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

objects?

##
Answer #1:

```
>>> import datetime
>>> first_time = datetime.datetime.now()
>>> later_time = datetime.datetime.now()
>>> difference = later_time - first_time
>>> seconds_in_day = 24 * 60 * 60
datetime.timedelta(0, 8, 562000)
>>> 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.

##
Answer #2:

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
```

##
Answer #3:

**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 200?s 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
```

##
Answer #4:

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)
```

You can convert `dd.days`

, `dd.seconds`

and `dd.microseconds`

to minutes.

##
Answer #5:

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.

##
Answer #6:

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)
```

##
Answer #7:

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)
```

##
Answer #8:

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
```