Example:
9:43 - 17:27 - how many hours and minutes elapsed between those times ?
Here is one approach to get the number of total minutes:
from datetime import datetime
s = '9:30 - 14:00 ; 14:30 - 16:30'
sum(((b-a).total_seconds()/60 for x in s.split(' ; ')
for a,b in [list(map(lambda t: datetime.strptime(t, '%H:%M'), x.split(' - ')))]))
Output: 390.0
If you know that the time periods will never span midnight, then you could simply split the time strings with time.split(":") and do the math yourself with the hours and minutes.
However, the correct solution would be to import the datetime module and calculate the timedelta.
This example could be condensed. I intentionally made it verbose without knowing exactly how you're getting your inputs:
from datetime import datetime
times = [
"9:30",
"14:00",
"14:30",
"16:30"
]
#Just using today's date to fill in the values with assumption all times are on the same day.
year = 2022
month = 6
day = 9
date_times = []
for time in times:
split_time = time.split(":")
hour = split_time[0]
minutes = split_time[1]
timestamp = datetime.datetime.today(year=year, month=month, day=day, hour=hour, min=minutes)
date_times.append(timestamp)
total_seconds = 0
for i in range(1, len(date_times), 2):
delta = date_times[i] - date_times[i-1] # The timedelta object returned will have days, seconds, milliseconds
total_seconds += delta.days * 86400 + delta.seconds
hours = total_seconds // 3600 # Integer division
minutes = round((total_seconds % 3600) / 60) # Change depending on if you want to round to nearest, or always up or down.
I'm given time in millis and I can convert that to date obj with
currentDate = datetime.datetime.fromtimestamp(time_stamp/1000.0).date()
I'm trying to get this time as a percentage of that given day so say its 12:00pm I'd like to get 50% as output.
Here's my attempt
currentDate = datetime.datetime.fromtimestamp(time_stamp/1000.0).date()
NextDay_Date = currentDate + datetime.timedelta(days=1)
start_of_next_day=NextDay_Date.replace(hour=00, minute=00)
difference=start_of_next_day.timestamp()*1000 -currentDate.timestamp()*1000
milis_in_a_day=86400000
percent_of_day=difference/milis_in_a_day
There has to be a more elegant solution.
You could just count the number of microseconds passed today, by just using the time part of the date.
from datetime import datetime
today = datetime.now()
microseconds_today = (
(
(
(today.hour * 60) + today.minute
) * 60 + today.second
) * 1_000_000 + today.microsecond
)
print(microseconds_today / 86_400_000_000)
You're basically on the right path, only thing I could think of to be a bit more "readable" is to use timedelta from datetime. Note the fully qualified names used below - datetime in particular can be a bit confusing to read, definitely suggest referencing the documentation: https://docs.python.org/3/library/datetime.html
import datetime
TOTAL_DAY_SECS = 86400.0
def timedelta_percentage(input_datetime):
d = input_datetime - datetime.datetime.combine(input_datetime.date(), datetime.time())
return d.total_seconds() / TOTAL_DAY_SECS
now = datetime.datetime.now()
print(f"'datetime.now()': {now}")
answer = round(timedelta_percentage(now) * 100, 2)
print(f"Percentage of day complete, based on 'now': {answer}%")
Output:
'datetime.now()': 2021-04-23 09:38:16.026000
Percentage of day complete, based on 'now': 40.16%
You could certainly adapt this to be more precise using microseconds, depending how granular of an answer you need.
I expect that if I substract two datetimes in Python, I'll get datetime with substracted days, weeks, etc...
Here is my sample. What I get are just substracted hours, minutes and seconds. Date variable is taken from database. On type() function returns datetime.datetime.
def elapsed_time(date):
"""
Custom filter that format time to "x time age".
:param date:
:return:
"""
if date is None:
return 'No time given'
now = datetime.datetime.now()
elapsed = (now - date).seconds
if elapsed < 60:
return '{} seconds ago'.format(elapsed)
elif elapsed < 3600:
return '{} minutes ago'.format(int(elapsed / 60))
elif elapsed < 86400:
return '{} hours ago'.format(int(elapsed / 3600))
else:
return '{} days ago'.format((elapsed / 86400))
My current example:
Given datetime is 2017-07-27 01:18:58.398231
Current datetime is 2017-07-31 20:23:36.095440
Result is 19 hours (68677 seconds)
The following line returns only the 'seconds' component of the difference and does not take into account the days/hours/minutes components of it.
elapsed = (now - date).seconds
What you need is to use total_seconds() instead of just seconds since that's what you're trying to compare in subsequent conditions. Use it as follows:
elapsed = (now - date).total_seconds()
The rest of the code remains the same and you will get your desired output.
If you subtract two datetime objects the result will be a timedelta
import datetime as dt
import time
t1 = dt.datetime.now()
time.sleep(4)
t2 = dt.datetime.now()
dt1 = t2 - t1
print(dt1)
print(dt1.total_seconds())
print(type(dt1))
dt2 = t1 - t2
print(dt2.total_seconds())
print(dt2)
print(type(dt2))
If the second timestep was earlier than the first one, the results can be irritating. See negative day in example.
dt.seconds is only a part of the result, you are looking for
timedelta.total_seconds()
How can I parse the time 004:32:55 into a datetime object? This:
datetime.strptime("004:32:55", "%H:%M:%S")
doesn't work becaush %H expects two digits. Any way to add the third digit?
Three options :
s = "004:32:55"
from datetime import datetime
datetime.strptime(s[1:], "%H:%M:%S") # from second 0 to the end
datetime.strptime(s, "0%H:%M:%S") # add 0 in formatting
from dateutil import parser # use dateutil
parser.parse(s)
There are 24 hours in a day so you can divide and get the modulus to figure out how many days and hours are in the time, then subtract the days, this needs some tweaking will get you started.
s = "04:32:55"
s = s.split(":",1)
hours, days = divmod(int(s[0]),24)
new_time = "{}:{}".format(hours,s[1])
past = datetime.now() - timedelta(days=days)
final = "{} {}".format(past.date().isoformat(),new_time)
print datetime.strptime(final,"%Y-%m-%d %H:%M:%S")
I chose a more pragmatic approach in the end and converted the time stamp to seconds:
hours = (0 if time_string.split(":")[0] == "000" else int(time_string.split(":")[0].lstrip("0")) * 3600)
mins = (0 if time_string.split(":")[1] == "00" else int(time_string.split(":")[1].lstrip("0")) * 60)
secs = (0 if time_string.split(":")[2] == "00" else int(time_string.split(":")[2].lstrip("0")))
return hours + mins + secs
Converting back to hours, minutes, and seconds is easy with datetime.timedelta(seconds=123).
EDIT:
A better solution (thanks to Ben):
hours = int(time_string.split(":")[0]) * 3600
mins = int(time_string.split(":")[1]) * 60
secs = int(time_string.split(":")[2])
return hours + mins + secs
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])
)