Question

我正试图想出一种以最好的pythonic方式实现这一目标的方法。现在，我能想到的唯一方法是暴力破解它。

用户以下列方式之一输入日期（通过命令行）（例如./mypy.py date ='20110909.00 23'）

date='20110909'
date='20110909.00 23'
date='20110909.00 20110909.23'

所有三个示例应该具有相同的结果，如果它填充列表（我可以排序）无关紧要，例如

['20110909.00', '20110909.23]

甚至两个已排序的单独变量，但在所有情况下都是YYYYMMDD.HH，并且需要确保它确实是日期而不是文本。

有什么想法吗？

谢谢。

+++++ EDIT +++++ 在插上这个之后，我想我需要先做很多日期检查/操作。这一切似乎都很有效。除了最后，我通过日期验证运行列表，每次都失败 - 即使它应该通过。

（我发布它） ./test.py date ='20110909.00 23'

（或日期的任何变化 - 即日期= '20 22'或日期='20110909'或日期='20110909.00 23'等）

import sys, re, time, datetime

now = datetime.datetime.now()
tempdate=[]
strfirstdate=None
strtempdate=None

temparg2 = sys.argv
del temparg2[0]
tempdate = temparg2[0].replace('date=','')
date = tempdate.split(' ');

tempdate=[]
date.sort(key=len, reverse=True)
result = None

# If no date is passed then create list according to [YYMMDD.HH, YYMMDD.HH]
if date[0] == 'None':
    tempdate.extend([now.strftime('%Y%m%d.00'), now.strftime('%Y%m%d.%H')])


# If length of date list is 1 than see if it is YYMMDD only or HH only, and create list according to [YYMMDD.HH, YYMMDD.HH]
elif len(date) == 1:
    if len(date[0]) == 8:
        tempdate.extend([ date[0] + '.00', date[0] + '.23'])
    elif len(date[0]) == 2:
        tempdate.extend([now.strftime('%Y%m%d') + '.' + date[0], now.strftime('%Y%m%d') + '.' + date[0]])
    else:
        tempdate.extend([date[0], date[0]])


# iterate through list, see if value is YYMMDD only or HH only or YYYYMMDD.HH, and create list accoring to [YYYYMMDD.HH, YYYYMMDD.HH] - maximum of 2 values
else:
    for _ in range(2):
        if len(date[_]) == 8:
            strfirstdate = date[0]
            tempdate.append([ date[_] + '.00'])
        elif len(date[_]) == 2:
            if _ == 0:  # both values passed could be hours only
                tempdate.append(now.strftime('%Y%m%d') + '.' + date[_])
            else:  # we must be at the 2nd value passed.
                if strfirstdate == None:
                    tempdate.append(now.strftime('%Y%m%d') + '.' + date[_])
                else:
                    tempdate.append(strfirstdate + '.' + date [_])
        else:
            strfirstdate = date[0][:8]
            tempdate.append(date[_])

tempdate.sort()


for s in tempdate:
    try:
        result = datetime.datetime.strptime(s, '%Y%m%d.%H')
    except:
        pass

if result is None:
    print 'Malformed date.'
else:
    print 'Date is fine.'

print tempdate

++++编辑2 ++++ 如果我删除底部（在tempdate.sort（）之后）并用它替换它。

strfirstdate = re.compile(r'([0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9]+\.[0-9][0-9])')
for s in tempdate:
    if re.match(strfirstdate, s):
        result = "validated"
    else:
        print "#####################"
        print "#####################"
        print "##  error in date  ##"
        print "#####################"
        print "#####################"
        exit

它将适当验证。

这整个方法似乎并不是非常pythonic。

Answer 1

您可以使用try...except创建一个掩码并对其进行解析，以确定日期字符串是否与多个掩码中的一个匹配。我有一个项目的代码，所以我稍微修改了它：

from time import mktime, strptime
from datetime import datetime

date = '20110909.00 20110909.23'.split(' ')[0]
result = None

for format in ['%Y%m%d', '%Y%m%d.%H']:
  try:
    result = datetime.strptime(date, format)
  except:
    pass

if result is None:
  print 'Malformed date.'
else:
  print 'Date is fine.'

Answer 2

当我尝试在我自己的解析中使用try..except代码示例时，我发现了一些问题，所以这里是我添加了修复的版本，我还解决了仅处理小时部分的问题：

from datetime import datetime

dates = ['20110909.00','20110909.23','13','20111212','20113131']

def dateTest(date):
  dateOk = False
  for format in ['%Y%m%d', '%Y%m%d.%H', '%H']:
    try:
      result = datetime.strptime(date, format)
      dateOk = (date == result.strftime(format)) # this makes sure the parsed date matches the original string
      if format == '%H': # this handles the hour only case
        date = '%s.%s' % (datetime.now().strftime('%Y%m%d'), date)
    except:
      pass

  if dateOk:
    print 'Date is fine.'
  else:
    print 'Malformed date.'
  return date

for date in dates:
  print date
  print dateTest(date)
  print ''

Answer 3

查看time模块。具体来说，请参阅time.strptime()函数。

在时间值和日期时间对象之间也可以轻松转换。

Answer 4

这对你有帮助吗？：

from datetime import datetime
import re

reg = re.compile('(\d{4})(\d\d)(\d\d)'
                 '(?:\.(\d\d)(\d\d)?(\d\d)? *'
                 '(?:(\d{4})(\d\d)(\d\d)\.)?(\d\d)(\d\d)?(\d\d)? *)?')

for x in ('20110909',
          '20110909.00 23',
          '20110909.00 74',
          '20110909.00 20110909.23',
          '20110909.00 19980412.23',
          '20110909.08 20110909.23',
          '20110935.08 20110909.23',
          '20110909.08 19970609.51'):
    print x

    gr = reg.match(x).groups('000')

    try:
        x1 = datetime(*map(int,gr[0:6]))

        if gr[6]=='000':

            if gr[9]=='000':
                x2 = x1

            else:
                y = map(int,gr[0:3] + gr[9:12])
                try:
                    x2 = datetime(*y)
                except:
                    x2 = "The second part isn't in range(0,25)"

        else:
            y = map(int,gr[6:12])
            try:
                x2 = datetime(*y)
            except:
                x2 = "The second part doesn't represent a real date"
    except:
        x1 = "The first part dosen't represent a real date"
        x2 = '--'

    print [str(x1),str(x2)],'\n'

结果

20110909
['2011-09-09 00:00:00', '2011-09-09 00:00:00'] 

20110909.00 23
['2011-09-09 00:00:00', '2011-09-09 23:00:00'] 

20110909.00 74
['2011-09-09 00:00:00', "The hour in the second part isn't in range(0,25)"] 

20110909.00 20110909.23
['2011-09-09 00:00:00', '2011-09-09 23:00:00'] 

20110909.00 19980412.23
['2011-09-09 00:00:00', '1998-04-12 23:00:00'] 

20110909.08 20110909.23
['2011-09-09 08:00:00', '2011-09-09 23:00:00'] 

20110935.08 20110909.23
["The first part dosen't represent a real date", '--'] 

20110909.08 19970609.51
['2011-09-09 08:00:00', "The second part doesn't represent a real date"]

请注意，对于无

的每个组，groups('000')将无替换为'000'

Python，日期验证

4 个答案: