将一年中的小数点转换为时间戳 [英] Convert a decimal day of the year to a timestamp
问题描述
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 23:14:41.065743
datetime
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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