Given two dates, I would like to generate a list of dates with a fixed time length in between one another using datetime, starting from the later date.
For instance, given 01/01/2018 and 01/09/2018 and time interval of 2 months the output would be:
[01/01/2018, 01/03/2018, 01/05/2018, 01/07/2018, 01/09/2018]
For an interval of 3 months:
[01/03/2018, 01/06/2018, 01/09/2018]
I cannot just subtract months using the .replace method on a datetime object since going from a 31 days month to a 30 days month would return an error.
I think relativedeleta module can help you on this - pip install python-dateutil
from dateutil.relativedelta import *
import datetime
date1 = datetime.datetime.strptime('01/01/2018', "%d/%m/%Y").date()
date2 = datetime.datetime.strptime('01/09/2018', "%d/%m/%Y").date()
f = [(date1 + relativedelta(months=i)).strftime("%d/%m/%Y") for i in range(date1.month, date2.month,2)]
Result will be - ['01/02/2018', '01/04/2018', '01/06/2018', '01/08/2018']
You did specify datetime, but if you're interested,
a time.localtime object can be broken down like so:
import time
secSinceEpoch = time.time()
currentTime = time.localtime(secSinceEpoch)
month = currentTime.tm_mon
day = currentTime.tm_mday
year = currentTime.tm_year
hour = currentTime.tm_hour
min = currentTime.tm_min
sec = currentTime.tm_sec
From here you could perform operations on specific parts of the date/time...
Related
I would like to know how many days are passed from a x ago to today
I wrote this:
from datetime import datetime
timestamp = 1629195530 # A month ago
before = datetime.fromtimestamp(timestamp)
daysBefore = before.strftime("%d")
now = datetime.now()
today = now.strftime("%d")
print(f"daysBefore {daysBefore} - today {today}")
daysPassed = int(today) - int(daysBefore)
But so it seems, daysBefore is returning the days of the month, I can't get my head around this :(
Exact format with date time hour minute accuracy
from datetime import datetime
timestamp = 1629195530 # A month ago
before = datetime.fromtimestamp(timestamp)
now = datetime.now()
print(now - before))
print(f"daysBefore {daysBefore} - today {today}")
The reason this doesn't work is that it gives the day of the month. For example 17th of July and 17th of August will give a difference of zero days.
Therefore the recommend method is as #abdul Niyas P M says, use the whole date.time format to subtract two dates and afterwards extract the days.
Your issue is due to this: strftime("%d")
You are converting you date to a string and then to an int to make the difference. You can just use the datetime to do this for you:
timestamp = 1629195530 # A month ago
before = datetime.fromtimestamp(timestamp)
now = datetime.now()
print(f"daysBefore {before} - today {now}")
daysPassed = now - before
print(daysPassed.days)
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!
How can I extract last three month names in Python? If I am running this today then I would like to see May, June and July as my result.
Easier way is to use "%B" using datetime and timedelta
from dateutil.relativedelta import relativedelta
from datetime import datetime
today = datetime.now()
for i in range(1,4):
print((today - relativedelta(months=i)).strftime('%B'))
Output:
July
June
May
One way is to use the python calendar module, and list slice a month name for a given, extracted datetime month.
.month_name returns a list of all the month names.
calendar is part of the standard library.
For timedelta, there isn't a month parameter because the length of a month is not a constant value, so use days, as an approximation.
See datetime for the available methods.
datetime is part of the python standard library, so doesn't require a separate installation.
Use .month to extract the month from the datetime.
from datetime import datetime, timedelta
import calendar
# get the time now
now = datetime.now()
# iterate through 3 different timedeltas as an example
for x in range(1, 4):
new = now - timedelta(days=31*x)
print(calendar.month_name[new.month])
[out]:
July
June
May
As mentioned in the answer by bigbounty, using .strftime with '%B' is a better option than using calendar
However, unlike the dateutil module, timedelta still doesn't have a month parameter.
The dateutil module provides powerful extensions to the standard datetime module and must be installed, and then imported.
# get the time now
now = datetime.now()
# iterate through 3 different timedeltas as an example
for x in range(1, 4):
new = now - timedelta(days=31*x)
print(new.strftime('%B'))
[out]:
July
June
May
Best way to do this is a combination of the date and calendar modules.
date.today().month will give you a numerical value for the current month (1-12)
calendar.month_name[x] will give you the name for the month represented by the number x
the % operator will be used to wrap around the index of the month_name object to avoid the pesky 0 index returning ''
Putting them together we have:
from datetime import date
from calendar import month_name
def previous_n_months(n):
current_month_idx = date.today().month - 1 # Value is now (0-11)
for i in range(1, n+1):
# The mod operator will wrap the negative index back to the positive one
previous_month_idx = (current_month_idx - i) % 12 #(0-11 scale)
m = int(previous_month_idx + 1)
print(month_name[m])
Example usage:
>>> previous_n_months(3)
July
June
May
I am making a program where I input start date to dataStart(example 21.10.2000) and then input int days dateEnd and I convert it to another date (example 3000 = 0008-02-20)... Now I need to count these dates together, but I didn't managed myself how to do that. Here is my code.
from datetime import date
start=str(input("type start date (DD.MM.YYYY)"))
end=int(input("how many days from it?"))
dataStart=start.split(".")
days=int(dataStart[0])
months=int(dataStart[1])
years=int(dataStart[2])
endYears=0
endMonths=0
endDays=0
dateStart = date(years, months, days)
while end>=365:
end-=365
endYears+=1
else:
while end>=30:
end-=30
endMonths+=1
else:
while end>=1:
end-=1
endDays+=1
dateEnd = date(endYears, endMonths, endDays)
For adding days into date, you need to user datetime.timedelta
start=str(input("type start date (DD.MM.YYYY)"))
end=int(input("how many days from it?"))
date = datetime.strptime(start, "%d.%m.%Y")
modified_date = date + timedelta(days=end)
print(datetime.strftime(modified_date, "%d.%m.%Y"))
You may use datetime.timedelta to add certain units of time to your datetime object.
See the answers here for code snippets: Adding 5 days to a date in Python
Alternatively, you may wish to use the third-party dateutil library if you need support for time additions in units larger than weeks. For example:
>>> from datetime import datetime
>>> from dateutil import relativedelta
>>> one_month_later = datetime(2017, 5, 1) + relativedelta.relativedelta(months=1)
>>> one_month_later
>>> datetime.datetime(2017, 6, 1, 0, 0)
It will be easier to convert to datetime using datetime.datetime.strptime and for the part about adding days just use datetime.timedelta.
Below is a small snippet on how to use it:
import datetime
start = "21.10.2000"
end = 8
dateStart = datetime.datetime.strptime(start, "%d.%m.%Y")
dateEnd = dateStart + datetime.timedelta(days=end)
dateEnd.date() # to get the date format of the endDate
If you have any doubts please look at the documentation python3/python2.
How can I subtract or add 100 years to a datetime field in the database in Django?
The date is in database, I just want to directly update the field without retrieving it out to calculate and then insert.
I would use the relativedelta function of the dateutil.relativedelta package, which will give you are more accurate 'n-years ago' calculation:
from dateutil.relativedelta import relativedelta
import datetime
years_ago = datetime.datetime.now() - relativedelta(years=5)
Then simply update the date field as others have shown here.
Use timedelta. Something like this should do the trick:
import datetime
years = 100
days_per_year = 365.24
hundred_years_later = my_object.date + datetime.timedelta(days=(years*days_per_year))
The .update() method on a Django query set allows you update all values without retrieving the object from the database. You can refer to the existing value using an F() object.
Unfortunately Python's timedelta doesn't work with years, so you'll have to work out 100 years expressed in days (it's 36524.25):
MyModel.objects.update(timestamp=F('timestamp')+timedelta(days=36524.25))
Though setting the number of days in a year as 365.25 (from (365+365+365+366)/4) perfectly offsets the difference-in-days error, it would sometimes lead to unwanted results as you might cause undesirable changes in attributes other than year, especially when you are adding/subtracting 1 or a few years.
If you want to just change the year while preventing changes in other datetime's attributes, just do the algebra on the year attribute like the following:
from datetime import datetime
d = my_obj.my_datetime_field
""" subtract 100 years. """
my_obj.my_datetime_field = datetime(d.year-100, d.month, d.day, d.hour, d.minute, d.second, d.microsecond, d.tzinfo)
my_obj.save()
Hope it helps!
Subtract year from today and use this format.
x = datetime.datetime(2020 - 100, 5, 17)
import datetime
datetime.date(datetime.date.today().year - 100, datetime.date.today().month, datetime.date.today().day)
I Know it's an old question, but I had the problem to find out a good one to solve my problem, I have created this: Use plus(+) or minus(-) to handle with:
import datetime # Don't forget to import it
def subadd_date(date,years):
''' Subtract or add Years to a specific date by pre add + or - '''
if isinstance(date,datetime.datetime) and isinstance(years,int):
day,month,year = date.day , date.month , date.year
#If you want to have HOUR, MINUTE, SECOND
#With TIME:
# day,month,year,hour,minute,second = date.day, date.month,date.year,date.hour,date.minute,date.second
py = year + years # The Past / Futur Year
new_date_str = "%s-%s-%s" % (day,month,py) # New Complete Date
# With TIME : new_date_str = "%s-%s-%s %s:%s:%s" % (month,day,py,hour,minute,second)
try:
new_date = datetime.datetime.strptime(new_date_str,"%d-%m-%Y")
except ValueError: # day is out of range for month (February 29th)
new_date_str = "%s-%s-%s" % (1,month+1,py) # New Complete Date : March 1st
new_date = datetime.datetime.strptime(new_date_str,"%d-%m-%Y")
return new_date
# With TIME : return datetime.datetime.strptime(new_date_str,"%d-%m-%Y %H:%M:%Y")
return None