如何将一年中某天的十进制表示形式转换为带有所有部分的时间戳,包括完整日期和时间?
How can I convert a decimal representation of a day in the year to a timestamp with all the parts, both the full date and the time?
例如,我的第一个小数点是 22.968530853511766 ,我希望它采用很好的时间戳格式。
For example, my first decimal is 22.968530853511766 and I want it in a nice timestamp format.
推荐答案使用 timedelta()对象,其值作为 days 参数;将其添加到上一年的12月31日午夜:
Use a timedelta() object with your value as the days parameter; add it to midnight December 31st of the previous year:
from datetime import datetime, timedelta epoch = datetime(datetime.now().year - 1, 12, 31) result = epoch + timedelta(days=your_decimal)Demo:
>>> from datetime import datetime, timedelta >>> epoch = datetime(datetime.now().year - 1, 12, 31) >>> epoch + timedelta(days=22.968530853511766) datetime.datetime(2015, 1, 22, 23, 14, 41, 65743) >>> print(epoch + timedelta(days=22.968530853511766)) 2015-01-22 23:14:41.065743datetime 对象可以使用 datetime.strftime()方法;我依赖演示中 print()调用的默认 str()转换。
A datetime object can be formatted any number of ways with the datetime.strftime() method; I relied on the default str() conversion as called by print() in the demo.
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将一年中的小数点转换为时间戳
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