Python Time + Timedelta Equivalent

python time + timedelta equivalent

The solution is in the link that you provided in your question:

datetime.combine(date.today(), time()) + timedelta(hours=1)

Full example:

from datetime import date, datetime, time, timedelta

dt = datetime.combine(date.today(), time(23, 55)) + timedelta(minutes=30)
print dt.time()

Output:

00:25:00

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

>>> 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.

What is the standard way to add N seconds to datetime.time in Python?

You can use full datetime variables with timedelta, and by providing a dummy date then using time to just get the time value.

For example:

import datetime
a = datetime.datetime(100,1,1,11,34,59)
b = a + datetime.timedelta(0,3) # days, seconds, then other fields.
print(a.time())
print(b.time())

results in the two values, three seconds apart:

11:34:59
11:35:02

You could also opt for the more readable

b = a + datetime.timedelta(seconds=3)

if you're so inclined.

---

If you're after a function that can do this, you can look into using addSecs below:

import datetime

def addSecs(tm, secs):
    fulldate = datetime.datetime(100, 1, 1, tm.hour, tm.minute, tm.second)
    fulldate = fulldate + datetime.timedelta(seconds=secs)
    return fulldate.time()

a = datetime.datetime.now().time()
b = addSecs(a, 300)
print(a)
print(b)

This outputs:

 09:11:55.775695
 09:16:55

How to create a DateTime equal to 15 minutes ago?

import datetime and then the magic timedelta stuff:

In [63]: datetime.datetime.now()
Out[63]: datetime.datetime(2010, 12, 27, 14, 39, 19, 700401)

In [64]: datetime.datetime.now() - datetime.timedelta(minutes=15)
Out[64]: datetime.datetime(2010, 12, 27, 14, 24, 21, 684435)

How do I get time from a datetime.timedelta object?

It's strange that Python returns the value as a datetime.timedelta. It probably should return a datetime.time. Anyway, it looks like it's returning the elapsed time since midnight (assuming the column in the table is 6:00 PM). In order to convert to a datetime.time, you can do the following::

value = datetime.timedelta(0, 64800)
(datetime.datetime.min + value).time()

datetime.datetime.min and datetime.time() are, of course, documented as part of the datetime module if you want more information.

A datetime.timedelta is, by the way, a representation of the difference between two datetime.datetime values. So if you subtract one datetime.datetime from another, you will get a datetime.timedelta. And if you add a datetime.datetime with a datetime.timedelta, you'll get a datetime.datetime. That's how the code above works.

Python - Date & Time Comparison using timestamps, timedelta

You should be able to use

tdelta.total_seconds()

to get the value you are looking for. This is because tdelta is a timedelta object, as is any difference between datetime objects.

A couple of notes:

1. Using strftime followed by strptime is superfluous. You should be able to get the current datetime with datetime.now.
2. Similarly, using time.ctime followed by strptime is more work than needed. You should be able to get the other datetime object with datetime.fromtimestamp.

So, your final code could be

now = datetime.now()
then = datetime.fromtimestamp(os.path.getmtime("x.cache"))
tdelta = now - then
seconds = tdelta.total_seconds()

How to construct a timedelta object from a simple string

For the first format (5hr34m56s), you should parse using regular expressions

Here is re-based solution:

import re
from datetime import timedelta

regex = re.compile(r'((?P<hours>\d+?)hr)?((?P<minutes>\d+?)m)?((?P<seconds>\d+?)s)?')

def parse_time(time_str):
    parts = regex.match(time_str)
    if not parts:
        return
    parts = parts.groupdict()
    time_params = {}
    for name, param in parts.items():
        if param:
            time_params[name] = int(param)
    return timedelta(**time_params)

>>> from parse_time import parse_time
>>> parse_time('12hr')
datetime.timedelta(0, 43200)
>>> parse_time('12hr5m10s')
datetime.timedelta(0, 43510)
>>> parse_time('12hr10s')
datetime.timedelta(0, 43210)
>>> parse_time('10s')
datetime.timedelta(0, 10)
>>> 

Formatting timedelta objects

But I was wondering if I can do it in a single line using any date time function like strftime.

As far as I can tell, there isn't a built-in method to timedelta that does that. If you're doing it often, you can create your own function, e.g.

def strfdelta(tdelta, fmt):
    d = {"days": tdelta.days}
    d["hours"], rem = divmod(tdelta.seconds, 3600)
    d["minutes"], d["seconds"] = divmod(rem, 60)
    return fmt.format(**d)

Usage:

>>> print strfdelta(delta_obj, "{days} days {hours}:{minutes}:{seconds}")
1 days 20:18:12
>>> print strfdelta(delta_obj, "{hours} hours and {minutes} to go")
20 hours and 18 to go

If you want to use a string format closer to the one used by strftime we can employ string.Template:

from string import Template

class DeltaTemplate(Template):
    delimiter = "%"

def strfdelta(tdelta, fmt):
    d = {"D": tdelta.days}
    d["H"], rem = divmod(tdelta.seconds, 3600)
    d["M"], d["S"] = divmod(rem, 60)
    t = DeltaTemplate(fmt)
    return t.substitute(**d)

Usage:

>>> print strfdelta(delta_obj, "%D days %H:%M:%S")
1 days 20:18:12
>>> print strfdelta(delta_obj, "%H hours and %M to go")
20 hours and 18 to go

---

The totalSeconds value is shown as 13374 instead of 99774. I.e. it's ignoring the "day" value.

Note in the example above that you can use timedelta.days to get the "day" value.

Alternatively, from Python 2.7 onwards, timedelta has a total_seconds() method which return the total number of seconds contained in the duration.

---

Related Topics