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])
)
Related
I have a method that converts a string to a datetime object using strptime("%I:%M %p"), I only want the hours in 24 and minutes without any dates, because I will get the difference between this time and another time. The problem is that when I try to get the difference with total_seconds(), it gets difference in negative because the date in the strptime is "1900-01-01". Does any one have any ideas how to solve this?
My Code:
fTime = datetime.strptime(time, "%I:%M %p")
if 0 < (fTime - datetime.now()).total_seconds() <= 3600:
return True
You can take one of two approaches: strip the date out of now, or add the current date to fTime. The first approach makes little sense, since you can't compare time objects like that anyway.
To convert fTime to a proper datetime, datetime.combine it with date.today():
fDate = datetime.combine(date.today(), fTime.time())
return 0 < (fDate - datetime.now()).total_seconds() <= 3600
Alternatively, you can replace the date portion:
today = date.today()
fDate = fTime.replace(year=today.year, month=today.month, day=today.day)
Personally, I would go with combine because it's less awkward code.
I want to add hours to a datetime and use:
date = date_object + datetime.timedelta(hours=6)
Now I want to add a time:
time='-7:00' (string) plus 4 hours.
I tried hours=time+4 but this doesn't work. I think I have to int the string like int(time) but this doesn't work either.
Better you parse your time like below and access datetime attributes for getting time components from the parsed datetime object
input_time = datetime.strptime(yourtimestring,'yourtimeformat')
input_seconds = input_time.second # for seconds
input_minutes = input_time.minute # for minutes
input_hours = input_time.hour # for hours
# Usage: input_time = datetime.strptime("07:00","%M:%S")
Rest you have datetime.timedelta method to compose the duration.
new_time = initial_datetime + datetime.timedelta(hours=input_hours,minutes=input_minutes,seconds=input_seconds)
See docs strptime
and datetime format
You need to convert to a datetime object in order to add timedelta to your current time, then return it back to just the time portion.
Using date.today() just uses the arbitrary current date and sets the time to the time you supply. This allows you to add over days and reset the clock to 00:00.
dt.time() prints out the result you were looking for.
from datetime import date, datetime, time, timedelta
dt = datetime.combine(date.today(), time(7, 00)) + timedelta(hours=4)
print dt.time()
Edit:
To get from a string time='7:00' to what you could split on the colon and then reference each.
this_time = this_time.split(':') # make it a list split at :
this_hour = this_time[0]
this_min = this_time[1]
Edit 2:
To put it all back together then:
from datetime import date, datetime, time, timedelta
this_time = '7:00'
this_time = this_time.split(':') # make it a list split at :
this_hour = int(this_time[0])
this_min = int(this_time[1])
dt = datetime.combine(date.today(), time(this_hour, this_min)) + timedelta(hours=4)
print dt.time()
If you already have a full date to use, as mentioned in the comments, you should convert it to a datetime using strptime. I think another answer walks through how to use it so I'm not going to put an example.
In this thread we have a good solution
How to calculate the time interval between two time strings
But how I can edit the output format?
From H:MM:SS to HH:MM, like 00:40 instead of 0:40:00
from datetime import datetime
from datetime import timedelta
T1 = input()
T2 = input()
format = '%H:%M'
tdiff = datetime.strptime(T1, format) - datetime.strptime(T2, format)
if tdiff.days < 0:
tdiff = timedelta(days = 0,
seconds = tdiff.seconds, microseconds = tdiff.microseconds)
print(tdiff)
EDIT: Thank you. This code below quite works, but the only problem is 7:45 instead of 07:45 for example. The code below removes the problem of seconds at the end. Now I only don't know how to force 0 is front of <10 hours format.
from datetime import datetime
from datetime import timedelta
T1 = input()
T2 = input()
format = '%H:%M'
tdiff = datetime.strptime(T1, format) - datetime.strptime(T2, format)
if tdiff.days < 0:
tdiff = timedelta(days = 0,
seconds = tdiff.seconds, microseconds = tdiff.microseconds)
print(str(tdiff).rstrip("0").rstrip(":"))
You're contradicting yourself by saying from H:MM:SS to HH:MM but you give an example of where you remove hours not seconds. Anyway, I'm guessing what you mean is that you want to remove any leading zeroes if they're empty (remove hours)?
In that case, there's two options. Either build the string yourself like so:
from datetime import datetime
from datetime import timedelta
T1 = '00:50'
T2 = '00:40'
format = '%H:%M'
tdiff = datetime.strptime(T1, format) - datetime.strptime(T2, format)
if tdiff.days < 0:
tdiff = timedelta(days = 0, seconds = tdiff.seconds, microseconds = tdiff.microseconds)
print(':'.join([block for block in str(tdiff).split(':') if block != '0']))
Where you skip single-digit zeroes (hours) but leave double-digit zeroes (seconds etc) intact.
Or you could do:
print(str(tdiff).lstrip("0").lstrip(":"))
Which simply strips the beginning. replace with .rstrip() if you need to strip the end.
And if you want to make sure there's always a two-digit representation in the first example, you could do:
print(':'.join(['{:02}'.format(int(block)) for block in str(tdiff).split(':') if block != '0']))
At this point, you're monkey-patching a problem that most likely could be solved in a neater and more efficient way, and your original problem description is not really in line with what you're asking/trying to fix. But this is one way of doing it.
If you want to remove seconds but keep hours intact, do something like:
print(':'.join(['{:02}'.format(int(block)) for block in str(tdiff).split(':')[:-1]]))
Python noob here
from datetime import datetime, time
now = datetime.now()
now_time = now.time()
if now_time >= time(10,30) and now_time <= time(13,30):
print "yes, within the interval"
I would like the timer to work between 10,30 AM today and 10 AM the next day. Changing time(13,30) to time(10,00) will not work, because I need to tell python 10,00 is the next day. I should use datetime function but don't know how. Any tips or examples appreciated.
The combine method on the datetime class will help you a lot, as will the timedelta class. Here's how you would use them:
from datetime import datetime, timedelta, date, time
today = date.today()
tomorrow = today + timedelta(days=1)
interval_start = datetime.combine(today, time(10,30))
interval_end = datetime.combine(tomorrow, time(10,00))
time_to_check = datetime.now() # Or any other datetime
if interval_start <= time_to_check <= interval_end:
print "Within the interval"
Notice how I did the comparison. Python lets you "nest" comparisons like that, which is usually more succinct than writing if start <= x and x <= end.
P.S. Read https://docs.python.org/2/library/datetime.html for more details about these classes.
Consider this:
from datetime import datetime, timedelta
now = datetime.now()
today_10 = now.replace(hour=10, minute=30)
tomorrow_10 = (now + timedelta(days=1)).replace(hour=10, minute=0)
if today_10 <= now <= tomorrow_10:
print "yes, within the interval"
The logic is to create 3 datetime objects: one for today 10 AM, one for right now and one for tomorrow 10 AM. Them simply checking for the condition.
An alternative to creating time objects for the sake of comparison is to simply query the hour and minute attributes:
now= datetime.now().time()
if now.hour<10 or now.hour>10 or (now.hour==10 and now.minute>30):
print('hooray')
I want to compare two times and if the new time is more than 2min then the if statement will print output, I can get the output of datetime.datetime.now() , but how do I check whether the old time is less than 2mins?
#!/usr/bin/env python
import datetime
from time import sleep
now = datetime.datetime.now()
sleep(2)
late = datetime.datetime.now()
constant = 2
diff = late-now
if diff <= constant:
print "True time is less than 2min"
else:
print "Time exceeds 2 mins"
any ideas?
UPDATED:
I am now storing the old date as string in file and then subtract it from current time, the old date is stored in the format
2011-12-16 16:14:50.800856
so when I do
now = "2011-12-16 16:14:50.838638"
sleep(2)
nnow = datetime.strptime(now, '%Y-%m-%d %H:%M:%S')
late = datetime.now()
diff = late-nnow
it gives me this error
ValueError: unconverted data remains: .838638
Subtracting two datetime instances returns a timedelta that has a total_seconds method:
contant = 2 * 60
diff = late-now
if diff.total_seconds() <= constant:
This is only an answer to the update since the answer from sje397 was perfect.
Use a format string like this to match the whole time string:
nnow = datetime.strptime(now, '%Y-%m-%d %H:%M:%S.%f')
The %f matches the microseconds after the dot. This is new since Python 2.6.
You could compare datetime objects by themselves:
from datetime import datetime, timedelta
ts = datetime.strptime("2011-12-16 16:14:50.838638Z", '%Y-%m-%d %H:%M:%S.%fZ')
ts += timedelta(minutes=2) # add 2 minutes
if datetime.utcnow() < ts:
print("time is less")
else:
print("time is more or equal")