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假设我有一个以以下五种方式之一格式化的日期列表：

Date_Type_1 = 2001 Apr 15
Date_Type_2 = 2001 Apr
Date_Type_3 = 2000 Spring
Date_Type_4 = 2000 Nov-Dec
Date_Type_5 = 2001

这样我的日期列表将如下所示。

Date_list = ["2001 Apr 15", "2004 May 15", "2011 Jan", "2011 Fall", "2000 Nov-Dec", "2012", "2000 Spring" ]

我现在想尝试格式化这些日期，以便将它们传递到 MySQL 数据库。我知道如何使用大量条件流的唯一方法。这是我将如何执行此操作的示例。我不会包括所有条件，因为它会占用太多空间。

for i in Date_list:
  year = i[:4]
  #This will be my conditional logic to define a month.
  #This is only an example.  Will not include all 12 months, 4 seasons, and various bi-months
  if "Apr" in i and "Mar-Apr" not in i:
     month = 4
  if "Mar-Apr" in i:
     month = 3
  if "May" in i and "May-Jun" not in i:
     month = 5
  if "Apr-May" in i:
     month = 4
  if "Spring" in i:
     month = 3
  #This will be conditional logic to define the day.
  #I would do this for each of the 31 days of the month.
  if "15" in i and "2015" not in i:
     day = 15

 date_return = datetime.datetime(year,month,day)
 date_format = date_return.date().isoformat

这个问题是我做了一些假设。我可以将季节定义为“春季/夏季..”并将双月刊（例如三月/四月）返回为特定月份。问题是，至少在定义天数的情况下，它不会捕获天数：

test_list = [2011 May, 2015 Apr 15]
for i in test_list:
  if "15" in i and "2015" not in i:
    day = 15

这不会赶上这一天。我想知道是否有更有效的方法来做到这一点？这种当前方法需要 50 多个条件语句来定义日/月。

4

2 回答 2

1

我认为你可以这样做：

>>> import datetime
>>> dates = ["2001 Apr 15", "2004 May 15", "2011 Jan", "2011 Fall", "2000 Nov-Dec", "2012", "2000 Spring" ]
>>>
>>> def convert(date_str):
...     tokens = date_str.split(' ')
...     if len(tokens) == 1:
...         date_time = datetime.datetime.strptime(date_str, '%Y')
...     elif len(tokens) == 3:
...         date_time = datetime.datetime.strptime(date_str, '%Y %b %d')
...     elif len(tokens) == 2 and '-' in tokens[1]:
...         date_str = date_str.split('-')[0]
...         date_time = datetime.datetime.strptime(date_str, '%Y %b')
...     else:
...         seasons = {
...             'spring': 'Mar',
...             'fall': 'Sep',
...         }
...         if tokens[1].lower() in seasons.keys():
...             date_str = '{} {}'.format(tokens[0], seasons[tokens[1].lower()])
...         date_time = datetime.datetime.strptime(date_str, '%Y %b')
...     return date_time.date().isoformat()
...
>>>
>>> for date_str in dates:
...     print '{} === {}'.format(date_str, convert(date_str))
...
2001 Apr 15 === 2001-04-15
2004 May 15 === 2004-05-15
2011 Jan === 2011-01-01
2011 Fall === 2011-09-01
2000 Nov-Dec === 2000-11-01
2012 === 2012-01-01
2000 Spring === 2000-03-01
于 2016-07-05T00:37:43.750 回答
1

您应该使用 Python 正则表达式模块re。这比试图弄乱 slice 和in.

import re

MONTHS = [ 'Jan', 'Feb', 'Mar', 'Apr', 'May', 'Jun', 'Jul', 
    'Aug', 'Sep', 'Oct', 'Nov', 'Dec']
SEASONS = { 'Spring': 'Mar', 'Summer': 'Jun', 'Fall': 'Sep', 
    'Autumn': 'Sep', 'Winter': 'Dec' }

_MONTHS_RE = '|'.join(MONTHS)
_SEASONS_RE = '|'.join(SEASONS)
DATE_RE = re.compile(r"""(?ax) # ASCII-only verbose mode
    (?P<year>20[0-9]{2}) # Year
    ( # followed by either...
        (?P<month>""" + _MONTHS_RE + r""") # a month name then...
        (   -(?P<endmonth>""" + _MONTHS_RE + r""") # a month range
        |   (?P<day>[1-9][0-9]?) # a day number
        )? # range and day are optional 
    |   (?P<season>""" + '|'.join(SEASONS) + r""") # or a season.
    )""")

def parse_date(datestr):
    m = DATE_RE.match(datestr)
    if m is None:
        return # Didn't match
    md = m.groupdict()
    year = int(md["year"])
    if "season" in md:
        month = SEASONS[md["season"]]
    else:
        month = md["month"]
        if "endmonth" in md:
            # handle month range here.
        day = int(md.get("month", 1))
    return year, month, day
    # Month is returned as a string; to get a number, use:
    return year, MONTHS.index(month) + 1, day

请注意，这并不能确保日期存在；它将接受"2099 Jun 50"或许多其他糟糕的日期。但我会将过滤作为练习留给读者。

于 2016-07-05T00:44:05.023 回答