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How can I convert time to seconds when the seconds are formatted in 3 digits?

Say I have this time
00:46:19,870
where it represents 46h 19m and 870 is 870/1000 of a minute (I think I can just get rid of the last part). How do I convert this to seconds?
I've tried
time.strptime('00:46:19,870'.split(',')[0],'%H:%M:%S')
but realized that it wouldn't work as it's using a format different than mine.
How can I convert 00:46:19,870 to 2779?

You are close, you can still use the datetime you just need to calculate the time delta. What you really have isn't really a date but what appears to be a stopwatch time time. You can still strip the time from that and you will notice that Python uses a default year, month, and day. You can use that default to figure out the delta in seconds:
from datetime import datetime
DEFAULT_DATE = (1900, 1, 1)
stopwatch = datetime.strptime('00:46:19,870', '%H:%M:%S,%f')
a_timedelta = stopwatch - datetime(*DEFAULT_DATE)
seconds = a_timedelta.total_seconds()
print(seconds)

Convert time (given in days since YYYY) into readable time format

I need help with converting time to a readable format. My time array (has 580 elements) is # of days since January 1st, 1900. How do I convert that to a normal time format (ie mm-dd-yyyy)?
For example, input time is 43,887 and output should read 02-27-2020 after adding it to Jan 1, 1900.
Thank you!

datetime.datetime and timedelta class can be helpful here.
from datetime import datetime, timedelta
def convert(inp, date1):
new_date = date1 + timedelta(inp)
return new_date.strftime("%m-%d-%Y")
date1 = datetime(1900, 1, 1)
print(convert(43887, date1))

You can use the datetime.datetime class can help you here. The following works, if those values are treated as integer days (you don't specify what they are).
from datetime import datetime
dt = datetime.fromordinal(43887)
dt.strftime('%d-%m-%Y')

Turn number representing total minutes into a datetime object in python

If I have a number representing a period I am interested in, for example the number 360 representing 360 minutes or 6 hours, how do I turn this into a datetime object such that I can perform the standard datetime object functions on it?
Similarly, if I have a datetime object in the format 00:30:00, representing 30 minutes, how do I turn that into a normal integer variable?

import datetime
t = datetime.timedelta(minutes=360)
This will create an object, t, that you can use with other datetime objects.
To answer the 2nd question you just edited in, you can use t.total_seconds() to return whatever your timedelta holds back into an integer in seconds. You'll have to do the conversion to minutes or hours manually though.

You may want to look at time deltas:
delta = datetime.timedelta(minutes=360)

If your time data is in '00:30:00' format then you should use strptime
>>> from datetime import datetime
>>> time = '00:30:00'
>>> datetime.strptime(time, '%H:%M:%S).time()
datetime.time(0, 30)
If your data is in 30 (integer) format
>>> from datetime import datetime, timedelta
>>> from time import strftime, gmtime
>>> minutes = timedelta(minutes=360)
>>> time = strftime('%H:%M:%S', gmtime(minutes.total_seconds()))
>>> datetime.strptime(time, '%H:%M:%S').time()
datetime.time(6, 0)

how to calculate timedelta python

What I am trying to do is to subtract 7 hours from a date. I searched stack overflow and found the answer on how to do it here. I then went to go read the documentation on timedelta because I was unable to understand what that line in the accepted answer does, rewritten here for ease:
from datetime import datetime
dt = datetime.strptime( date, '%Y-%m-%d %H:%M' )
dt_plus_25 = dt + datetime.timedelta( 0, 2*60*60 + 30*60 )
Unfortunately, even after reading the documentation I still do not understand how that line works.
What is the timedelta line doing? How does it work?
Additionally, before I found this stackoverflow post, I was working with time.struct_time tuples. I had a variable tm:
tm = time.strptime(...)
I was simply accessing the hour through tm.tm_hour and subtracting seven from it but this, for obvious reasons, does not work. This is why I am now trying to use datetime. tm now has the value
tm = datetime.strptime(...)
I'm assuming using datetime is the best way to subtract seven hours?
Note: subtracting seven hours because I want to go from UTC to US/Pacific timezone. Is there a built-in way to do this?

What is the timedelta line doing? How does it work?
It creates a timedelta object.
There are two meanings of "time".
"Point in Time" (i.e, date or datetime)
"Duration" or interval or "time delta"
A time delta is an interval, a duration, a span of time. You provided 3 values.
0 days.
2*60*60 + 30*60 seconds.

timedelta() generates an object representing an amount of time—the Greek letter delta is used in math to represent "difference". So to compute an addition or a subtraction of an amount of time, you take the starting time and add the change, or delta, that you want.
The specific call you've quoted is for generating the timedelta for 2.5 hours. The first parameter is days, and the second is seconds, so you have (0 days, 2.5 hours), and 2.5 hours in seconds is (2 hours * 60 minutes/hour * 60 seconds/minute) + (30 minutes * 60 seconds / minute).
For your case, you have a negative time delta of 0 days, 7 hours, so you'd write:
timedelta(0, -7 * 60 * 60)
... or timedelta(0, -7 * 3600) or whatever makes it clear to you what you're doing.

Note: subtracting seven hours because I want to go from UTC to US/Pacific timezone. Is there a built-in way to do this?
Yes there is: datetime has built-in timezone conversion capabilities. If you get your datetime object using something like this:
tm = datetime.strptime(date_string, '%Y-%m-%d %H:%M')
it will not have any particular timezone "attached" to it at first, but you can give it a timezone using
tm_utc = tm.replace(tzinfo=pytz.UTC)
Then you can convert it to US/Pacific with
tm_pacific = tm_utc.astimezone(pytz.all_timezones('US/Pacific'))
I'd suggest doing this instead of subtracting seven hours manually because it makes it clear that you're keeping the actual time the same, just converting it to a different timezone, whereas if you manually subtracted seven hours, it looks more like you're actually trying to get a time seven hours in the past. Besides, the timezone conversion properly handles oddities like daylight savings time.
To do this you will need to install the pytz package, which is not included in the Python standard library.

Python fraction of seconds

What is the best way to handle portions of a second in Python? The datetime library is excellent, but as far as I can tell it cannot handle any unit less than a second.

In the datetime module, the datetime, time, and timedelta classes all have the smallest resolution of microseconds:
>>> from datetime import datetime, timedelta
>>> now = datetime.now()
>>> now
datetime.datetime(2009, 12, 4, 23, 3, 27, 343000)
>>> now.microsecond
343000
if you want to display a datetime with fractional seconds, just insert a decimal point and strip trailing zeros:
>>> now.strftime("%Y-%m-%d %H:%M:%S.%f").rstrip('0')
'2009-12-04 23:03:27.343'
the datetime and time classes only accept integer input and hours, minutes and seconds must be between 0 to 59 and microseconds must be between 0 and 999999. The timedelta class, however, will accept floating point values with fractions and do all the proper modulo arithmetic for you:
>>> span = timedelta(seconds=3662.567)
>>> span
datetime.timedelta(0, 3662, 567000)
The basic components of timedelta are day, second and microsecond (0, 3662, 567000 above), but the constructor will also accept milliseconds, hours and weeks. All inputs may be integers or floats (positive or negative). All arguments are converted to the base units and then normalized so that 0 <= seconds < 60 and 0 <= microseconds < 1000000.
You can add or subtract the span to a datetime or time instance or to another span. Fool around with it, you can probably easily come up with some functions or classes to do exaxtly what you want. You could probably do all your date/time processing using timedelta instances relative to some fixed datetime, say basetime = datetime(2000,1,1,0,0,0), then convert to a datetime or time instance for display or storage.

A different, non mentioned approach which I like:
from datetime import datetime
from time import sleep
t0 = datetime.now()
sleep(3)
t1 = datetime.now()
tdelta = t1 - t0
print(tdelta.total_seconds())
# will print something near (but not exactly 3)
# 3.0067

To get a better answer you'll need to specify your question further, but this should show at least how datetime can handle microseconds:
>>> from datetime import datetime
>>> t=datetime.now()
>>> t.microsecond
519943

NumPy 1.4 (in release candidate stage) has support for its own Date and DateArray objects. The one advantage is that it supports frequencies smaller than femtoseconds: http://projects.scipy.org/numpy/browser/trunk/doc/neps/datetime-proposal.rst
Otherwise I would go with the regular datetime subsecond frequencies.