Python-解析并将字符串转换为时间戳 [英] Python - Parsing and converting string into timestamp
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问题描述
我有以下格式的字符串: 2017-02-14T09:51:46.000-0600
I have string in next format: 2017-02-14T09:51:46.000-0600
解析字符串并将其转换为时间戳的最佳方法是什么?我可以选择使用正则表达式或编写自己的函数进行解析,但是有任何内置方法可以帮助我吗?
What is the best approach to parse and convert string into timestamp? I have options to use regular expression or to write my own function for parsing, but are there any builtin methods which can help me?
推荐答案
第一部分将创建日期时间对象:
First part would be creating datetime object:
from datetime import datetime
date_string = "2017-02-14T09:51:46.000-0600"
# I'm using date_string[:-9] to skip ".000-0600"
format_date = datetime.strptime(date_string, '%Y-%m-%dT%H:%M:%S.%f%z'))
以下格式日期为:
print(format_date)
2017-02-14 09:51:46
时间戳是:
print(format_date.timestamp())
1487062306.0
这里很少澄清,在python 参考页上,您可以看到有关的定义我使用的'%Y-%m-%dT%H:%M:%S.%f%z')
格式说明符.
Little clarification here, on python reference page, you can see definition for '%Y-%m-%dT%H:%M:%S.%f%z')
format specifiers I used.
- %Y :以世纪作为十进制数字的年份,例如1970、1988、2001、2013
- %m :月份为零填充的十进制数字(例如01、02,...,12)
- %d :每月的一天,以零填充的十进制数字(例如01、02,...,31)
- %H :小时(24小时制)为零填充的十进制数字(例如00、01,...,23)
- %M :以零填充的十进制数字表示(例如00、01,...,59)
- %S :第二个,为零填充的十进制数字(例如00、01,...,59)
- %f :微秒,十进制数字,左侧零填充(000000,000001,...,999999)
- %z :UTC偏移量,格式为+ HHMM或-HHMM,如果对象是幼稚对象,则为空字符串(空或+ 0000,-0400,+ 1030)
- %Y: Year with century as a decimal number, e.g. 1970, 1988, 2001, 2013
- %m: Month as a zero-padded decimal number (e.g. 01, 02, ..., 12)
- %d: Day of the month as a zero-padded decimal number (e.g. 01, 02, ..., 31)
- %H: Hour (24-hour clock) as a zero-padded decimal number (e.g 00, 01, ..., 23)
- %M: Minute as a zero-padded decimal number (e.g 00, 01, ..., 59)
- %S: Second as a zero-padded decimal number (e.g. 00, 01, ..., 59)
- %f: Microsecond as a decimal number, zero-padded on the left (000000, 000001, ..., 999999)
- %z: UTC offset in the form +HHMM or -HHMM, empty string if the the object is naive, (empty or +0000, -0400, +1030)
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