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Python不规范的日期字符串处理类

Jun 06, 2016 am 11:31 AM
python

我分析了形如19920203、199203、1992.02.03、1992.02、1992-02-03、1992-02、920203时间格式特征,列出了正则表达式如下:

代码如下:


^((?:19|20)?\d{2})[-.]?((?:[0-1]?|1)[0-9])[-.]?((?:[0-3]?|[1-3])[0-9])?$


当然这个表达式还不是很完善,只能做简单的切割,不能判断日期的合法性,关于日期是否合法,我还是交给Python的时间功能来处理吧。

根据上面的正则表达式,我写的DateParser类如下:

代码如下:


import re
import datetime

# ***************************************************
# *
# * Description: 非标准的日期字符串处理
# * Author: wangye 
# *
# ***************************************************
class DateParser(object):

    def __init__(self):
        self.pattern = re.compile(
        r'^((?:19|20)?\d{2})[-.]?((?:[0-1]?|1)[0-9])[-.]?((?:[0-3]?|[1-3])[0-9])?$'
        )

    def __cutDate(self, date, flags):
        y = date.year
        m = date.month if flags[1] else 1
        d = date.day if flags[2] else 1
        return datetime.date(y, m, d)

    def __mergeFlags(self, flags1, flags2):
        l = []
        length = min(len(flags1), len(flags2))
        for i in range(0, length):
            if flags1[i] and flags2[i]:
                l.append(True)
            else:
                l.append(False)
        return l

    def parse(self, strdate):
        """
        描述:时间解析方法。
        参数:strdate 要分析的时间字符串,比如目标时间类型datetime(1992, 2, 3)
              可以被解析的是下述字符串之一:
            19920203
            199203
            1992.02.03
            1992.02
            1992-02-03
            1992-02
            920203
        返回值:
            如果成功
            元组(datetime, flags),其中datetime表示转换完成的合法时间,
        flags是标志位,表示有效位数,比如199202实际转换了年和月,日没有,
        但是本函数将默认返回1日,但是flags将表示为(True, True, False),
        前面两个True分别表示年和月被转换,最后一个False表示日没有被转换。
            如果失败
            返回None。
        """
        m = self.pattern.match(strdate)
        flags = [False, False, False]
        if m:
            matches = list(m.groups())
            flags = list(map(lambda x:True if x!=None else False, matches))
            results = list(map(lambda x:int(x) if x!=None else 1, matches))
            # results = list(map(lambda x:1 if x==None else x, results))
            if results[0]                if results[0]>9:
                    results[0] += 1900
                else:
                    results[0] += 2000

            return (datetime.date(results[0], results[1], results[2]), flags)
        else:
            return None

    def convert(self, strdate, format):
        """
        描述:转换日期为指定格式。
        参数:strdate 同parse方法的strdate参数。
              format Python时间格式标识,同datetime.date.strftime格式化标识。
        返回值:
            如果成功,返回指定format格式的时间字符串。
            如果失败,返回None。
        """
        date = self.parse(strdate)
        if date:
            date = date[0]
            return datetime.date.strftime(date, format)
        else:
            return None

    def compare(self, strdate1, strdate2):
        """
        描述:比较两个日期。
        参数:strdate1 和 strdate2 同parse方法的strdate参数
        返回值:
            可以是下列值之一
            -4  strdate1 无效,  strdate2 有效
            -3  strdate1 有效,  strdate2 无效
            -2  strdate1 和 strdate2 无效
            -1  strdate1              0  strdate1 = strdate2
             1  strdate1 > strdate2
        """
        date1,flags1 = self.parse(strdate1)
        date2,flags2 = self.parse(strdate2)

        if date1 == None and date2 != None:
            return -4
        if date1 != None and date2 == None:
            return -3
        elif date1 == None and date2 == None:
            return -2

        flags = self.__mergeFlags(flags1, flags2)
        date1 = self.__cutDate(date1, flags)
        date2 = self.__cutDate(date2, flags)

        if date1>date2:
            return 1
        elif date1            return -1
        else:
            return 0

下面举几个例子供大家参考:

代码如下:


>>> DateParser().parse("19860126")
(datetime.date(1986, 1, 26), [True, True, True])
>>> DateParser().parse("199111")
(datetime.date(1991, 11, 1), [True, True, False])
>>> DateParser().parse("1991")
(datetime.date(1919, 9, 1), [True, True, True])
>>> DateParser().parse("8511")
(datetime.date(1985, 11, 1), [True, True, False])
>>> DateParser().convert("19911101", "%Y * %m * %d")
'1991 * 11 * 01'
>>> DateParser().convert("1990.1.01", "%Y.%m.%d")
'1990.01.01'
>>> DateParser().compare("1992.2", "19922")
0
>>> DateParser().compare("1992.2", "1956.03.1")
1

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