Question

我是Python和编程的完全初学者。我为 Spotify's Best Before puzzle 制作了一个程序。它被接受了。我在互联网上看了一眼，看了解问题的其他解决方案，我见过的每个人都有几个模块，包括日历模块。我知道这可能是一个很好的解决方案，但我想把自己作为一种练习。

我非常感谢所有提示和提示，但主要是不必导入代码。主要是printer(a)和det dataMaker()需要修改。

normYear = [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]
leapYear = [31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]
answerList = []

u''' Check if any of the integers are years '''

def yearCheck():
    for x in xrange(0, 3):
        a = dataList[x]
        if len(a) > 2:
            if not len(a) == 4 and int(a) in xrange(2000,3000):
                if int(a) in xrange(100,1000):
                    dataList[x] = int(a) + 2000
                else:
                    print data + u" is illegal"

u''' Make integers and sort '''

def integer():
    for x in xrange(0, 3):
        dataList[x] = int(dataList[x])
    dataList.sort()

u''' Check for possible leap years '''

def leapYears():
    global leapList
    leapList = []
    for x in xrange(0, 3):
        if dataList[x] % 4 == 0:
            if dataList[x] % 100 == 0:
                if dataList[x] % 400 == 0:
                   leapList.append(x)
            else:
                leapList.append(x)

u''' Changes year type '''

def defYear(a):
    global xYear
    if a in leapList:
        xYear = leapYear
    else:
        xYear = normYear

u''' Printer '''

def printer(a):
    if dataList[a] < 2000:
        dataList[a] += 2000
    year = dataList[a]
    del dataList[a]
    if not dataList[0] == 0:
        month = dataList.pop(0)
        day = dataList.pop(0)
        answerList.append(unicode(year))
        answerList.append(unicode(u'%02d' % month))
        answerList.append(unicode(u'%02d' % day))
        print u'-'.join(answerList)
    else:
        print data + u" is illegal"

u''' Looks for legal dates, first [Y<M<D] then [M<Y<D] then [M,D,Y] '''

def dateMaker():
    for x in xrange(0,4):
        defYear(x)
        if x == 0:
            if dataList[1] <= 12 and dataList[2] <= xYear[dataList[1]-1]:
                printer(x)
                break
        elif x == 1:
            if dataList[0] <= 12 and dataList[2] <= xYear[dataList[0]-1]:
                printer(x)
                break
        elif x == 2:
            if dataList[0] <= 12 and dataList[1] <= xYear[dataList[0]-1]:
                printer(x)
                break
        else:
            print data + u" is illegal"

u''' Program '''

data = raw_input()
dataList = data.split(u"/")
yearCheck()
integer()
leapYears()
dateMaker()

Answer 1

我会考虑尝试使用类重新实现它。即使作为练习，这也许是一种很好的做法。

您还应该考虑：

利用字符串，列表等的可迭代特性。您似乎正在考虑循环，就像人们期望在C中看到的那样。
试图摆脱嵌套if语句。遵循逻辑是很困难的。（见：http://eflorenzano.com/blog/2012/01/01/reducing-code-nesting/）

我会考虑按照以下方式实现您的前两个功能（可能还有其他方法可以改进它）：

data = '8/5/32'
data_list = data.split('/')

def yearCheck(data_list):
    # Years may be truncated to two digits and may in that case
    # also omit the leading 0 (if there is one), so 2000 could 
    # be given as "2000", "00" or "0" (but not as an empty string).
    # Further examples:
    #   if 2099, could be given as 99
    #   if 2005, could be given as 05 or 5
    #   199 will not happen i.e. doesn't say that years may be 
    #       truncated to three digits
    for index, item in enumerate(data_list):
        if len(item) > 4:
            # e.g. 30000
            print item, '- Data is invalid'
            return
        if len(item) == 4 and int(item) not in xrange(2000, 3000):
            # e.g. 3015
            print item, '- Data is invalid'
            return        
        if len(item) == 3:
            # e.g. 199
            print item, '- Data is invalid'   
            return
        if len(item) < 3 and int(item) in xrange(32, 100):
            data_list[index] = int(item) + 2000
    return data_list

def integer(data_list):
    int_data_list = [int(item) for item in data_list]
    return int_data_list.sort()

yearCheck(data_list)
integer(data_list)
print data_list

简化Python程序

1 个答案: