如何将以毫秒为单位的包含7位数字的日期字符串转换为Python中的日期 [英] How do I convert a date string containing 7 digits in milliseconds into a date in Python

问题描述

当千位位数为6位时，％f起作用，但如果位数超过6位，则抛出错误。我有一个临时解决方案，将第7位数字硬编码为0，但是还有更好的方法吗？目前以下作品

When the milliseonds have 6 digits, %f works, but it throws an error if there are more than 6 digits. I have a temporary solution by hardcoding a 0 for the 7th digit, but is there a better way to do? Currently the below works

print (datetime.datetime.strptime(('2014-11-19 00:00:00.0000000').strip(), '%Y-%m-%d  %H:%M:%S.0%f')).date()

推荐答案

根据 datetime.strptime（） https://docs.python.org/2/library/datetime.html ＃strftime-and-strptime-behavior ，技术说明（4），％f仅接受1-6个字符：

According to the documentation of datetime.strptime() https://docs.python.org/2/library/datetime.html#strftime-and-strptime-behavior, technical note (4), %f accepts only 1 - 6 chars:

％f是C标准
中格式字符集的扩展（但在datetime对象中单独实现，因此始终有
可用）。与strptime（）方法一起使用时，％f指令
接受右侧的1到6位数字和零填充。

%f is an extension to the set of format characters in the C standard (but implemented separately in datetime objects, and therefore always available). When used with the strptime() method, the %f directive accepts from one to six digits and zero pads on the right.

我认为您没有正确解决问题。您不应该在字符串前添加零，而应将所有内容都丢弃到6以上（这对时间贡献意义不大）。

I do NOT believe you have solved the problem correctly. You should not prefix the string with zero, but instead drop everything past 6 (which is less significant to contributing to time).

类似这样的事情：

s='2014-11-19 00:00:00.0000000'
print (datetime.datetime.strptime((s[:26]).strip(), '%Y-%m-%d  %H:%M:%S.%f')).date()

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