I would like to add a certain time to a formatted time string in python. For this I tried the following
from datetime import datetime, timedelta
timestemp_Original = '2021-07-13T00:15:00Z'
timestemp_Added1 = '2021-07-13T00:15:00Z' + timedelta(minutes=15)
timestemp_Added2 = timestemp_Original + datetime.timedelta(hours=0, minutes=15)
but this leads to error messages (I took it from here How to add hours to current time in python and Add time to datetime). Can aynone tell me how to do this?
First, You need to convert str to datetime with a specific format of string_date and then use timedelta(minutes=15).
from datetime import datetime, timedelta
timestemp_Original = '2021-07-13T00:15:00Z'
timestemp_Added1 = datetime.strptime(timestemp_Original, "%Y-%m-%dT%H:%M:%SZ") + timedelta(minutes=15)
print(timestemp_Added1)
# If you want to get as original format
print(timestemp_Added1.strftime("%Y-%m-%dT%H:%M:%SZ"))
# 2021-07-13T00:30:00Z
2021-07-13 00:30:00
This question already has answers here:
Subtract hours and minutes from time
(3 answers)
Closed last year.
My initial string looks like following:
a1 = "06:00:00"
a2 = "01:00:00"
I want to set the time back by two hours.
How to get the following output (in string format)?
a1_new = "04:00:00"
a2_new = "23:00:00"
Here you go!
from datetime import datetime, timedelta
a1 = "06:00:00"
x = datetime.strptime(a1, "%H:%M:%S") - timedelta(hours=2, minutes=0)
y = x.strftime("%H:%M:%S")
print(y)
Steps:
Convert HMS into a DateTime Object
Minus 2 hours from this
Convert the result into a String that only contains Hour Minute & Second
from datetime import datetime
from datetime import timedelta
time_fmt = "%H:%M:%S"
a1_new = datetime.strptime(a1, time_fmt) - timedelta(hours = 2)
a1_new = a1_new.strftime("%H:%M:%S")
print(a1_new)
'08:00:00'
I am assuming here that you only need a simple 24-hour clock.
s = "01:00:00"
h, m, s = s.split(":")
new_hours = (int(h) - 2) % 24
result = ':'.join((str(new_hours).zfill(2), m, s))
convert to datetime:
import datetime
a1 = "06:00:00"
obj = datetime.datetime.strptime(a1,"%H:%M:%S")
obj.replace(hour=obj.hour-2) #hours = hours - 2
tostr = obj.hour+":"+obj.min+":"+obj.second
print(tostr)
If your strings are always going to follow that exact format and you don't want to use datetime, here's a different way to do it: You could split the strings by their colons to isolate the hours, then work on them that way before joining back to a string.
a1 = "06:00:00"
parts = a1.split(":") # split by colons
hour = (int(parts[0]) - 2) % 24 # isolate hour, convert to int, and subtract hours, and clamp to our 0-23 bounds
parts[0] = f"{hour:02}" # :02 in an f-string specifies that you want to zero-pad that string up to a maximum of 2 characters
a1_new = ":".join(parts) # rejoin string to get new time
If there's any uncertainty in the format of the string however, this completely falls apart.
Convert to datetime, subtract timedelta, convert to string.
from datetime import datetime, timedelta
olds = ["06:00:00", "01:00:00"]
objs = [datetime.strptime(t, "%H:%M:%S") - timedelta(hours=2) for t in olds]
news = [t.strftime("%H:%M:%S") for t in objs]
You can use datetime and benefit from the parameters of datetime.timedelta:
from datetime import datetime, timedelta
def subtime(t, **kwargs):
return (datetime.strptime(t, "%H:%M:%S") # convert to datetime
- timedelta(**kwargs) # subtract parameters passed to function
).strftime("%H:%M:%S") # format as text again
subtime('01:00:00', hours=2)
# '23:00:00'
subtime('01:00:00', hours=2, minutes=62)
# '21:58:00'
Kindly help below my query:
I got an estimated time from API server like below:
2019-09-25T20:11:23+08:00
it seems like iso 8601 standard with timezone.
I would like to know how to calculate how many days, hours, minutes and seconds left from above value to the current time.
import datetime
Receved_time_frim_API = "2019-09-25T20:11:23+08:00"
Current_time = datetime.datetime.now()
left_days =
left_hour =
left_min =
left_sec =
Your time string contains timezone info. According to https://stackoverflow.com/a/13182163/12112986 it's easy to convert it to datetime object in python 3.7
import datetime
received = datetime.datetime.fromisoformat(Receved_time_frim_API)
In previous versions there is no easy oneliner to convert string with timezone to datetime object. If you're using earlier python version, you can try something crude, like
>>> date, timezone = Receved_time_frim_API.split("+")
>>> tz_hours, tz_minutes = timezone.split(":")
>>> date = datetime.datetime.strptime(date, "%Y-%m-%dT%H:%M:%S")
>>> date -= datetime.timedelta(hours=int(tz_hours))
>>> date -= datetime.timedelta(minutes=int(tz_minutes))
Note that this will work only in case of positive timezones
To substract two datetime objects use
td = date - Current_time
left_days = td.days
left_hour = td.seconds // 3600
left_min = (td.seconds//60)%60
left_sec = td.seconds % 60
Okay first you need to parse the Receved_time_frim_API into datetime format:
from dateutil import parser
Receved_time_frim_API = parser.parse("2019-09-25T20:11:23+08:00")
But you can't just substract this from your Current_time, because datetime.now() is not aware of a timezone:
from datetime import timezone
Current_time = datetime.datetime.now().replace(tzinfo=timezone.utc)
print (Current_time-Receved_time_frim_API)
The result is a datetime.timedelta
I have a dataset file with a time variable in "seconds since 1981-01-01 00:00:00".
What I need is to convert this time into calendar date (YYYY-MM-DD HH:mm:ss).
I've seen a lot of different ways to do this for time since epoch (1970) (timestamp, calendar.timegm, etc) but I'm failing to do this with a different reference date.
I thought of doing something that is not pretty but it works:
time = 1054425600
st = (datetime.datetime(1981,1,1,0,0)-datetime.datetime(1970,1,1)).total_seconds()
datetime.datetime.fromtimestamp(st+time)
But any other way of doing this will be welcome!
You could try this:
from datetime import datetime, timedelta
t0 = datetime(1981, 1, 1)
seconds = 1000000
dt = t0 + timedelta(seconds=seconds)
print dt
# 1981-01-12 13:46:40
Here t0 is set to a datetime object representing the "epoch" of 1981-01-01. This is your reference datetime to which you can add a timedelta object initialised with an arbitrary number of seconds. The result is a datetime object representing the required date time.
N.B. this assumes UTC time
I have two times, a start and a stop time, in the format of 10:33:26 (HH:MM:SS). I need the difference between the two times. I've been looking through documentation for Python and searching online and I would imagine it would have something to do with the datetime and/or time modules. I can't get it to work properly and keep finding only how to do this when a date is involved.
Ultimately, I need to calculate the averages of multiple time durations. I got the time differences to work and I'm storing them in a list. I now need to calculate the average. I'm using regular expressions to parse out the original times and then doing the differences.
For the averaging, should I convert to seconds and then average?
Yes, definitely datetime is what you need here. Specifically, the datetime.strptime() method, which parses a string into a datetime object.
from datetime import datetime
s1 = '10:33:26'
s2 = '11:15:49' # for example
FMT = '%H:%M:%S'
tdelta = datetime.strptime(s2, FMT) - datetime.strptime(s1, FMT)
That gets you a timedelta object that contains the difference between the two times. You can do whatever you want with that, e.g. converting it to seconds or adding it to another datetime.
This will return a negative result if the end time is earlier than the start time, for example s1 = 12:00:00 and s2 = 05:00:00. If you want the code to assume the interval crosses midnight in this case (i.e. it should assume the end time is never earlier than the start time), you can add the following lines to the above code:
if tdelta.days < 0:
tdelta = timedelta(
days=0,
seconds=tdelta.seconds,
microseconds=tdelta.microseconds
)
(of course you need to include from datetime import timedelta somewhere). Thanks to J.F. Sebastian for pointing out this use case.
Try this -- it's efficient for timing short-term events. If something takes more than an hour, then the final display probably will want some friendly formatting.
import time
start = time.time()
time.sleep(10) # or do something more productive
done = time.time()
elapsed = done - start
print(elapsed)
The time difference is returned as the number of elapsed seconds.
Here's a solution that supports finding the difference even if the end time is less than the start time (over midnight interval) such as 23:55:00-00:25:00 (a half an hour duration):
#!/usr/bin/env python
from datetime import datetime, time as datetime_time, timedelta
def time_diff(start, end):
if isinstance(start, datetime_time): # convert to datetime
assert isinstance(end, datetime_time)
start, end = [datetime.combine(datetime.min, t) for t in [start, end]]
if start <= end: # e.g., 10:33:26-11:15:49
return end - start
else: # end < start e.g., 23:55:00-00:25:00
end += timedelta(1) # +day
assert end > start
return end - start
for time_range in ['10:33:26-11:15:49', '23:55:00-00:25:00']:
s, e = [datetime.strptime(t, '%H:%M:%S') for t in time_range.split('-')]
print(time_diff(s, e))
assert time_diff(s, e) == time_diff(s.time(), e.time())
Output
0:42:23
0:30:00
time_diff() returns a timedelta object that you can pass (as a part of the sequence) to a mean() function directly e.g.:
#!/usr/bin/env python
from datetime import timedelta
def mean(data, start=timedelta(0)):
"""Find arithmetic average."""
return sum(data, start) / len(data)
data = [timedelta(minutes=42, seconds=23), # 0:42:23
timedelta(minutes=30)] # 0:30:00
print(repr(mean(data)))
# -> datetime.timedelta(0, 2171, 500000) # days, seconds, microseconds
The mean() result is also timedelta() object that you can convert to seconds (td.total_seconds() method (since Python 2.7)), hours (td / timedelta(hours=1) (Python 3)), etc.
This site says to try:
import datetime as dt
start="09:35:23"
end="10:23:00"
start_dt = dt.datetime.strptime(start, '%H:%M:%S')
end_dt = dt.datetime.strptime(end, '%H:%M:%S')
diff = (end_dt - start_dt)
diff.seconds/60
This forum uses time.mktime()
Structure that represent time difference in Python is called timedelta. If you have start_time and end_time as datetime types you can calculate the difference using - operator like:
diff = end_time - start_time
you should do this before converting to particualr string format (eg. before start_time.strftime(...)). In case you have already string representation you need to convert it back to time/datetime by using strptime method.
I like how this guy does it — https://amalgjose.com/2015/02/19/python-code-for-calculating-the-difference-between-two-time-stamps.
Not sure if it has some cons.
But looks neat for me :)
from datetime import datetime
from dateutil.relativedelta import relativedelta
t_a = datetime.now()
t_b = datetime.now()
def diff(t_a, t_b):
t_diff = relativedelta(t_b, t_a) # later/end time comes first!
return '{h}h {m}m {s}s'.format(h=t_diff.hours, m=t_diff.minutes, s=t_diff.seconds)
Regarding to the question you still need to use datetime.strptime() as others said earlier.
Try this
import datetime
import time
start_time = datetime.datetime.now().time().strftime('%H:%M:%S')
time.sleep(5)
end_time = datetime.datetime.now().time().strftime('%H:%M:%S')
total_time=(datetime.datetime.strptime(end_time,'%H:%M:%S') - datetime.datetime.strptime(start_time,'%H:%M:%S'))
print total_time
OUTPUT :
0:00:05
import datetime as dt
from dateutil.relativedelta import relativedelta
start = "09:35:23"
end = "10:23:00"
start_dt = dt.datetime.strptime(start, "%H:%M:%S")
end_dt = dt.datetime.strptime(end, "%H:%M:%S")
timedelta_obj = relativedelta(start_dt, end_dt)
print(
timedelta_obj.years,
timedelta_obj.months,
timedelta_obj.days,
timedelta_obj.hours,
timedelta_obj.minutes,
timedelta_obj.seconds,
)
result:
0 0 0 0 -47 -37
Both time and datetime have a date component.
Normally if you are just dealing with the time part you'd supply a default date. If you are just interested in the difference and know that both times are on the same day then construct a datetime for each with the day set to today and subtract the start from the stop time to get the interval (timedelta).
Take a look at the datetime module and the timedelta objects. You should end up constructing a datetime object for the start and stop times, and when you subtract them, you get a timedelta.
you can use pendulum:
import pendulum
t1 = pendulum.parse("10:33:26")
t2 = pendulum.parse("10:43:36")
period = t2 - t1
print(period.seconds)
would output:
610
import datetime
day = int(input("day[1,2,3,..31]: "))
month = int(input("Month[1,2,3,...12]: "))
year = int(input("year[0~2020]: "))
start_date = datetime.date(year, month, day)
day = int(input("day[1,2,3,..31]: "))
month = int(input("Month[1,2,3,...12]: "))
year = int(input("year[0~2020]: "))
end_date = datetime.date(year, month, day)
time_difference = end_date - start_date
age = time_difference.days
print("Total days: " + str(age))
Concise if you are just interested in the time elapsed that is under 24 hours. You can format the output as needed in the return statement :
import datetime
def elapsed_interval(start,end):
elapsed = end - start
min,secs=divmod(elapsed.days * 86400 + elapsed.seconds, 60)
hour, minutes = divmod(min, 60)
return '%.2d:%.2d:%.2d' % (hour,minutes,secs)
if __name__ == '__main__':
time_start=datetime.datetime.now()
""" do your process """
time_end=datetime.datetime.now()
total_time=elapsed_interval(time_start,time_end)
Usually, you have more than one case to deal with and perhaps have it in a pd.DataFrame(data) format. Then:
import pandas as pd
df['duration'] = pd.to_datetime(df['stop time']) - pd.to_datetime(df['start time'])
gives you the time difference without any manual conversion.
Taken from Convert DataFrame column type from string to datetime.
If you are lazy and do not mind the overhead of pandas, then you could do this even for just one entry.
Here is the code if the string contains days also [-1 day 32:43:02]:
print(
(int(time.replace('-', '').split(' ')[0]) * 24) * 60
+ (int(time.split(' ')[-1].split(':')[0]) * 60)
+ int(time.split(' ')[-1].split(':')[1])
)