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Python : Is there a way to get the names of last three Month Names?

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

Python calculate the number of year in date column

I've recently start coding with Python, and I'm struggling to calculate the number of years between the current date and a given date.
Dataframe
I would like to calculate the number of year for each column.
I tried this but it's not working:
def Number_of_years(d1,d2):
if d1 is not None:
return relativedelta(d2,d1).years
for col in df.select_dtypes(include=['datetime64[ns]']):
df[col]=Number_of_years(df[col],date.today())
Can anyone help me find a solution to this?

I see that the format of dates is day/month/year.
Given this format is same for all the grids, you can parse the date using the datetime module like so:
from datetime import datetime # import module
def numberOfYears(element):
# parse the date string according to the fixed format
date = datetime.strptime(element, '%d/%m/%Y')
# return the difference in the years
return datetime.today().year - date.year
# make things more interesting by vectorizing this function
function = np.vectorize(numberOfYears)
# This returns a numpy array containing difference between years.
# call this for each column, and you should be good
difference = function(df.Date_creation)

You code is basically right, but you're operating over a pandas series so you can't just call relativedelta directly:
def number_of_years(d1,d2):
return relativedelta(d2,d1).years
for col in df.select_dtypes(include=['datetime64[ns]']):
df[col]= df[col].apply(lambda d: number_of_years(x, date.today()))

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

Datetime difference to return Days only?

$ cat .t.py
import re
from datetime import datetime as dtt
oldestDate = dateComp = dtt.strptime('1.1.1001', '%d.%m.%Y')
dateComp = dtt.strptime('11.1.2011', '%d.%m.%Y')
ind = re.sub(" days,.*", "", str((dateComp - oldestDate)))
print ind
print dateComp - oldestDate
$ python .t.py
368905
368905 days, 0:00:00
How can I get days only without the regex code-smell? The problem escalate because I need to use the index in many locations. So some cleaner way to do this?

Don't use str() so soon. The result you get back from subtracting one datetime from another is a timedelta object, which has a .days property that you can read.
(dateComp - oldestDate).days
Note that reading only the .days property will mean that it will round down the difference - if you instead want to round to the nearest number of days, you'll need to add some logic to check the .seconds property to see whether it's closer to 0 or 86400.

Is there a consistent way to enumerate days/weeks/months between two dates?

I have two datetime objects; a start date and an end date. I need to enumerate the days, weeks and months between the two, inclusive.
Ideally the results would be in datetime form, though any compatible form is fine. Weeks and months are represented by a date corresponding to the first day of the week/month, where Monday is the first day of a week, as in ISO-8601. This means that the result may contain a date earlier than the start date.
For example, given 2010-11-28 to 2010-12-01, the results would be as follows:
days: 2010-11-28, 2010-11-29, 2010-11-30, 2010-12-01
weeks: 2010-11-22, 2010-11-29
months: 2010-11-01, 2010-12-01
I realize that the list of days is by itself straightforward, but I'd like a clean and consistent solution that uses a similar approach for all three. It seems like the calendar module should be useful, but I'm not seeing a good way to use it for this purpose.

Using dateutil:
import datetime
import dateutil.rrule as drrule
import dateutil.relativedelta as drel
import pprint
def dt2d(date):
'''
Convert a datetime.datetime to datetime.date object
'''
return datetime.date(date.year,date.month,date.day)
def enumerate_dates(start,end):
days=map(dt2d,drrule.rrule(drrule.DAILY, dtstart=start, until=end))
# Find the Monday on or before start
start_week=start+drel.relativedelta(weekday=drel.MO(-1))
end_week=end+drel.relativedelta(weekday=drel.MO(-1))
weeks=map(dt2d,drrule.rrule(drrule.WEEKLY, dtstart=start_week, until=end_week))
# Find the first day of the month
start_month=start.replace(day=1)
end_month=end.replace(day=1)
months=map(dt2d,drrule.rrule(drrule.MONTHLY, dtstart=start_month, until=end_month))
return days,weeks,months
if __name__=='__main__':
days,weeks,months=enumerate_dates(datetime.date(2010,11,28),
datetime.date(2010,12,01))
print('''\
days: {d}
weeks: {w}
months: {m}'''.format(d=map(str,days),w=map(str,weeks),m=map(str,months)))
yields
days: ['2010-11-28', '2010-11-29', '2010-11-30', '2010-12-01']
weeks: ['2010-11-22', '2010-11-29']
months: ['2010-11-01', '2010-12-01']