字符串转换为日期时间对象，秒数以浮点数表示

Question

>>> a = "2016-03-22 12:33:45.7565"
>>> datetime.datetime.strptime(a, "%Y-%m-%d %H:%M:%f")
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "/usr/local/Cellar/python/2.7.11/Frameworks/Python.framework/Versions/2.7/lib/python2.7/_strptime.py", line 328, in _strptime
    data_string[found.end():])
ValueError: unconverted data remains: .7565
>>> datetime.datetime.strptime(a, "%Y-%m-%d %H:%M:%S")
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "/usr/local/Cellar/python/2.7.11/Frameworks/Python.framework/Versions/2.7/lib/python2.7/_strptime.py", line 328, in _strptime
    data_string[found.end():])
ValueError: unconverted data remains: .7565

我曾经使用过%S和%f，但如果这是类型float，我怎样处理秒钟呢？

任何帮助将不胜感激。

- Hackaholic

Answer 1

小数点后的数字是微秒，并与秒数分开格式化：

>>> a = "2016-03-22 12:33:45.7565"
>>> datetime.datetime.strptime(a, "%Y-%m-%d %H:%M:%S.%f")
datetime.datetime(2016, 3, 22, 12, 33, 45, 756500)

- snakecharmerb

Answer 2

这里介绍一种替代方法 - 使用dateutil解析器来完成此任务:

>>> from dateutil.parser import parse
>>> a = "2016-03-22 12:33:45.7565"
>>> parse(a)
datetime.datetime(2016, 3, 22, 12, 33, 45, 756500)

- alecxe

Answer 3

In [18]: a = "2016-03-22 12:33:45.7565"

In [19]: datetime.datetime.strptime(a, "%Y-%m-%d %H:%M:%S.%f")
Out[19]: datetime.datetime(2016, 3, 22, 12, 33, 45, 756500)

- Pavan