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How to Split a substract of a date in python

My code is the following:
date = datetime.datetime.now()- datetime.datetime.now()
print date
h, m , s = str(date).split(':')
When I print h the result is:
-1 day, 23
How do I get only the hour (the 23) from the substract using datetime?
Thanks.

If you subtract the current date from a past date, you would get a negative timedelta value.
You can get the seconds with td.seconds and corresponding hour value via just dividing by 3600.
from datetime import datetime
import time
date1 = datetime.now()
time.sleep(3)
date2 = datetime.now()
# timedelta object
td = date2 - date1
print(td.days, td.seconds // 3600, td.seconds)
# 0 0 3

You're not too far off but you should just ask your question as opposed to a question with a "real scenario" later as those are often two very different questions. That way you get an answer to your actual question.
All that said, rather than going through a lot of hoop-jumping with splitting the datetime object, assigning it to a variable which you then later use look for what you need in, it's better to just know what DateTime can do since that can be such a common part of your coding. You would also do well to look at timedelta (which is part of datetime) and if you use pandas, timestamp.
from datetime import datetime
date = datetime.now()
print(date)
print(date.hour)

I can get you the hour of datetime.datetime.now()
You could try indexing a list of a string of datetime.datetime.now():
print(list(str(datetime.datetime.now()))[11] + list(str(datetime.datetime.now()))[12])
Output (in my case when tested):
09
Hope I am of help!

Converting a UTC Time to Epoch format

I'm trying to convert a specific UTC time & date to seconds since the Epoch; however, I have not been able to figure out how to do this since I am not in the UTC time zone.
I tried using the datetime module to convert a date & time to seconds since the epoch, but python has been using my system's local time so the returned value is off by 7 hours. I understand that I could simply subtract 7*60 so that it would work in my time zone; however, I need this to work in multiple time zones without hardcoding the time change into my program.
This works except it uses the system time (MST), but I am looking for a solution that is specifically UTC time. Note the variables defined here represent an example of a time in UTC that I am trying to convert to seconds since the epoch.
import datetime
year=2019
month=5
day=9
hour=21
minute=45
Time=datetime.datetime(year, month, day, hour, minute).timestamp()
print(Time)
Output:
1557463500.0
Desired output (7 hours earlier):
1557438300.0

import datetime
year=2019
month=5
day=9
hour=21
minute=45
e = datetime.datetime(1970, 1, 1, 0, 0)
t = datetime.datetime(year, month, day, hour, minute)
print((t-e).total_seconds())

You can caclulate the timestamp of epoch date using datetime.datetime.utcfromtimestamp(0).timestamp() and then subtract your current timestamp from it
import datetime
year=2019
month=5
day=9
hour=21
minute=45
#Date timestamp
dt_timestamp=datetime.datetime(year, month, day, hour, minute).timestamp()
#Epoch timestamp
epoch = datetime.datetime.utcfromtimestamp(0).timestamp()
#Epoch
print(dt_timestamp-epoch)
The output will be
1557438300.0

How to use 24 hour time series data as a predictive feature

I am just wondering how best to approach using this 24 hour time format as a predictive feature. My thoughts were to bin it into 24 categories for each hour of the day. Is there an easy way to convert this object into a python datetime object that would make binning easier or how would you advise handling this feature? Thanks :)
df['Duration']
0 2:50
1 7:25
2 19:00
3 5:25
4 4:45
5 2:25
df['Duration'].dtype
dtype('O')

The best solution will depend on what you hope to get from your model. In many cases it makes sense to convert it to total number of seconds (or minutes or hours) since some epoch. To convert your data to seconds since 00:00, you can use:
from datetime import datetime
t_str = "2:50"
t_delta = datetime.strptime(t_str, "%H:%M") - datetime(1900, 1, 1)
seconds = t_delta.total_seconds()
hours = seconds/60**2
print(seconds)
# 10200.0
Using Python's datetime class will not support time values over 23:59. Since it appears that your data may actually be a duration, you may want to represent it as an instance of Python's timedelta class.
from datetime import timedelta
h, m = map(int, t_str.split(sep=':'))
t_delta = timedelta(hours=h, minutes=m)
# Get total number of seconds
seconds = t_delta.total_seconds()

You can use datetime to create a useable datetime string
>>> from datetime import datetime
>>> x = datetime(2019, 1, 1, 0).strftime('%Y-%m-%d %H:%M:%S')
>>> # Use that for your timestring then you can reverse it nicely back into a datetime object
>>> d = datetime.strptime('2019-01-01 00:00:00', '%Y-%m-%d %H:%M:%S')
Of course you can use any valid format string.

You should calculate the time in seconds or minutes or hours from some initial time like the 1st time. Then you can make an x-y scatter plot of the data since the x-axis (time) is now numbers.

Python - How to convert datetime data using toordinal considering the time

Let's assume that I have the following data:
25/01/2000 05:50
When I convert it using datetime.toordinal, it returns this value:
730144
That's nice, but this value just considers the date itself. I also want it to consider the hour and minutes (05:50). How can I do it using datetime?
EDIT:
I want to convert a whole Pandas Series.

An ordinal date is by definition only considering the year and day of year, i.e. its resolution is 1 day.
You can get the microseconds / milliseconds (depending on your platform) from epoch using
datetime.datetime.strptime('25/01/2000 05:50', '%d/%m/%Y %H:%M').timestamp()
for a pandas series you can do
s = pd.Series(['25/01/2000 05:50', '25/01/2000 05:50', '25/01/2000 05:50'])
s = pd.to_datetime(s) # make sure you're dealing with datetime instances
s.apply(lambda v: v.timestamp())

If you use python 3.x. You can get date with time in seconds from 1/1/1970 00:00
from datetime import datetime
dt = datetime.today() # Get timezone naive now
seconds = dt.timestamp()

Converting days since epoch to date

How can one convert a serial date number, representing the number of days since epoch (1970), to the corresponding date string? I have seen multiple posts showing how to go from string to date number, but I haven't been able to find any posts on how to do the reverse.
For example, 15951 corresponds to "2013-09-02".
>>> import datetime
>>> (datetime.datetime(2013, 9, 2) - datetime.datetime(1970,1,1)).days + 1
15951
(The + 1 because whatever generated these date numbers followed the convention that Jan 1, 1970 = 1.)
TL;DR: Looking for something to do the following:
>>> serial_date_to_string(15951) # arg is number of days since 1970
"2013-09-02"
This is different from Python: Converting Epoch time into the datetime because I am starting with days since 1970. I not sure if you can just multiply by 86,400 due to leap seconds, etc.

Use the datetime package as follows:
import datetime
def serial_date_to_string(srl_no):
new_date = datetime.datetime(1970,1,1,0,0) + datetime.timedelta(srl_no - 1)
return new_date.strftime("%Y-%m-%d")
This is a function which returns the string as required.
So:
serial_date_to_string(15951)
Returns
>> "2013-09-02"

And for a Pandas Dataframe:
df["date"] = pd.to_datetime(df["date"], unit="d")
... assuming that the "date" column contains values like 18687 which is days from Unix Epoch of 1970-01-01 to 2021-03-01.
Also handles seconds and milliseconds since Unix Epoch, use unit="s" and unit="ms" respectively.
Also see my other answer with the exact reverse.