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Timestamp to iso

I have a dataframe with timestamp of different formats one with 05-28-2022 14:05:30 and one with 06-04-2022 03:04:13.002 both I want to convert into iso format how can I do that?
input output
05-28-2022 14:05:30 -> 2022-05-28T14:05:30.000+0000
06-04-2022 03:04:13.002 -> 2022-06-04T03:04:13.002+0000

You can use strptime() + strftime(). Here is an example:
from datetime import datetime
import pytz
# parse str to instance
first = datetime.strptime('05-28-2022 14:05:30', '%m-%d-%Y %H:%M:%S')
first = first.replace(tzinfo=pytz.UTC)
print(first.strftime('%Y-%m-%dT%H:%M:%S.%f%z'))
print(f'{first.isoformat()}')
second = datetime.strptime('06-04-2022 03:04:13.002', '%m-%d-%Y %H:%M:%S.%f')
second = second.replace(tzinfo=pytz.UTC)
print(second.strftime('%Y-%m-%dT%H:%M:%S.%f%z'))
print(second.isoformat())
# 2022-05-28T14:05:30.000000+0000
# 2022-05-28T14:05:30+00:00
# 2022-06-04T03:04:13.002000+0000
# 2022-06-04T03:04:13.002000+00:00
See datetime docs. Also you can use other packages for dates processing / formatting:
iso8601
pendulum
dateutil
arrow
Example with dataframe:
import pandas as pd
import pytz
from datetime import datetime
df = pd.DataFrame({'date': ['05-28-2022 14:05:30', '06-04-2022 03:04:13.002']})
def convert_date(x):
dt_format = '%m-%d-%Y %H:%M:%S.%f' if x.rfind('.', 1) > -1 else '%m-%d-%Y %H:%M:%S'
dt = datetime.strptime(x, dt_format).replace(tzinfo=pytz.UTC)
return dt.strftime('%Y-%m-%dT%H:%M:%S.%f%z')
df['new_date'] = df['date'].apply(convert_date)
print(df)
date new_date
0 05-28-2022 14:05:30 2022-05-28T14:05:30.000000+0000
1 06-04-2022 03:04:13.002 2022-06-04T03:04:13.002000+0000

subtracting dates in python

i have an dataframe with dates and would like to get the time between the first date and the last date, when i run the code below
df.sort_values('timestamp', inplace=True)
firstDay = df.iloc[0]['timestamp']
lastDay = df.iloc[len(df)-1]['timestamp']
print(firstDay)
print(lastDay)
it provides the following formate of the dates :
2016-09-24 17:42:27.839496
2017-01-18 10:24:08.629327
and I'm trying to get the different between them but they're in the str format, and I've been having trouble converting them to a form where i can get the difference

here you go :o)
import datetime
from datetime import date
from datetime import datetime
import pandas as pd
date_format_str = '%Y-%m-%d %H:%M:%S.%f'
date_1 = '2016-09-24 17:42:27.839496'
date_2 = '2017-01-18 10:24:08.629327'
start = datetime.strptime(date_1, date_format_str)
end = datetime.strptime(date_2, date_format_str)
diff = end - start
# Get interval between two timstamps as timedelta object
diff_in_hours = diff.total_seconds() / 3600
print(diff_in_hours)
# get the difference between two dates as timedelta object
diff = end.date() - start.date()
print(diff.days)
Pandas
import datetime
from datetime import date
from datetime import datetime
import pandas as pd
date_1 = '2016-09-24 17:42:27.839496'
date_2 = '2017-01-18 10:24:08.629327'
start = pd.to_datetime(date_1, format='%Y-%m-%d %H:%M:%S.%f')
end = pd.to_datetime(date_2, format='%Y-%m-%d %H:%M:%S.%f')
# get the difference between two datetimes as timedelta object
diff = end - start
print(diff.days)

Converting timedate string with timezone to utc python

I have a a basic string + another string which reresents timezone. I'm trying to return the UTC equivalent..
timezone = "America/New_York"
scheduleDate = "2021-09-21 21:00:00"
dtimestamp = datetime.datetime.strptime(scheduleDate, '%Y-%m-%d %H:%M:%S').astimezone(
pytz.timezone(timezone)).timestamp()
sdate = datetime.datetime.utcfromtimestamp(dtimestamp).strftime('%Y-%m-%dT%H:%M:%SZ')
print(sdate)

This works, but not sure if there's a better approach for this. I haven't looked too closely at the pytz docs before.
from datetime import datetime
import pytz
timezone = "America/New_York"
scheduleDate = "2021-09-21 21:00:00"
# Convert to naive datetime
dt_naive: datetime = datetime.strptime(scheduleDate, '%Y-%m-%d %H:%M:%S')
# localize datetime with ET timezone
dt: datetime = pytz.timezone(timezone).localize(dt_naive)
# normalize datetime to UTC time
dt_utc: datetime = pytz.utc.normalize(dt)
print(str(dt_utc))

How to fix date formatting using python3

I have data with the date format as follows:
date_format = 190410
year = 19
month = 04
date = 10
I want to change the date format, to be like this:
date_format = 10-04-2019
How do I solve this problem?

>>> import datetime
>>> date = 190410
>>> datetime.datetime.strptime(str(date), "%y%m%d").strftime("%d-%m-%Y")
'10-04-2019'
datetime.strptime() takes a data string and a format, and turns that into datetime object, and datetime objects have a method called strftime that turns datetime objects to string with given format. You can look what %y %m %d %Y are from here.

This is what you want(Notice that you have to change your format)
import datetime
date_format = '2019-04-10'
date_time_obj = datetime.datetime.strptime(date_format, '%Y-%m-%d')
print(date_time_obj)
Here is an other example
import datetime
date_time_str = '2018-06-29 08:15:27.243860'
date_time_obj = datetime.datetime.strptime(date_time_str, '%Y-%m-%d %H:%M:%S.%f')
print('Date:', date_time_obj.date())
print('Time:', date_time_obj.time())
print('Date-time:', date_time_obj)
You can also do this
from datetime import datetime, timedelta
s = "20120213"
# you could also import date instead of datetime and use that.
date = datetime(year=int(s[0:4]), month=int(s[4:6]), day=int(s[6:8]))
print(date)
There are many ways to achieve what you want.

How to convert a timedelta object into a datetime object

What is the proper way to convert a timedelta object into a datetime object?
I immediately think of something like datetime(0)+deltaObj, but that's not very nice... Isn't there a toDateTime() function or something of the sort?

It doesn't make sense to convert a timedelta into a datetime, but it does make sense to pick an initial or starting datetime and add or subtract a timedelta from that.
>>> import datetime
>>> today = datetime.datetime.today()
>>> today
datetime.datetime(2010, 3, 9, 18, 25, 19, 474362)
>>> today + datetime.timedelta(days=1)
datetime.datetime(2010, 3, 10, 18, 25, 19, 474362)

Since a datetime represents a time within a single day, your timedelta should be less than 24 hours (86400 seconds), even though timedeltas are not subject to this constraint.
import datetime
seconds = 86399
td = datetime.timedelta(seconds=seconds)
print(td)
dt = datetime.datetime.strptime(str(td), "%H:%M:%S")
print(dt)
23:59:59
1900-01-01 23:59:59
If you don't want a default date and know the date of your timedelta:
date = "05/15/2020"
dt2 = datetime.datetime.strptime("{} {}".format(date, td), "%m/%d/%Y %H:%M:%S")
print(dt2)
2020-05-15 23:59:59

I found that I could take the .total_seconds() and use that to create a new time object (or datetime object if needed).
import time
import datetime
start_dt_obj = datetime.datetime.fromtimestamp(start_timestamp)
stop_dt_obj = datetime.datetime.fromtimestamp(stop_timestamp)
delta = stop_dt_obj - start_dt_obj
delta_as_time_obj = time.gmtime(delta.total_seconds())
This allows you to do something like:
print('The duration was {0}'.format(
time.strftime('%H:%M', delta_as_time_obj)
)

Improving #sadpanduar answer with example on converting one column in pandas.DataFrame:
from datetime import timedelta
import time
def seconds_to_datetime(seconds, format='%Y-%m-%d %H:%M:%S'):
td = timedelta(seconds=seconds)
time_obj = time.gmtime(td.total_seconds())
return time.strftime(format, time_obj)
df = pd.read_csv(CSV_PATH)
df['TIMESTAMP_COLUMN'] = df['TIMESTAMP_COLUMN'].apply(seconds_to_datetime)

import datetime`enter code here
lastDownloadedDate = datetime.date(2022,8,4)
print('lastDownloadedDate: ', lastDownloadedDate)
fdate = lastDownloadedDate + datetime.timedelta(days=1)
fdate = datetime.datetime.strptime(str(fdate), "%Y-%m-%d")
fdate = datetime.date(fdate.year, fdate.month, fdate.day)
print('fdate: ', dt3)`