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Transform date while loop into for loop

I have following while loop in Python:
start_date: datetime.date = utcnow().date() - datetime.timedelta(
days=30
)
end_date: datetime.date = utcnow().date()
DELTA_1D: datetime.timedelta = datetime.timedelta(days=1)
while start_date <= end_date:
# some code
start_date += DELTA_1D
But I want to transform it to a for loop. What's the best implementation for this?

Work out he number of days between by simply subtracting
from datetime import date, timedelta
d0 = date(2008, 8, 18)
d1 = date(2008, 9, 26)
delta = d1 - d0
for i in range(0,delta.days):
date_to_handle = d0 + timedelta(days=i)

Convert date format yyyy-m-d into yyyy-mm-dd on Python

In my table, I have different types of dates just with numbers and in this two formats:
yyyy-m-d
yyyy-mm-dd
Some values, as the month for example, don't have the zero in the case of months under 10 and I need it to create a condition to chose elements by the latest date.
I want that all of them have the same format:
yyyy-mm-dd
Any pythonic way to solve that?
For the moment I am using this:
if line.startswith('# Date: '):
#date = 2014-5-28
d = line.strip().split(':')[-1].split('-').replace(' ','')
if len(d[0]) == 4:
year = str(d[0])
elif len(d[1]) < 2:
month = '0'+ str(d[1])
elif len(d[2]< 2):
day = '0'+ str(d[1])
date = year + month + day

You can use the python inbuilt datetime module
import datetime
date1 = "2018-1-1"
date2 = "2018-01-01"
datetime_object = datetime.datetime.strptime(date1, "%Y-%m-%d")
datetime_object2 = datetime.datetime.strptime(date2, "%Y-%m-%d")
print datetime_object.strftime("%Y-%m-%d")
print datetime_object2.strftime("%Y-%m-%d")
Result:
2018-01-01
2018-01-01

Try the below code !
You have to import the date time file .
Input :
import datetime
date1 = datetime.datetime.strptime("2015-1-3", "%Y-%m-%d").strftime("%d-%m-%Y")
print(date1)
today = datetime.date.today().strftime("%d-%m-%Y")
print(today)
Output :
03-01-2015
17-01-2018

You can try:
>>> d = "2018-1-1"
>>> d_list = d.split("-")
>>> d_list
['2018', '1', '1']
>>> if len(d_list[1]) < 2:
d_list[1] = "0"+d_list[1]
>>> if len(d_list[2]) < 2:
d_list[2] = "0"+d_list[2]
>>> d_list
['2018', '01', '01']

This helps
import datetime
d = datetime.datetime.strptime('2014-5-28', '%Y-%m-%d')
d.strftime('%Y-%m-%d')

This should work as well:
from datetime import datetime
d1 = "2001-1-1"
d2 = "2001-01-01"
d1 = datetime.strptime(d1, '%Y-%m-%d')
d1 = d1.strftime('%Y-%m-%d')
print(d1)
d2 = datetime.strptime(d2, '%Y-%m-%d')
d2 = d2.strftime('%Y-%m-%d')
print(d2)
Results:
2001-01-01
2001-01-01

May be this will help:
Data:
de = ["2018-1-1", "2018-02-1", "2017-3-29"]
Function:
from datetime import datetime
def format_date(d):
"""Format string representing date to format YYYY-MM-DD"""
dl = d.split("-")
return '{:%Y-%m-%d}'.format(datetime(int(dl[0]),int(dl[1]),int(dl[2])))
print([format_date(i) for i in de])
Result:
['2018-1-1', '2018-02-1', '2017-3-29']

Weeks difference between two dates in python

How do I get the differences between two valid dates in weeks. I have googled many, but none are the one that I have been looking for
Say I have two dates:
02-Dec-2016 and 10-Jan-2017.
I want it to provide me with output like following
02-Dec-2016 - 04-Dec-2016 (2 days) (2 days before monday comes)
05-Dec-2016 - 08-Jan-2017 (5 weeks) (starts from monday-sunday)
08-Jan-2017 - 10-Jan-2017 (2 days) (2 days after monday has gone)

This is what you actualy want:
import datetime
def diff(d1, d2):
result = []
delta = datetime.timedelta(days=0)
day = datetime.timedelta(days=1)
while d1.weekday() != 0:
d1 += day
delta += day
result.append((d1 - delta, d1 - day))
weeks, days = divmod((d2 - d1).days, 7)
d3 = d1 + datetime.timedelta(weeks=weeks)
d4 = d3 + datetime.timedelta(days=days)
result.append((d1, d3 - day))
result.append((d3, d4))
return result
d1 = datetime.date(2016, 12, 2)
d2 = datetime.date(2017, 01, 10)
for i,j in diff(d1,d2):
print '{} - {} ({} days)'.format(datetime.datetime.strftime(i, "%d-%b-%Y"), datetime.datetime.strftime(j, "%d-%b-%Y"), (j-i).days + 1)
# 02-Dec-2016 - 04-Dec-2016 (3 days)
# 05-Dec-2016 - 08-Jan-2017 (35 days)
# 09-Jan-2017 - 10-Jan-2017 (2 days)

It is somewhat surprising how complicated it is to compute the difference between two times in Python. The following code is for differences in minutes, but you can modify this to weeks or other attributes.
# Compute the difference between two time values
import datetime
df = pd.DataFrame({'ATime1' : ['8/26/2016 10:00','8/26/2016 10:01','8/26/2016 10:02','8/26/2016 10:03'], 'BBB' : [10,20,30,40],'CCC' : [100,50,-30,-50], 'ATime2' : ['8/26/2016 10:01','8/26/2016 10:02','8/26/2016 10:03','8/26/2016 10:04']})
s1 = pd.Series(df['ATime1']) # Select one column of the dataframe and convert to a Series
s2 = pd.Series(df['ATime2'])
s1 = pd.to_datetime(s1) # Convert the Series object values to datetime values
s2 = pd.to_datetime(s2)
m1 = s1.dt.minute # Select minutes from the datetime values
m2 = s2.dt.minute
t1 = m1.loc[1] # Select the first minutes value in the column
t2 = m2.loc[1]
t1 = int(t1) # Convert minutes to integer
t2 = int(t2)
diff = t2 - t1
if t2 > t1:
print "ATime2 starts later than Atime1 by ", diff, " minute(s)."
else:
print "ATime1 starts later than Atime2 by ", diff, " minute(s)."
print t1, t2

Hour deltatime mistake

now = datetime.now()
d1 = datetime(now.year, now.month, now.day, now.hour, now.minute, 0)
if now.minute in xrange(46, 60):
res = 0
print now.hour
print now.hour+1
d1 = d1 + timedelta(hours=now.hour+1)
print d1
now.hour prints, for example, 15. Second line shows up 16 but d1 after adding timedelta becomes: 2012-07-21 07:57:00. This date is next day at 7am.
What's wrong?

You create a timedelta with a value of 16 hours and then add that to d1. d1 (if already at 15 hours) plus 16 hours will be 07 hours the next day.
You don't say what you are trying to achieve but if you are trying to increment by one hour then you should use d1 = d1 + timedelta(hours=1);

How do I find the time difference between two datetime objects in python?

How do I tell the time difference in minutes between two datetime objects?

>>> import datetime
>>> first_time = datetime.datetime.now()
>>> later_time = datetime.datetime.now()
>>> difference = later_time - first_time
datetime.timedelta(0, 8, 562000)
>>> seconds_in_day = 24 * 60 * 60
>>> divmod(difference.days * seconds_in_day + difference.seconds, 60)
(0, 8) # 0 minutes, 8 seconds
Subtracting the later time from the first time difference = later_time - first_time creates a datetime object that only holds the difference.
In the example above it is 0 minutes, 8 seconds and 562000 microseconds.

Using datetime example
>>> from datetime import datetime
>>> then = datetime(2012, 3, 5, 23, 8, 15) # Random date in the past
>>> now = datetime.now() # Now
>>> duration = now - then # For build-in functions
>>> duration_in_s = duration.total_seconds() # Total number of seconds between dates
Duration in years
>>> years = divmod(duration_in_s, 31536000)[0] # Seconds in a year=365*24*60*60 = 31536000.
Duration in days
>>> days = duration.days # Build-in datetime function
>>> days = divmod(duration_in_s, 86400)[0] # Seconds in a day = 86400
Duration in hours
>>> hours = divmod(duration_in_s, 3600)[0] # Seconds in an hour = 3600
Duration in minutes
>>> minutes = divmod(duration_in_s, 60)[0] # Seconds in a minute = 60
Duration in seconds
[!] See warning about using duration in seconds in the bottom of this post
>>> seconds = duration.seconds # Build-in datetime function
>>> seconds = duration_in_s
Duration in microseconds
[!] See warning about using duration in microseconds in the bottom of this post
>>> microseconds = duration.microseconds # Build-in datetime function
Total duration between the two dates
>>> days = divmod(duration_in_s, 86400) # Get days (without [0]!)
>>> hours = divmod(days[1], 3600) # Use remainder of days to calc hours
>>> minutes = divmod(hours[1], 60) # Use remainder of hours to calc minutes
>>> seconds = divmod(minutes[1], 1) # Use remainder of minutes to calc seconds
>>> print("Time between dates: %d days, %d hours, %d minutes and %d seconds" % (days[0], hours[0], minutes[0], seconds[0]))
or simply:
>>> print(now - then)
Edit 2019
Since this answer has gained traction, I'll add a function, which might simplify the usage for some
from datetime import datetime
def getDuration(then, now = datetime.now(), interval = "default"):
# Returns a duration as specified by variable interval
# Functions, except totalDuration, returns [quotient, remainder]
duration = now - then # For build-in functions
duration_in_s = duration.total_seconds()
def years():
return divmod(duration_in_s, 31536000) # Seconds in a year=31536000.
def days(seconds = None):
return divmod(seconds if seconds != None else duration_in_s, 86400) # Seconds in a day = 86400
def hours(seconds = None):
return divmod(seconds if seconds != None else duration_in_s, 3600) # Seconds in an hour = 3600
def minutes(seconds = None):
return divmod(seconds if seconds != None else duration_in_s, 60) # Seconds in a minute = 60
def seconds(seconds = None):
if seconds != None:
return divmod(seconds, 1)
return duration_in_s
def totalDuration():
y = years()
d = days(y[1]) # Use remainder to calculate next variable
h = hours(d[1])
m = minutes(h[1])
s = seconds(m[1])
return "Time between dates: {} years, {} days, {} hours, {} minutes and {} seconds".format(int(y[0]), int(d[0]), int(h[0]), int(m[0]), int(s[0]))
return {
'years': int(years()[0]),
'days': int(days()[0]),
'hours': int(hours()[0]),
'minutes': int(minutes()[0]),
'seconds': int(seconds()),
'default': totalDuration()
}[interval]
# Example usage
then = datetime(2012, 3, 5, 23, 8, 15)
now = datetime.now()
print(getDuration(then)) # E.g. Time between dates: 7 years, 208 days, 21 hours, 19 minutes and 15 seconds
print(getDuration(then, now, 'years')) # Prints duration in years
print(getDuration(then, now, 'days')) # days
print(getDuration(then, now, 'hours')) # hours
print(getDuration(then, now, 'minutes')) # minutes
print(getDuration(then, now, 'seconds')) # seconds
Warning: Caveat about built-in .seconds and .microseconds
datetime.seconds and datetime.microseconds are capped to [0,86400) and [0,10^6) respectively.
They should be used carefully if timedelta is bigger than the max returned value.
Examples:
end is 1h and 200μs after start:
>>> start = datetime(2020,12,31,22,0,0,500)
>>> end = datetime(2020,12,31,23,0,0,700)
>>> delta = end - start
>>> delta.microseconds
RESULT: 200
EXPECTED: 3600000200
end is 1d and 1h after start:
>>> start = datetime(2020,12,30,22,0,0)
>>> end = datetime(2020,12,31,23,0,0)
>>> delta = end - start
>>> delta.seconds
RESULT: 3600
EXPECTED: 90000

New at Python 2.7 is the timedelta instance method .total_seconds(). From the Python docs, this is equivalent to (td.microseconds + (td.seconds + td.days * 24 * 3600) * 10**6) / 10**6.
Reference: http://docs.python.org/2/library/datetime.html#datetime.timedelta.total_seconds
>>> import datetime
>>> time1 = datetime.datetime.now()
>>> time2 = datetime.datetime.now() # waited a few minutes before pressing enter
>>> elapsedTime = time2 - time1
>>> elapsedTime
datetime.timedelta(0, 125, 749430)
>>> divmod(elapsedTime.total_seconds(), 60)
(2.0, 5.749430000000004) # divmod returns quotient and remainder
# 2 minutes, 5.74943 seconds

Just subtract one from the other. You get a timedelta object with the difference.
>>> import datetime
>>> d1 = datetime.datetime.now()
>>> d2 = datetime.datetime.now() # after a 5-second or so pause
>>> d2 - d1
datetime.timedelta(0, 5, 203000)
>>> dd = d2 - d1
>>> print (dd.days) # get days
>>> print (dd.seconds) # get seconds
>>> print (dd.microseconds) # get microseconds
>>> print (int(round(dd.total_seconds()/60, 0))) # get minutes

If a, b are datetime objects then to find the time difference between them in Python 3:
from datetime import timedelta
time_difference = a - b
time_difference_in_minutes = time_difference / timedelta(minutes=1)
On earlier Python versions:
time_difference_in_minutes = time_difference.total_seconds() / 60
If a, b are naive datetime objects such as returned by datetime.now() then the result may be wrong if the objects represent local time with different UTC offsets e.g., around DST transitions or for past/future dates. More details: Find if 24 hrs have passed between datetimes - Python.
To get reliable results, use UTC time or timezone-aware datetime objects.

Use divmod:
now = int(time.time()) # epoch seconds
then = now - 90000 # some time in the past
d = divmod(now-then,86400) # days
h = divmod(d[1],3600) # hours
m = divmod(h[1],60) # minutes
s = m[1] # seconds
print '%d days, %d hours, %d minutes, %d seconds' % (d[0],h[0],m[0],s)

To just find the number of days: timedelta has a 'days' attribute. You can simply query that.
>>>from datetime import datetime, timedelta
>>>d1 = datetime(2015, 9, 12, 13, 9, 45)
>>>d2 = datetime(2015, 8, 29, 21, 10, 12)
>>>d3 = d1- d2
>>>print d3
13 days, 15:59:33
>>>print d3.days
13

Just thought it might be useful to mention formatting as well in regards to timedelta. strptime() parses a string representing a time according to a format.
from datetime import datetime
datetimeFormat = '%Y/%m/%d %H:%M:%S.%f'
time1 = '2016/03/16 10:01:28.585'
time2 = '2016/03/16 09:56:28.067'
time_dif = datetime.strptime(time1, datetimeFormat) - datetime.strptime(time2,datetimeFormat)
print(time_dif)
This will output:
0:05:00.518000

This is how I get the number of hours that elapsed between two datetime.datetime objects:
before = datetime.datetime.now()
after = datetime.datetime.now()
hours = math.floor(((after - before).seconds) / 3600)

To get the hour, minute and second, you can do this
>>> import datetime
>>> first_time = datetime.datetime.now()
>>> later_time = datetime.datetime.now()
>>> difference = later_time - first_time
>>> m, s = divmod(difference.total_seconds(), 60)
>>> print("H:M:S is {}:{}:{}".format(m//60, m%60, s))

I use somethign like this :
from datetime import datetime
def check_time_difference(t1: datetime, t2: datetime):
t1_date = datetime(
t1.year,
t1.month,
t1.day,
t1.hour,
t1.minute,
t1.second)
t2_date = datetime(
t2.year,
t2.month,
t2.day,
t2.hour,
t2.minute,
t2.second)
t_elapsed = t1_date - t2_date
return t_elapsed
# usage
f = "%Y-%m-%d %H:%M:%S+01:00"
t1 = datetime.strptime("2018-03-07 22:56:57+01:00", f)
t2 = datetime.strptime("2018-03-07 22:48:05+01:00", f)
elapsed_time = check_time_difference(t1, t2)
print(elapsed_time)
#return : 0:08:52

This will give the difference in seconds (then just divide by 60 to get minutes):
import time
import datetime
t_start = datetime.datetime.now()
time.sleep(10)
t_end = datetime.datetime.now()
elapsedTime = (t_end - t_start )
print(elapsedTime.total_seconds())
outputs:
10.009222
This is the simplest way in my opinion, and you don't need to worry about precision or overflow.
For instance, using elapsedTime.seconds you lose a lot of precision (it returns an integer). Also, elapsedTime.microseconds is capped at 10^6, as this answer pointed out. So, for example, for a 10 second sleep(), elapsedTime.microseconds gives 8325 (which is wrong, should be around 10,000,000).

this is to find the difference between current time and 9.30 am
t=datetime.now()-datetime.now().replace(hour=9,minute=30)

Based on #Attaque great answer, I propose a shorter simplified version of the datetime difference calculator:
seconds_mapping = {
'y': 31536000,
'm': 2628002.88, # this is approximate, 365 / 12; use with caution
'w': 604800,
'd': 86400,
'h': 3600,
'min': 60,
's': 1,
'mil': 0.001,
}
def get_duration(d1, d2, interval, with_reminder=False):
if with_reminder:
return divmod((d2 - d1).total_seconds(), seconds_mapping[interval])
else:
return (d2 - d1).total_seconds() / seconds_mapping[interval]
I've changed it to avoid declaring repetetive functions, removed the pretty print default interval and added support for milliseconds, weeks and ISO months (bare in mind months are just approximate, based on assumption that each month is equal to 365/12).
Which produces:
d1 = datetime(2011, 3, 1, 1, 1, 1, 1000)
d2 = datetime(2011, 4, 1, 1, 1, 1, 2500)
print(get_duration(d1, d2, 'y', True)) # => (0.0, 2678400.0015)
print(get_duration(d1, d2, 'm', True)) # => (1.0, 50397.12149999989)
print(get_duration(d1, d2, 'w', True)) # => (4.0, 259200.00149999978)
print(get_duration(d1, d2, 'd', True)) # => (31.0, 0.0014999997802078724)
print(get_duration(d1, d2, 'h', True)) # => (744.0, 0.0014999997802078724)
print(get_duration(d1, d2, 'min', True)) # => (44640.0, 0.0014999997802078724)
print(get_duration(d1, d2, 's', True)) # => (2678400.0, 0.0014999997802078724)
print(get_duration(d1, d2, 'mil', True)) # => (2678400001.0, 0.0004999997244524721)
print(get_duration(d1, d2, 'y', False)) # => 0.08493150689687975
print(get_duration(d1, d2, 'm', False)) # => 1.019176965856293
print(get_duration(d1, d2, 'w', False)) # => 4.428571431051587
print(get_duration(d1, d2, 'd', False)) # => 31.00000001736111
print(get_duration(d1, d2, 'h', False)) # => 744.0000004166666
print(get_duration(d1, d2, 'min', False)) # => 44640.000024999994
print(get_duration(d1, d2, 's', False)) # => 2678400.0015
print(get_duration(d1, d2, 'mil', False)) # => 2678400001.4999995

This is my approach using mktime.
from datetime import datetime, timedelta
from time import mktime
yesterday = datetime.now() - timedelta(days=1)
today = datetime.now()
difference_in_seconds = abs(mktime(yesterday.timetuple()) - mktime(today.timetuple()))
difference_in_minutes = difference_in_seconds / 60

In Other ways to get difference between date;
import dateutil.parser
import datetime
last_sent_date = "" # date string
timeDifference = current_date - dateutil.parser.parse(last_sent_date)
time_difference_in_minutes = (int(timeDifference.days) * 24 * 60) + int((timeDifference.seconds) / 60)
So get output in Min.
Thanks

I have used time differences for continuous integration tests to check and improve my functions. Here is simple code if somebody need it
from datetime import datetime
class TimeLogger:
time_cursor = None
def pin_time(self):
global time_cursor
time_cursor = datetime.now()
def log(self, text=None) -> float:
global time_cursor
if not time_cursor:
time_cursor = datetime.now()
now = datetime.now()
t_delta = now - time_cursor
seconds = t_delta.total_seconds()
result = str(now) + ' tl -----------> %.5f' % seconds
if text:
result += " " + text
print(result)
self.pin_time()
return seconds
time_logger = TimeLogger()
Using:
from .tests_time_logger import time_logger
class Tests(TestCase):
def test_workflow(self):
time_logger.pin_time()
... my functions here ...
time_logger.log()
... other function(s) ...
time_logger.log(text='Tests finished')
and i have something like that in log output
2019-12-20 17:19:23.635297 tl -----------> 0.00007
2019-12-20 17:19:28.147656 tl -----------> 4.51234 Tests finished

You may find this fast snippet useful in not so much long time intervals:
from datetime import datetime as dttm
time_ago = dttm(2017, 3, 1, 1, 1, 1, 1348)
delta = dttm.now() - time_ago
days = delta.days # can be converted into years which complicates a bit…
hours, minutes, seconds = map(int, delta.__format__('').split('.')[0].split(' ')[-1].split(':'))
tested on Python v.3.8.6

Here is an answer that is easy to generalise or turn into a function and which is reasonable compact and easy to follow.
ts_start=datetime(2020, 12, 1, 3, 9, 45)
ts_end=datetime.now()
ts_diff=ts_end-ts_start
secs=ts_diff.total_seconds()
days,secs=divmod(secs,secs_per_day:=60*60*24)
hrs,secs=divmod(secs,secs_per_hr:=60*60)
mins,secs=divmod(secs,secs_per_min:=60)
secs=round(secs, 2)
answer='Duration={} days, {} hrs, {} mins and {} secs'.format(int(days),int(hrs),int(mins),secs)
print(answer)
It gives an answer in the form Duration=270 days, 10 hrs, 32 mins and 42.13 secs

This might help someone, find is expired or not with this method its calculating with days. There are dt.seconds and dt.microseconds also available
from datetime import datetime
# updated_at = "2022-10-20T07:18:56.950563"
def is_expired(updated_at):
expires_in = 7 #days
datetime_format = '%Y-%m-%dT%H:%M:%S.%f'
time_difference = datetime.now() - datetime.strptime(updated_at, datetime_format)
return True if time_difference.days > expires_in else False

import datetime
date = datetime.date(1, 1, 1)
#combine a dummy date to the time
datetime1 = datetime.datetime.combine(date, start_time)
datetime2 = datetime.datetime.combine(date, stop_time)
#compute the difference
time_elapsed = datetime1 - datetime2
start_time --> start time for datetime object
end_time--> end time for datetime object
we cannot directly subtract the datetime.time objects
hence we need to add a random date to it (we use combine)
or you can use the "today" instead of (1,1,1)
hope this helps