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How to convert object datetime.time.now() to object datetime.timedelta()?

I'm using the datetime.time.now() for the current time, i.e. I want to perform an operation that counts in the totals of the hours (e.g. 1h:45min - 0h:50min). I cannot convert the current time to the datetime.timedelta object.

There is no datetime.time.now() — you must mean datetime.now() which returns a datetime instance which has a year, month, and day as well as the time of day. If you want a different time on the same day you can use its attributes to construct one.
If you subtract two datetimes the result is a timedelta.
You can also subtract an arbitrary timedelta from a datetime (resulting in another datetime).
Note that timedelta instances only have the attributes days, seconds, and microseconds, so if you want to know how long they are in hours and minutes, you have to manually calculate them.
Here's an example of doing all of these things.
from datetime import datetime, timedelta
now = datetime.now() # Current time.
# Construct a time on the same day.
sunrise = datetime(now.year, now.month, now.day, hour=6, minute=58)
if sunrise > now: # Future
delta = sunrise - now
when = 'will be in'
ago = ''
else: # Past
delta = now - sunrise
when = 'happened'
ago = 'ago'
days = delta.days
seconds = delta.seconds
hours = delta.seconds//3600
minutes = (delta.seconds//60) % 60
print(f'sunrise {when} {hours} hours {minutes} minutes {ago}')
print(f'30 minutes before sunrise today is {sunrise - timedelta(minutes=30)}')

I think I've found it; I wanted to compare the current time with the sunrise and sunset that Python itself retrieved.
I've done it this way now (so the next one can do it too)
import datetime as dt
DTN = dt.datetime.now()
H = int(DTN .strftime("%H"))
M = int(DTN .strftime("%M"))
S = int(DTN .strftime("%S"))
t1 = dt.timedelta(hours= H, minutes= M, seconds=S)
t2 = dt.timedelta(hours= 1, minutes= 0, seconds=0)
if t1 > t2:
timeCal = t1-t2 }
elif t1<t2:
timeCal = t2-t1
else:
timeCal = t1+t2
print(timeCal)
actual time = 20:00:00
result: 19:00:00

Fetch hourly data

I am calculating the hour delta as below
ct = time.time() * 1000 // milliseconds
pt = ct - (1000*60*60*1) // 1 hour
The above code will give me the hour timestamp. How can I use the above code to get the past 20 days data hour by hour?

from datetime import datetime, timedelta
times = [datetime.now() - timedelta(hours=n) for n in range(24*20)] #±smh like this

scaling current time to a range in python

I want to scale the current time in a range. Like my start time is 08:00 hours and end time is 19:00 hours which is a 11 hr period. scaled value of my start time should be 0 and scaled value of my end time should be 11. Times like 10:00 and 10:30 should be scaled as 1 and 1.5 respectively.My code
from datetime import datetime
now = datetime.now()
current_time = now.strftime("%H:%M")
start_time = '08:00'
end_time = '19:00'
The above code is reading the variable as strings and I'm not able to perform calculations on them. Can some one help me achieving the scaling.
Thanks

Change now.strftime("%H:%M") to now.strftime("%H%M")
This way you can convert it to an integer, so you have 1000 instead of 10:00 and you can use addition and subtraction

You can extract hours and minutes from your time using datetime.hour and datetime.minute.
To get the scaled value you want, for the range you specified (08:00 to 19:00), you can have :
from datetime import datetime
now = datetime.now()
if(8 <= now.hour <= 19):
scaled_time = now.hour - 8
scaled_time += now.minute / 60
Hope it's help.
Edit: you can change 8 and 19 to variables that contain your start and end time, so you can easily modify them

You can divide timedelta objects obtained from subtracting two datetime objects.
>>> from datetime import datetime
>>> start = datetime(2019, 12, day=19, hour=8, minute=0)
>>> end = datetime(2019, 12, day=19, hour=19, minute=0)
>>> now = datetime.now()
>>> print((now - start) / (end - start))
0.7824158481565656

8 to 19 is exactly 11 hours, so you don't need to scale it just use current time minus to 8 o'clock. so problem is down to change string to hours in decimal.
from datetime import datetime
now = datetime.now()
current_time = now.strftime("%H:%M")
start_time = '08:00'
end_time = '19:00'
def conv(s):
l = s.split(':')
h = int(l[0])
m = int(l[1])
return h+m/60
st = conv(start_time)
# et = conv(end_time)
n = conv(current_time)
scaled = n-st
print(scaled)

calculate difference between two time in hour

I want to calculate difference between two time in hours using django in sql db the time are stored in timefield.
I tried this:
def DesigInfo(request): # attendance summary
emplist = models.staff.objects.values('empId', 'name')
fDate = request.POST.get('fromDate')
tDate = request.POST.get('toDate')
if request.GET.get('empId_id'):
sel = attendance.objects.filter(empId_id=request.GET.get('empId_id'),)
for i in sel:
# print i.
# print i.outTime
# print i.inTime.hour,i.inTime.minute,i.inTime.second - i.outTime.hour,i.outTime.minute,i.outTime.second
ss = i.inTime.hour
ss1 = 12 - ss
mm = i.outTime.hour
mm1 = (12 + mm) - 12
print ss1 + mm1

Since i.inTime and i.outTime are time objects you cannot simply subtract them. A good approach is to convert them to datetime adding the date part (use today() but it is irrelevant to the difference), then subtract obtaining a timedelta object.
delta = datetime.combine(date.today(), i.outTime) - datetime.combine(date.today(), i.inTime)
(Look here: subtract two times in python)
Then if you want to express delta in hours:
delta_hours = delta.days * 24 + delta.seconds / 3600.0
A timedelta object has 3 properties representing 3 different resolutions for time differences (days, seconds and microseconds). In the last expression I avoided to add the microseconds but I suppose it is not relevant in your case. If it is also add delta.microseconds / 3600000000.0
Note that simply dividing seconds by 3600 would have returned only the integer part of hours avoiding fractions. It depends on your business rules how to round it up (round, floor, ceil or leave the fractional part as I did)

Using datetime objects: https://docs.python.org/2/library/datetime.html
A good stack overflow post on the topic How to get current time in Python
from datetime import datetime
now = datetime.now()
# wait some time
then = ... some time
# diff is a datetime.timedelta instance
diff = then - now
diff_hours = diff.seconds / 3600

You might want to play with this codes:
from datetime import datetime
#set the date and time format
date_format = "%m-%d-%Y %H:%M:%S"
#convert string to actual date and time
time1 = datetime.strptime('8-01-2008 00:00:00', date_format)
time2 = datetime.strptime('8-02-2008 01:30:00', date_format)
#find the difference between two dates
diff = time2 - time1
''' days and overall hours between two dates '''
print ('Days & Overall hours from the above two dates')
#print days
days = diff.days
print (str(days) + ' day(s)')
#print overall hours
days_to_hours = days * 24
diff_btw_two_times = (diff.seconds) / 3600
overall_hours = days_to_hours + diff_btw_two_times
print (str(overall_hours) + ' hours');
''' now print only the time difference '''
''' between two times (date is ignored) '''
print ('\nTime difference between two times (date is not considered)')
#like days there is no hours in python
#but it has seconds, finding hours from seconds is easy
#just divide it by 3600
hours = (diff.seconds) / 3600
print (str(hours) + ' Hours')
#same for minutes just divide the seconds by 60
minutes = (diff.seconds) / 60
print (str(minutes) + ' Minutes')
#to print seconds, you know already ;)
print (str(diff.seconds) + ' secs')

The easiest way through I achieve is the comment of Zac given above. I was using relativedelta like this
from dateutil import relativedelta
difference = relativedelta.relativedelta( date1, date2)
no_of_hours = difference.hours
but it did not give me correct result when the days changes. So, I used the approach expressed above:
no_of_hours = (difference.days * 24) + (difference.seconds / 3600)
Please note that you will be getting negative value if date2 is greater than date1. So, you need to swipe the position of date variables in relativedelta.

Python time format with three-digit hour

How can I parse the time 004:32:55 into a datetime object? This:
datetime.strptime("004:32:55", "%H:%M:%S")
doesn't work becaush %H expects two digits. Any way to add the third digit?

Three options :
s = "004:32:55"
from datetime import datetime
datetime.strptime(s[1:], "%H:%M:%S") # from second 0 to the end
datetime.strptime(s, "0%H:%M:%S") # add 0 in formatting
from dateutil import parser # use dateutil
parser.parse(s)
There are 24 hours in a day so you can divide and get the modulus to figure out how many days and hours are in the time, then subtract the days, this needs some tweaking will get you started.
s = "04:32:55"
s = s.split(":",1)
hours, days = divmod(int(s[0]),24)
new_time = "{}:{}".format(hours,s[1])
past = datetime.now() - timedelta(days=days)
final = "{} {}".format(past.date().isoformat(),new_time)
print datetime.strptime(final,"%Y-%m-%d %H:%M:%S")

I chose a more pragmatic approach in the end and converted the time stamp to seconds:
hours = (0 if time_string.split(":")[0] == "000" else int(time_string.split(":")[0].lstrip("0")) * 3600)
mins = (0 if time_string.split(":")[1] == "00" else int(time_string.split(":")[1].lstrip("0")) * 60)
secs = (0 if time_string.split(":")[2] == "00" else int(time_string.split(":")[2].lstrip("0")))
return hours + mins + secs
Converting back to hours, minutes, and seconds is easy with datetime.timedelta(seconds=123).
EDIT:
A better solution (thanks to Ben):
hours = int(time_string.split(":")[0]) * 3600
mins = int(time_string.split(":")[1]) * 60
secs = int(time_string.split(":")[2])
return hours + mins + secs