处理parsedatetime的不同结果

Question

我花了大约15年左右的时间在Perl工作并且只是偶尔学习python.

我无法理解如何处理来自parsedatetime的Calendar.parse()的解析方法的两种不同类型的结果

鉴于此脚本:

#!/usr/bin/python

import parsedatetime.parsedatetime as pdt
import parsedatetime.parsedatetime_consts as pdc
import sys
import os

# create an instance of Constants class so we can override some of the defaults

c = pdc.Constants()

# create an instance of the Calendar class and pass in our Constants # object instead of letting it create a default

p = pdt.Calendar(c)

while True:
 reply = raw_input('Enter text:')
 if reply == 'stop': 
  break
 else:
  result = p.parse(reply)
  print result
  print

这个示例运行:

输入文字:明天
(time.struct_time(tm_year = 2009,tm_mon = 11,tm_mday = 28,tm_hour = 9,tm_min = 0,tm_sec = 0,tm_wday = 5,tm_yday = 332,tm_isdst = -1),1)

输入文字:11/28
((2009,11,28,14,42,55,4,331,0 ),1)

我无法弄清楚如何获得输出,以便我可以如此使用结果:

print result[0].tm_mon, result[0].tm_mday

这在输入为"11/28"的情况下不起作用,因为输出只是一个元组而不是struct_time.

可能是一件简单的事......但不适合这个新手.从我的角度来看,Calendar.parse()的输出是不可预测的并且难以使用.任何帮助赞赏.蒂亚.

Answer 1

我知道这是一个老问题,但我昨天遇到了这个问题,这里的答案是不完整的(在parse()返回日期时间的情况下会失败).

从parsedatetime文档:

parse()返回一个元组(结果,类型),其中type指定以下之一:

   0 = not parsed at all
   1 = parsed as a date (of type struct_time)
   2 = parsed as a time (of type struct_time)
   3 = parsed as a datetime (of type datetime.datetime)

这有点奇怪,也许不是最清楚的方法,但它起作用并且非常有用.

这里有一小段代码可以将它返回的内容转换为适当的python日期时间:

import parsedatetime.parsedatetime as pdt

def datetimeFromString( s ):

    c = pdt.Calendar()
    result, what = c.parse( s )

    dt = None

    # what was returned (see http://code-bear.com/code/parsedatetime/docs/)
    # 0 = failed to parse
    # 1 = date (with current time, as a struct_time)
    # 2 = time (with current date, as a struct_time)
    # 3 = datetime
    if what in (1,2):
        # result is struct_time
        dt = datetime.datetime( *result[:6] )
    elif what == 3:
        # result is a datetime
        dt = result

    if dt is None:
        # Failed to parse
        raise ValueError, ("Don't understand date '"+s+"'")

    return dt