('Please enter a year: '))

        year = int(year)


        if year < 1982:
            print("Year is out of range")
        elif year > 2048:
            print("Year out of range")
        else:
            date = easterCal(year)

        if 22 <= date <= 31:
            print('Easter date for year {0} is March,{1}'.format(year , date))

        elif  32 <= date <= 56: 
            print('Easter date for year {0} is April,{1}'.format(year,date-31))

        else:
            print('Incorrect.')

    except NameError:
        print('Please enter date in numbers')
    except:
        print('There was a problem, try again.')

if __name__ == '__main__':
    easterii()

Answer 1

这一行肯定存在一个问题：

year = eval(input('Please enter a year: '))

您为什么使用eval？只需使用year = int(input('Please enter a year: '))即可。如果您正确输入年份，您的代码将起作用（我对其进行了测试）。

Answer 2

如果您允许异常传播，那么您将获得回溯：

# input year out of range, e.g. 1952
Traceback (most recent call last):
  File "...\test.py", line 32, in <module>
    main()
  File "...\test.py", line 23, in main
    if 22 <= date <= 31:
UnboundLocalError: local variable 'date' referenced before assignment

原因是，在您的代码中，如果年份不正确，您不能计算date，但仍会在if条款中引用它。

main的正确代码是：

def main():
    year = int(input('Please enter a year: '))

    if 1982 < year < 2048:
        date = easterCal(year)
        if 22 <= date <= 31:
            print('Easter date for year {0} is March,{1}'.format(year , date))
        elif  32 <= date <= 56: 
            print('Easter date for year {0} is April,{1}'.format(year, date-31))
    else:
        print("Year is out of range")

请注意我已移除try-catch阻止，如果需要，请随时重新添加。

Answer 3

问题是你必须为eval（）提供一个字符串，如代码引发的异常所述：

eval() arg 1 must be a string or code object

在这一行

year = eval(input('Please enter a year: '))

你可以str(eval(input()))

请注意，您可以通过执行

来打印例外

except Exception, e:
  prtint e

Answer 4

def dateEaster(year):
    if year >= 1900 and year <= 2099:
        a = year % 19
        b = year % 4
        c = year % 7
        d = (19 * a + 24) % 30
        e = (2 * b + 4 * c + 6 * d +5) % 7
        dateofeaster = 22 + d + e
        if year == 1954 or year == 1981 or year == 2049 or year == 2076:
            dateofeaster = dateofeaster - 7

        if dateofeaster > 31:
            dateofeaster = dateofeaster - 31
            print("April", dateofeaster)
        else:
            print("March", dateofeaster)
    else:
        print("There is an error")

year = int(input("Enter a year"))
dateEaster(year)

Answer 5

您的代码问题是eval。您的代码中很少有冗余，例如在分配之前引用了局部变量date。因此，明确将date声明为global，您可以将if和elif合并在一起。

def easterCal(year):
    a = year % 19
    b = year % 4
    c = year % 7
    d = ((19 * a) + 24) % 30
    e = ((2 * b) + (4 * c) + (6 * d) + 5) % 7
    date = 22 + d + e
    return date


def main():
    global date
    try:
        year = int(input('Please enter a year: '))
        if 1982 < year > 2048:
            print("Year is out of range")
        else:
            date = easterCal(year)

        if 22 <= date <= 31:
            print('Easter date for year {0} is March,{1}'.format(year, date))

        elif 32 <= date <= 56:
            print('Easter date for year {0} is April,{1}'.format(year, date - 31))

        else:
            print('Incorrect.')

    except NameError:
        print('Please enter date in numbers')
    except:
        print('There was a problem, try again.')


if __name__ == '__main__':
    main()

Python问题

5 个答案: