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?
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6 Answers
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)
4 Comments
S.Lott
+1: Epochal dates. March 8th is my daughter's birthday, that's a better epochal date than March 9th. Unix folks like Jan 1, 1970. Don't know why, that's not my daughter's birthday.
HuLu ViCa
I am trying to do this:
period['start_date'] = (self.start_date + timedelta(days=roundtrip_component.day-1)).strftime("%d/%m/%Y"), and I get this error: TypeError: must be str, not datetime.timedelta. So I guess in this case it makes sense to change the type timedelta to dateGeorgi Ivanov Dimitrov
It makes a lot of sense to convert it I have StartTime and EndTime of trips which are both in time delta format and I am trying to figure out why and how to put them in time format. On the phpmyadmin is normal time and when I download it from a tunnel and use it in python, they are both timedelta and I can not do anything with them
Bill Gale
Not helpful. Do not redefine a question and then answer it. I have a timedelta I need to get into a datetime so I can format it.
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
Comments
I share your pain bro
import datetime
from dateutil.relativedelta import relativedelta
# Both variables datetime
birthday = datetime.datetime(2000, 5, 24)
today = datetime.datetime.today()
# Time you've been alive in timedelta format
time_alive = today - birthday
print(time_alive)
>>> 8609 days, 18:47:53.341251
# Get only the time to concatenate later
hours = (datetime.datetime.min + time_alive).time()
print(hours)
>>> 18:47:53.341251
# Date that you've been alive
age = datetime.datetime.combine(datetime.datetime.fromordinal(time_alive.days) - relativedelta(years=1), hours)
print(age)
>>> 0023-07-27 18:47:53.341251
Comments
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)
)
1 Comment
Unknow0059
that's a time object not a datetime object ( as asked )
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)`
1 Comment
Community
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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)
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