I have a datetime object t which is in UTC time but attached with Alaska timezone offset(-08:00):
t
-> datetime.datetime(2019, 3, 23, 15, 49, 34, tzinfo=tzfile('/usr/share/zoneinfo/America/Anchorage'))
I try to convert t to a datetime tuple in Alaska localtime. I do like this but I still get the UTC time.
time.strptime(str(t),"%Y-%m-%d %H:%M:%S%z")
-> time.struct_time(tm_year=2019, tm_mon=3, tm_mday=23, tm_hour=15, tm_min=49, tm_sec=34,\
tm_wday=5, tm_yday=82, tm_isdst=-1)
So the time.strptime() will not process the offset, even I input it in the format(%z means the offset)?
What should I do to get:
-> time.struct_time(tm_year=2019, tm_mon=3, tm_mday=23, tm_hour=7, tm_min=49, tm_sec=34,\
tm_wday=5, tm_yday=82, tm_isdst=-1)
(tm_hour I want is 7 = 15 - 8)
Thanks so much!
Related
I am wondering why the delta is 3601 seconds when it should be 1 second:
from datetime import datetime
from dateutil import tz
# Create eastern timezone
eastern = tz.gettz('America/New_York')
# 2017-03-12 01:59:59 in Eastern Time (EST)
spring_ahead_159am = datetime(2017, 3, 12, 1, 59, 59, tzinfo = eastern)
# 2017-03-12 03:00:00 in Eastern Time (EDT)
spring_ahead_3am = datetime(2017, 3, 12, 3, 0, 0, tzinfo = eastern)
(spring_ahead_3am - spring_ahead_159am).seconds
3601
Please explain me what I am doing wrong? Why it's not giving me 1 second diff?
P.S:
Please explain why this will do work easily:
Let:
EST = timezone(timedelta(hours=-5))
EDT = timezone(timedelta(hours=-4))
spring_ahead_159am = datetime(2017, 3, 12, 1, 59, 59, tzinfo = EST)
spring_ahead_3am = datetime(2017, 3, 12, 3, 0, 0, tzinfo = EDT)
(spring_ahead_3am - spring_ahead_159am).seconds
1
This is a little quirk of datetime objects. Quoting from the documentation:
Subtraction of a datetime from a datetime is defined only if both operands are naive, or if both are aware. If one is aware and the other is naive, TypeError is raised.
If both are naive, or both are aware and have the same tzinfo attribute, the tzinfo attributes are ignored, and the result is a timedelta object t such that datetime2 + t == datetime1. No time zone adjustments are done in this case.
You can certain get what you want by subtracting their DST components:
>>> spring_ahead_3am - spring_ahead_3am.dst() - spring_ahead_159am + spring_ahead_159am.dst()
datetime.timedelta(seconds=1)
Update
Consider using this function, which does the Right Thing across timezones and DST changes:
def datesub( dt1, dt2 ):
return datetime.timedelta( seconds=dt2.timestamp()-dt1.timestamp() )
I have been analyzing the JSON file generated using chrome://webrtc-internal, while running webrtc.
I'm trying to find the matching timezone of the startTime in the following result:
'RTCVideoSource_6-width': {
'startTime': '2021-04-14T07:09:33.163Z',
'endTime': '2021-04-14T07:14:12.161Z',
'values': '[1920,1920,1920,1920,1920,1920,1920,1920,1920]'},
I tried UTC timezone to find the corresponding timestamp in my timezone but didn't succeed.
yourdate = dateutil.parser.parse(RTCVideoSource_6-width['startTime'])
startTime_utc = yourdate.replace(tzinfo=pytz.UTC)
startTime_hk=startTime_utc.astimezone(pytz.timezone("Asia/Hong_Kong")) #astimezone method
#since 1970
startAge_utc = datetime.datetime(1970,1,1).replace(tzinfo=pytz.UTC)
startAge_hk=startAge_utc.astimezone(pytz.timezone("Asia/Hong_Kong")) #astimezone method
start_in_seconds = int((startTime_hk- startAge_hk).total_seconds())
When I compared now() as a timestamp in seconds, I got values with a relatively big difference,
1618384173 and 1618383543.
What is the timezone used in chrome://webrtc-internal stats?
Is my conversion method correct?
See ISO8601: you have UTC date/time (Z = zulu = UTC). See How do I parse an ISO 8601-formatted date? how to parse. Unix time (seconds since 1970-01-01) on the other hand always (should) refer to UTC (e.g. startTime_utc and startTime_hk must give the same timestamp!). Note: datetime objects have a timestamp method.
EX:
from datetime import datetime
from zoneinfo import ZoneInfo # Python 3.9+
d = {'startTime': '2021-04-14T07:09:33.163Z',
'endTime': '2021-04-14T07:14:12.161Z',
'values': '[1920,1920,1920,1920,1920,1920,1920,1920,1920]'}
startTime_utc = datetime.fromisoformat(d['startTime'].replace('Z', '+00:00'))
print(repr(startTime_utc))
# datetime.datetime(2021, 4, 14, 7, 9, 33, 163000, tzinfo=datetime.timezone.utc)
print(startTime_utc.timestamp())
# 1618384173.163
startTime_HK = startTime_utc.astimezone(ZoneInfo('Asia/Hong_Kong'))
print(repr(startTime_HK))
# datetime.datetime(2021, 4, 14, 15, 9, 33, 163000, tzinfo=zoneinfo.ZoneInfo(key='Asia/Hong_Kong'))
print(startTime_HK.timestamp())
# 1618384173.163
I need to convert DateTime from WxWidget
Fri Sep 28 00:00:00 2018
To another format like :
28/09/18 20:35:00
But i can't find anything to do
Use wx.DateTime Format()
i.e.
d=wx.DateTime.Now()
d
<wx.DateTime: "Sat Sep 29 10:35:28 2018">
d.Format('%d/%m/%y %H:%M:%S')
'29/09/18 10:35:28'
For the format options see datetime manual pages https://docs.python.org/3/library/datetime.html#strftime-and-strptime-behavior
For wx.DateTime options, see https://wxpython.org/Phoenix/docs/html/wx.DateTime.html
To generate an actual python datetime (as per your question heading):
f = d.Format('%d/%m/%y %H:%M:%S')
x = datetime.datetime.strptime(f,'%d/%m/%y %H:%M:%S')
x
datetime.datetime(2018, 9, 29, 10, 35, 28)
I am creating a naive datetime object which contains no timezone information, but I know that it is always in UTC. I want to calculate the time difference between local time and UTC for any timezone I define, which would take into account DST as well.
What I am thus doing is the following:
from datetime import datetime, timedelta
now = datetime(2018,3,27,15,20) #Create a naive datetime object
now_utc = timezone("UTC").localize(now) # Add UTC timezone information to it
now_madrid = now_utc.astimezone(timezone("Europe/Madrid")) # convert to Madrid timezone
Normally, if I was to calculate the time difference between two datetime objects I would subtract them. But when I try diff = now_utc - now_madrid , the result is the following:
In [275]: now_utc-now_madrid
Out[275]: datetime.timedelta(0)
Can anyone please tell me, how I could find the timezone difference in this case? Thank you in advance :)
Time delta is "0" because Madrid is 2 hours head of UTC.
now_utc-now_madrid
When you subtract it like this means it's in another timezone behind UTC.
There maybe a more efficient methods out there but this is how I would do it.
>>> import pytz
>>> import datetime
>>> utc_now = pytz.utc.localize(datetime.datetime.utcnow ())
>>> utc_now
datetime.datetime(2018, 3, 27, 19, 11, 19, 659257, tzinfo=<UTC>)
>>> madrid_now = utc_now.astimezone(pytz.timezone ('Europe/Madrid'))
>>> madrid_now
datetime.datetime(2018, 3, 27, 21, 11, 19, 659257, tzinfo=<DstTzInfo 'Europe/Madrid' CEST+2:00:00 DST>)
As you can see, the timezone conversion is already provided by tzinfo CEST+2:00:00 DST.
If you need to do any arithmetic on that value then try:
>>> utc_now + timedelta (hours = 2) # Since Madrid is +2:00:00 hours
datetime.datetime(2018, 3, 27, 21, 11, 19, 659257, tzinfo=<UTC>)
Or extract the offset timezone differences from strftime.
>>> madrid_now.strftime ('%z')
'+0200'
You can find more reading here:
http://pytz.sourceforge.net/
I've got a pandas dataframe with two datetime columns and I would like to calculate the timedelta between the columns in "business minutes". It's easy to add business timedeltas using the offsets method, but I can't seem to find something built in that returns a timedelta in business days, hours, minutes, seconds. I'm very new to Python so it's very likely I'm missing something.
Thanks,
Nick
I don't think there's a way in numpy/pandas, but you can do it with python lib businesstime:
>>> datetime(2013, 12, 26, 5) - datetime(2013, 12, 23, 12)
datetime.timedelta(2, 61200)
>>> bt = businesstime.BusinessTime(holidays=businesstime.holidays.usa.USFederalHolidays())
>>> bt.businesstimedelta(datetime(2013, 12, 23, 12), datetime(2013, 12, 26, 5))
datetime.timedelta(1, 18000)