I have a program (sar command line utility) which outputs it's lines with time column. I parse this file with my python script and I would like to convert sar's 02:31:33 PM into epochs e.g. 1377181906 (current year, month and day with hours, minutes and seconds from abovementioned string). How can this done in a less cumbersome way? I tried to do this by myself, but stuck with time/datetime and herd of their methods.
Here's one way to do it:
read the string into datetime using strptime
set year, month, day of the datetime object to current date's year, month and day via replace
convert datetime into unix timestamp via calendar.timegm
>>> from datetime import datetime
>>> import calendar
>>> dt = datetime.strptime("02:31:33 PM", "%I:%M:%S %p")
>>> dt_now = datetime.now()
>>> dt = dt.replace(year=dt_now.year, month=dt_now.month, day=dt_now.day)
>>> calendar.timegm(dt.utctimetuple())
1377138693
Note that in python >= 3.3, you can get the timestamp from a datetime by calling dt.timestamp().
Also see:
Python Create unix timestamp five minutes in the future
An another way to have epoch time is to use mktime from time module and pass time tuple of date, so you can do this:
>>> from datetime import datetime
>>> from time import mktime
>>> dt = datetime.strptime("02:31:33 PM", "%H:%M:%S %p")
>>> dt_now = datetime.now()
>>> dt = dt.replace(year=dt_now.year, month=dt_now.month, day=dt_now.day)
>>> int(mktime(dt.timetuple()))
1377131493
Related
I am using Python in dynamo and I am facing a problem.
I have to convert date time into integer so I could further process
it
I have tried some codes but they are not helpful.
If you'd like to convert the datetime to a unix timestamp (number of seconds elapsed since Jan 1, 1970), then you can do
>>> import datetime as dt
>>> ts = dt.datetime.now()
>>> print(int(ts.timestamp())
1588967243
Maybe you want to get timestamp?
import time
import datetime
s = "01/12/2011"
time.mktime(datetime.datetime.strptime(s, "%d/%m/%Y").timetuple())
Result: 1322697600.0
With the datetime module, I can get the current time, like so:
>>> datetime.now().strftime('%Y-%m-%d %H:%M:%S')
'2017-08-29 23:01:32'
I have access to the time at which a file was created, in the same format:
>>> data['created']
'2017-08-29 20:59:09'
Is there a way, using the datetime module, that I can calculate the time between the two, in hours?
Performing subtraction on two datetime objects will result in a timedelta. You can use datetime.strptime to get that second datetime object, access the seconds attribute of that timedelta and calculate the hours from there:
from datetime import datetime
...
file_created = datetime.strptime(data['created'], '%Y-%m-%d %H:%M:%S')
difference = (datetime.now() - file_created).seconds
print("Hours since creation: " + str(difference // 3600)) # 3600 seconds in 1 hour
As an input to an API request I need to get yesterday's date as a string in the format YYYY-MM-DD. I have a working version which is:
yesterday = datetime.date.fromordinal(datetime.date.today().toordinal()-1)
report_date = str(yesterday.year) + \
('-' if len(str(yesterday.month)) == 2 else '-0') + str(yesterday.month) + \
('-' if len(str(yesterday.day)) == 2 else '-0') + str(yesterday.day)
There must be a more elegant way to do this, interested for educational purposes as much as anything else!
You Just need to subtract one day from today's date. In Python datetime.timedelta object lets you create specific spans of time as a timedelta object.
datetime.timedelta(1) gives you the duration of "one day" and is subtractable from a datetime object. After you subtracted the objects you can use datetime.strftime in order to convert the result --which is a date object-- to string format based on your format of choice:
>>> from datetime import datetime, timedelta
>>> yesterday = datetime.now() - timedelta(1)
>>> type(yesterday)
>>> datetime.datetime
>>> datetime.strftime(yesterday, '%Y-%m-%d')
'2015-05-26'
Note that instead of calling the datetime.strftime function, you can also directly use strftime method of datetime objects:
>>> (datetime.now() - timedelta(1)).strftime('%Y-%m-%d')
'2015-05-26'
As a function:
from datetime import datetime, timedelta
def yesterday(frmt='%Y-%m-%d', string=True):
yesterday = datetime.now() - timedelta(1)
if string:
return yesterday.strftime(frmt)
return yesterday
example:
In [10]: yesterday()
Out[10]: '2022-05-13'
In [11]: yesterday(string=False)
Out[11]: datetime.datetime(2022, 5, 13, 12, 34, 31, 701270)
An alternative answer that uses today() method to calculate current date and then subtracts one using timedelta(). Rest of the steps remain the same.
https://docs.python.org/3.7/library/datetime.html#timedelta-objects
from datetime import date, timedelta
today = date.today()
yesterday = today - timedelta(days = 1)
print(today)
print(yesterday)
Output:
2019-06-14
2019-06-13
>>> import datetime
>>> datetime.date.fromordinal(datetime.date.today().toordinal()-1).strftime("%F")
'2015-05-26'
Calling .isoformat() on a date object will give you YYYY-MM-DD
from datetime import date, timedelta
(date.today() - timedelta(1)).isoformat()
I'm trying to use only import datetime based on this answer.
import datetime
oneday = datetime.timedelta(days=1)
yesterday = datetime.date.today() - oneday
I'm pulling a timestamp that looks like this - 2014-02-03T19:24:07Z
I'm trying to calculate the number of days since January 1.
I was able to convert it to datetime using
yourdate = dateutil.parser.parse(timestamp)
But now I'm trying to parse it and grab individual elements, such as the month & day.
Is there a way to convert it to strptime so I can select each element?
Just access the month, day using year, month, day attributes..
>>> import dateutil.parser
>>> yourdate = dateutil.parser.parse('2014-02-03T19:24:07Z')
>>> yourdate.year
2014
>>> yourdate.month
2
>>> yourdate.day
3
Just to be a little more complete:
>>> from dateutil.parser import parse
>>> from datetime import datetime
>>> import pytz
>>> d = parse('2014-02-03T19:24:07Z')
>>> other = datetime(year=2014, month=1, day=1, tzinfo=pytz.utc)
>>> (d-other).days
33
You have to make sure the other datetime is timezone aware if you're creating it with datetime as opposed to the datetime you're parsing with dateutil.
There's no need for converting. The resulting datetime.datetime object has all necessary properties which you can access directly. For example:
>>> import dateutil.parser
>>> timestamp="2014-02-03T19:24:07Z"
>>> yourdate = dateutil.parser.parse(timestamp)
>>> yourdate.day
3
>>> yourdate.month
2
See: https://docs.python.org/2/library/datetime.html#datetime-objects
if you want to calculate:
import dateutil.parser
yourdate = dateutil.parser.parse('2014-02-03T19:24:07Z')
startdate = dateutil.parser.parse('2014-01-01T00:00:00Z')
print (yourdate - startdate)
Another way to solve without the dateutil module:
import datetime
# start date for comparision
start = datetime.date(2014, 1, 1)
# timestamp as string
datefmt = "%Y-%m-%dT%H:%M:%SZ"
current = "2014-02-03T19:24:07Z"
# convert timestamp string to date, dropping time
end = datetime.datetime.strptime(current, datefmt).date()
# compare dates and get number of days from timedelta object
days = (end - start).days
This assumes you don't care about time (including timezones).
Does time.time() in the Python time module return the system's time or the time in UTC?
The time.time() function returns the number of seconds since the epoch, as a float. Note that "the epoch" is defined as the start of January 1st, 1970 in UTC. So the epoch is defined in terms of UTC and establishes a global moment in time. No matter where on Earth you are, "seconds past epoch" (time.time()) returns the same value at the same moment.
Here is some sample output I ran on my computer, converting it to a string as well.
>>> import time
>>> ts = time.time()
>>> ts
1355563265.81
>>> import datetime
>>> datetime.datetime.fromtimestamp(ts).strftime('%Y-%m-%d %H:%M:%S')
'2012-12-15 01:21:05'
>>>
The ts variable is the time returned in seconds. I then converted it to a human-readable string using the datetime library.
This is for the text form of a timestamp that can be used in your text files. (The title of the question was different in the past, so the introduction to this answer was changed to clarify how it could be interpreted as the time. [updated 2016-01-14])
You can get the timestamp as a string using the .now() or .utcnow() of the datetime.datetime:
>>> import datetime
>>> print datetime.datetime.utcnow()
2012-12-15 10:14:51.898000
The now differs from utcnow as expected -- otherwise they work the same way:
>>> print datetime.datetime.now()
2012-12-15 11:15:09.205000
You can render the timestamp to the string explicitly:
>>> str(datetime.datetime.now())
'2012-12-15 11:15:24.984000'
Or you can be even more explicit to format the timestamp the way you like:
>>> datetime.datetime.now().strftime("%A, %d. %B %Y %I:%M%p")
'Saturday, 15. December 2012 11:19AM'
If you want the ISO format, use the .isoformat() method of the object:
>>> datetime.datetime.now().isoformat()
'2013-11-18T08:18:31.809000'
You can use these in variables for calculations and printing without conversions.
>>> ts = datetime.datetime.now()
>>> tf = datetime.datetime.now()
>>> te = tf - ts
>>> print ts
2015-04-21 12:02:19.209915
>>> print tf
2015-04-21 12:02:30.449895
>>> print te
0:00:11.239980
Based on the answer from #squiguy, to get a true timestamp I would type cast it from float.
>>> import time
>>> ts = int(time.time())
>>> print(ts)
1389177318
At least that's the concept.
The answer could be neither or both.
neither: time.time() returns approximately the number of seconds elapsed since the Epoch. The result doesn't depend on timezone so it is neither UTC nor local time. Here's POSIX defintion for "Seconds Since the Epoch".
both: time.time() doesn't require your system's clock to be synchronized so it reflects its value (though it has nothing to do with local timezone). Different computers may get different results at the same time. On the other hand if your computer time is synchronized then it is easy to get UTC time from the timestamp (if we ignore leap seconds):
from datetime import datetime
utc_dt = datetime.utcfromtimestamp(timestamp)
On how to get timestamps from UTC time in various Python versions, see How can I get a date converted to seconds since epoch according to UTC?
To get a local timestamp using datetime library, Python 3.x
#wanted format: year-month-day hour:minute:seconds
from datetime import datetime
# get time now
dt = datetime.now()
# format it to a string
timeStamp = dt.strftime('%Y-%m-%d %H:%M:%S')
# print it to screen
print(timeStamp)
I eventually settled for:
>>> import time
>>> time.mktime(time.gmtime())
1509467455.0
There is no such thing as an "epoch" in a specific timezone. The epoch is well-defined as a specific moment in time, so if you change the timezone, the time itself changes as well. Specifically, this time is Jan 1 1970 00:00:00 UTC. So time.time() returns the number of seconds since the epoch.
timestamp is always time in utc, but when you call datetime.datetime.fromtimestamp it returns you time in your local timezone corresponding to this timestamp, so result depend of your locale.
>>> import time, datetime
>>> time.time()
1564494136.0434234
>>> datetime.datetime.now()
datetime.datetime(2019, 7, 30, 16, 42, 3, 899179)
>>> datetime.datetime.fromtimestamp(time.time())
datetime.datetime(2019, 7, 30, 16, 43, 12, 4610)
There exist nice library arrow with different behaviour. In same case it returns you time object with UTC timezone.
>>> import arrow
>>> arrow.now()
<Arrow [2019-07-30T16:43:27.868760+03:00]>
>>> arrow.get(time.time())
<Arrow [2019-07-30T13:43:56.565342+00:00]>
time.time() return the unix timestamp.
you could use datetime library to get local time or UTC time.
import datetime
local_time = datetime.datetime.now()
print(local_time.strftime('%Y%m%d %H%M%S'))
utc_time = datetime.datetime.utcnow()
print(utc_time.strftime('%Y%m%d %H%M%S'))