当前位置：文江博客话题详情

将 xs:duration 数据类型解析为 Python datetime.timedelta 对象？

发布于 2024-08-31 09:26:45 字数 589 浏览 6 评论 0原文

根据标题，我正在尝试解析包含 xs:duration 数据类型。我想将其转换为Python < code>timedelta 对象，然后我可以在进一步的计算中使用它。

是否有任何内置方法可以执行此操作，类似于 strptime() 函数？如果没有，实现这一目标的最佳方法是什么？

原文

分享到QQ

分享到微博

如果你对这篇内容有疑问，欢迎到本站社区发帖提问参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。

发布评论

需要登录才能够评论，你可以免费注册一个本站的账号。

马蹄踏│碎落叶 2024-09-07 09:26:45

鉴于我已经有一个关于我在问题中提出的问题的工作示例，为了完整起见，我将其发布在这里。如果有更好的答案出现，我会接受。

period = '-P14D'
regex  = re.compile('(?P<sign>-?)P(?:(?P<years>\d+)Y)?(?:(?P<months>\d+)M)?(?:(?P<days>\d+)D)?(?:T(?:(?P<hours>\d+)H)?(?:(?P<minutes>\d+)M)?(?:(?P<seconds>\d+)S)?)?')

# Fetch the match groups with default value of 0 (not None)
duration = regex.match(period).groupdict(0)

# Create the timedelta object from extracted groups
delta = timedelta(days=int(duration['days']) + (int(duration['months']) * 30) + (int(duration['years']) * 365),
                  hours=int(duration['hours']),
                  minutes=int(duration['minutes']),
                  seconds=int(duration['seconds']))

if duration['sign'] == "-":
    delta *= -1

这可行，但无法正确处理月份长度或闰年。就我而言，这不是问题，但值得牢记。

Seeing as I already have a working example of what I asked in the question, I'll post it here for completeness. If any better answers come up I'll accept.

period = '-P14D'
regex  = re.compile('(?P<sign>-?)P(?:(?P<years>\d+)Y)?(?:(?P<months>\d+)M)?(?:(?P<days>\d+)D)?(?:T(?:(?P<hours>\d+)H)?(?:(?P<minutes>\d+)M)?(?:(?P<seconds>\d+)S)?)?')

# Fetch the match groups with default value of 0 (not None)
duration = regex.match(period).groupdict(0)

# Create the timedelta object from extracted groups
delta = timedelta(days=int(duration['days']) + (int(duration['months']) * 30) + (int(duration['years']) * 365),
                  hours=int(duration['hours']),
                  minutes=int(duration['minutes']),
                  seconds=int(duration['seconds']))

if duration['sign'] == "-":
    delta *= -1

This works, but won't handle the month lengths or leap years correctly. For my purposes this isn't an issue, but it is worth keeping in mind.

回复收藏 0 原文