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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!

Formatting a string in Python is giving me a file not found error

I am new to python and trying to work with Pandas to do some work with several .csv files that have predictable names, Log_(yyyy/mm/dd).
What I'm planning is simple enough, but opening the file is giving me problems.
today = date.today()
m,d,y = today.month, today.day, today.year
file_name = 'Log_{}-{}-{}'.format(y,m,d)
pd.read_csv(file_name)
This will give me an error, but this works
file_name = 'Log_2015-01-10'
pd.read_csv(file_name)
They print the same thing, and str(file_name) doesn't fix the issue.

You have two problems: your tuple assignment swaps the day and year values, and you need to zero-pad values below 10. You are actually producing the string 'Log_10-1-2015', not 'Log_2014-01-10'.
Formatting a date into a string is easiest done by leaving formatting to the date object rather than extracting the individual components yourself:
today = date.today()
file_name = 'Log_{:%Y-%m-%d}'.format(today)
The % fields are strftime() formatting instructions that use zero-padding by default.
Demo:
>>> from datetime import date
>>> today = date.today()
>>> 'Log_{:%Y-%m-%d}'.format(today)
'Log_2015-01-11'
Yes, it is already the 11th in my timezone. :-)

Django filter by datetime, after converting string to datetime

I'm trying to filter my query by a datetime. This datetime is the datetime for the value range the customer wants to know information for. I'm trying to set it to the first of the month selected by the customer. I pass the month number convert it to the correct string format and then convert to a datetime object because simply looking for the string object was returning no values and Django's documentation says you need to do it like:
pub_date__gte=datetime(2005, 1, 30)
Code:
if 'billing-report' in request.POST:
customer_id = int(post_data['selected_customer'])
This is the code I use to get the selected customer date and turn it into a tupple
if 'billing-report' in request.POST:
customer_id = int(post_data['selected_customer'])
selected_date = int(post_data['month'])
if selected_date < 10:
selected_date = '0'+str(selected_date)
year = datetime.now()
year = year.year
query_date = str(year) + '-' + str(selected_date) + '-01'
query_date_filter = datetime.strptime(query_date, "%Y-%m-%d")
compute_usages = ComputeUsages.objects.filter(customer_id = customer_id).filter(values_date = query_date_filter)
django debug shows: datetime.datetime(2014, 10, 1, 0, 0)
query_date looks like: 2014-07-01 before it is converted
.
No error but no data is returned
I used to use:
compute_usages = ComputeUsages.objects.filter(customer_id = customer_id).filter(values_date = datetime(query_date_filter))
which was causing the error. I'm sorry for changing my question as it evolved that is why I'm re-including what I was doing before so the comments make sense.

Almost all of that code is irrelevant to your question.
I don't understand why you are calling datetime on query_date. That is already a datetime, as you know because you converted it to one with strptime earlier. So there's no need for any more conversion:
ComputeUsages.objects.filter(customer_id=customer_id).filter(values_date=query_date)

Well after spending sometime exploring setting the query filter to datetime(year, month, day) I came to the realization that django doesn't convert it to a neutral datetime format it has to match exactly. Also my data in the database had the year, day, month.
Learning point:
You have to use the datetime() exactly how it is in the database django does not convert to a neutral format and compare. I assumed it was like writing a query and saying to_date or to_timestamp where the db will take your format and convert it to a neutral format to compare against the rest of the db.
Here is the correct way
compute_usages = ComputeUsages.objects.filter(customer_id = customer_id).filter(values_date = datetime(year, day, selected_month))

How to set a variable to be "Today's" date in Python/Pandas

I am trying to set a variable to equal today's date.
I looked this up and found a related article:
Set today date as default value in the model
However, this didn't particularly answer my question.
I used the suggested:
dt.date.today
But after
import datetime as dt
date = dt.date.today
print date
<built-in method today of type object at 0x000000001E2658B0>
Df['Date'] = date
I didn't get what I actually wanted which as a clean date format of today's date...in Month/Day/Year.
How can I create a variable of today's day in order for me to input that variable in a DataFrame?

You mention you are using Pandas (in your title). If so, there is no need to use an external library, you can just use to_datetime
>>> pandas.to_datetime('today').normalize()
Timestamp('2015-10-14 00:00:00')
This will always return today's date at midnight, irrespective of the actual time, and can be directly used in pandas to do comparisons etc. Pandas always includes 00:00:00 in its datetimes.
Replacing today with now would give you the date in UTC instead of local time; note that in neither case is the tzinfo (timezone) added.
In pandas versions prior to 0.23.x, normalize may not have been necessary to remove the non-midnight timestamp.

If you want a string mm/dd/yyyy instead of the datetime object, you can use strftime (string format time):
>>> dt.datetime.today().strftime("%m/%d/%Y")
# ^ note parentheses
'02/12/2014'

Using pandas: pd.Timestamp("today").strftime("%m/%d/%Y")

pd.datetime.now().strftime("%d/%m/%Y")
this will give output as '11/02/2019'
you can use add time if you want
pd.datetime.now().strftime("%d/%m/%Y %I:%M:%S")
this will give output as '11/02/2019 11:08:26'
strftime formats

You can also look into pandas.Timestamp, which includes methods like .now and .today.
Unlike pandas.to_datetime('now'), pandas.Timestamp.now() won't default to UTC:
import pandas as pd
pd.Timestamp.now() # will return California time
# Timestamp('2018-12-19 09:17:07.693648')
pd.to_datetime('now') # will return UTC time
# Timestamp('2018-12-19 17:17:08')

i got the same problem so tried so many things
but finally this is the solution.
import time
print (time.strftime("%d/%m/%Y"))

simply just use pd.Timestamp.now()
for example:
input: pd.Timestamp.now()
output: Timestamp('2022-01-12 14:43:05.521896')
I know all you want is Timestamp('2022-01-12') you don't anything after
thus we could use replace to remove hour, minutes , second and microsecond
here:
input: pd.Timestamp.now().replace(hour=0, minute=0, second=0, microsecond=0)
output: Timestamp('2022-01-12 00:00:00')
but looks too complicated right, here is a simple way use normalize
input: pd.Timestamp.now().normalize()
output: Timestamp('2022-01-12 00:00:00')

Easy solution in Python3+:
import time
todaysdate = time.strftime("%d/%m/%Y")
#with '.' isntead of '/'
todaysdate = time.strftime("%d.%m.%Y")

import datetime
def today_date():
'''
utils:
get the datetime of today
'''
date=datetime.datetime.now().date()
date=pd.to_datetime(date)
return date
Df['Date'] = today_date()
this could be safely used in pandas dataframes.

There are already quite a few good answers, but to answer the more general question about "any" period:
Use the function for time periods in pandas. For Day, use 'D', for month 'M' etc.:
>pd.Timestamp.now().to_period('D')
Period('2021-03-26', 'D')
>p = pd.Timestamp.now().to_period('D')
>p.to_timestamp().strftime("%Y-%m-%d")
'2021-03-26'
note: If you need to consider UTC, you can use: pd.Timestamp.utcnow().tz_localize(None).to_period('D')...

From your solution that you have you can use:
import pandas as pd
pd.to_datetime(date)
using the date variable that you use

Attempting to insert an integer from a list into datetime object

What I am trying to accomplish is very simple: creating a loop from a range (pretty self explanatory below) that will insert the month into the datetime object. I know %d requires an integer, and I know that 'month' type is int...so I'm kind of stuck as to why I can't substitute my month variable. Here is my code:
all_months=range(1,13)
for month in all_months:
month_start = (datetime.date(2010,'%d',1))%month
next_month_begin= datetime.date(2010,'%d',1)%(month+1)
month_end=next_month_begin - timedelta(days=1)
print month_start
print month_end
What am I doing wrong?
All help appreciated! Thanks

There are a few things that you need to fix here.
EDIT: First, be careful with your range, since you are using month+1 to create next_month_begin, you do not want this to be greater than 12 or you will get an error.
Next, when you are trying to create the date object you are passing the month in as a string when you use (datetime.date(2010,'%d',1))%month. Your code probably throwing this error TypeError: an integer is required.
You need to give it the integer representing the month, not a string of the integer (there is a difference between 1 and '1'). This is also a simple fix, since you have variable named month that is already an integer, just use that instead of making a string. So you code should be something like:
month_start = datetime.date(2010,month,1)
I think you can figure out how to apply this to your next_month_begin assignment.
The last problem is that you need to use datetime.timedelta to tell Python to look in the datetime module for the timedelta() function -- your program would currently give you an error saying that timedelta is not defined.
Let me know if you have any problems applying these fixes. Be sure to include what the error you may be getting as well.

You've got other answers, but here's a way to get the last day of the month. Adding 31 days will get you into the next month regardless of the number of days in the current month, then moving back to the first and subtracting a day will give the ending date.
import datetime
for month in range(1,13):
month_start = datetime.date(2010,month,1)
into_next_month = month_start + datetime.timedelta(days=31)
month_end = into_next_month.replace(day=1) - datetime.timedelta(days=1)
print month_start,month_end

month is a variable and you can use it to create the datetime object. I think you want to do the following:
month_start = datetime.date(2010, month, 1)
next_month_begin = datetime.date(2010, month+1, 1)
That will work, because datetime.date() requires 3 integer arguments. '%d' % month would instead format the integer month as string. '%04d' % 3 for example would format the number 3 with 4 digits and leading zeros. But it's important to know, that even the (nearly unformatted) string "3" is different to the number 3 in Python.
And you can't write datetime(...) % 3 because the % operator will only work when used on a format string like the previous "%03d" % 3 example and not on a datetime object.
But other types might also accept the % operator (not including datetime objects). For example, integers accept the % operator to get the remainder of a division: 3 % 2 # returns 1. But there, the meaning of % is completely different, because the meaning of the operator depends on the types involved. For example, try 3 + 2 and "3" + "2". There, the meaning of + differs (integer addition vs. string concatenation), because the types are different too.

Check out the calendar module (http://docs.python.org/library/calendar.html).
It has batteries included for this sort of thing...
You could just do:
from calendar import Calendar
def start_and_end_days(year, month):
cal = Calendar()
month_days = [day for day in cal.itermonthdays(year, month) if day.month == month]
first_day = month_days[0]
last_day = month_days[-1]
return (first_day, last_day)