Have the following string:
"date Thursday June 03 12:02:56 2017"
What would be the proper way of convert it to epoch time?
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What exactly is that suppose to represent? A string? – juanpa.arrivillaga Jun 08 '17 at 20:08
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@juanpa.arrivillaga Sorry, yes. Just clarified it – Jun 08 '17 at 20:09
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Just a start, not a solution, but this thread may help: https://stackoverflow.com/questions/311627/how-to-print-date-in-a-regular-format-in-python – Christopher Goings Jun 08 '17 at 20:10
2 Answers
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You can use datetime.strptime() to parse your date and then just do delta with the epoch:
from datetime import datetime as dt
epoch = dt(1970, 1, 1)
date = "date Thursday June 03 12:02:56 2017"
epoch_time = int((dt.strptime(date, "date %A %B %d %H:%M:%S %Y") - epoch).total_seconds())
# 1496491376
zwer
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1@juanpa.arrivillaga - only on Python 3.x, and the OP clearly tagged his question with `python-2.7`... – zwer Jun 08 '17 at 20:18
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Oh nice, I was totally unaware that Python 2 datetime objects lacked a timestamp method. – juanpa.arrivillaga Jun 08 '17 at 20:19
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Another solution is to create a time.struct_time structure by parsing with time.strptime() and then pass it into calendar.timegm() to convert to epoch time.
import time
import calendar
timestr = "date Thursday June 03 12:02:56 2017"
calendar.timegm(time.strptime(timestr, "date %A %B %d %H:%M:%S %Y"))
# returns 1496491376
alkasm
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