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How to get last Friday's date with Python?

Is there a way to use Python to get last Friday's date and store it in a variable, regardless of which day of the week the program runs?
That is, if I run the program on Monday June 19th 2021 or Thursday June 22nd 2021, it always returns the previus Friday as a date variable: 2021-07-16.

To get the day of the week as an int we use datetime.datetime.today().weekday() and to subtract days from a datetime we use datetime.today() - timedelta(days=days_to_subtract) now we can make a dictionary linking the day of the week to the number of days to subtract and use that dictionary to make a subtraction:
from datetime import datetime, timedelta
d = {0:3,1:4,2:5,3:6,4:0,5:1,6:2}
lastfriday = datetime.today()-timedelta(days=d[datetime.today().weekday()])

Your question is similar to this one
Here's a starting point:
import datetime
def get_last_friday():
current_time = datetime.datetime.now()
last_friday = (current_time.date()
- datetime.timedelta(days=current_time.weekday())
+ datetime.timedelta(days=4, weeks=-1))
return last_friday

Get first and last day of given week number in python

I need a function which returns the first and last day respectively the Monday and Sunday of a given week number (and a year).
There is a difficulty for weeks where Jan 1st is not a Monday so I cannot use the standard datetime.datetime.strptime().

Here's the solution:
import calendar
import datetime
from datetime import timedelta
def get_start_and_end_date_from_calendar_week(year, calendar_week):
monday = datetime.datetime.strptime(f'{year}-{calendar_week}-1', "%Y-%W-%w").date()
return monday, monday + datetime.timedelta(days=6.9)
I extended the great logic of this post.
Update
There is a pitfall with the first calendar week. Certain countries handle the first week number differently. For example in Germany, if the first week in January has less than 4 days, it is counted as the last week of the year before. There's an overview at Wikipedia.

This other version found here is flawless and simple: https://www.pythonprogramming.in/how-to-get-start-and-end-of-week-data-from-a-given-date.html
##
# Python's program to get start and end of week
from datetime import datetime, timedelta
date_str = '2018-01-14'
date_obj = datetime.strptime(date_str, '%Y-%m-%d')
start_of_week = date_obj - timedelta(days=date_obj.weekday()) # Monday
end_of_week = start_of_week + timedelta(days=6) # Sunday
print(start_of_week)
print(end_of_week)

Get date from week number

Please what's wrong with my code:
import datetime
d = "2013-W26"
r = datetime.datetime.strptime(d, "%Y-W%W")
print(r)
Display "2013-01-01 00:00:00", Thanks.

A week number is not enough to generate a date; you need a day of the week as well. Add a default:
import datetime
d = "2013-W26"
r = datetime.datetime.strptime(d + '-1', "%Y-W%W-%w")
print(r)
The -1 and -%w pattern tells the parser to pick the Monday in that week. This outputs:
2013-07-01 00:00:00
%W uses Monday as the first day of the week. While you can pick your own weekday, you may get unexpected results if you deviate from that.
See the strftime() and strptime() behaviour section in the documentation, footnote 4:
When used with the strptime() method, %U and %W are only used in calculations when the day of the week and the year are specified.
Note, if your week number is a ISO week date, you'll want to use %G-W%V-%u instead! Those directives require Python 3.6 or newer.

In Python 3.8 there is the handy datetime.date.fromisocalendar:
>>> from datetime import date
>>> date.fromisocalendar(2020, 1, 1) # (year, week, day of week)
datetime.date(2019, 12, 30, 0, 0)
In older Python versions (3.7-) the calculation can use the information from datetime.date.isocalendar to figure out the week ISO8601 compliant weeks:
from datetime import date, timedelta
def monday_of_calenderweek(year, week):
first = date(year, 1, 1)
base = 1 if first.isocalendar()[1] == 1 else 8
return first + timedelta(days=base - first.isocalendar()[2] + 7 * (week - 1))
Both works also with datetime.datetime.

To complete the other answers - if you are using ISO week numbers, this string is appropriate (to get the Monday of a given ISO week number):
import datetime
d = '2013-W26'
r = datetime.datetime.strptime(d + '-1', '%G-W%V-%u')
print(r)
%G, %V, %u are ISO equivalents of %Y, %W, %w, so this outputs:
2013-06-24 00:00:00
Availabe in Python 3.6+; from docs.

import datetime
res = datetime.datetime.strptime("2018 W30 w1", "%Y %W w%w")
print res
Adding of 1 as week day will yield exact current week start. Adding of timedelta(days=6) will gives you the week end.
datetime.datetime(2018, 7, 23)

If anyone is looking for a simple function that returns all working days (Mo-Fr) dates from a week number consider this (based on accepted answer)
import datetime
def weeknum_to_dates(weeknum):
return [datetime.datetime.strptime("2021-W"+ str(weeknum) + str(x), "%Y-W%W-%w").strftime('%d.%m.%Y') for x in range(-5,0)]
weeknum_to_dates(37)
Output:
['17.09.2021', '16.09.2021', '15.09.2021', '14.09.2021', '13.09.2021']

In case you have the yearly number of week, just add the number of weeks to the first day of the year.
>>> import datetime
>>> from dateutil.relativedelta import relativedelta
>>> week = 40
>>> year = 2019
>>> date = datetime.date(year,1,1)+relativedelta(weeks=+week)
>>> date
datetime.date(2019, 10, 8)

Another solution which worked for me that accepts series data as opposed to strptime only accepting single string values:
#fw_to_date
import datetime
import pandas as pd
# fw is input in format 'YYYY-WW'
# Add weekday number to string 1 = Monday
fw = fw + '-1'
# dt is output column
# Use %G-%V-%w if input is in ISO format
dt = pd.to_datetime(fw, format='%Y-%W-%w', errors='coerce')

Here's a handy function including the issue with zero-week.

Python week number generating

I am trying to generate the number of the week based on the date that user input. I am aware of this following function that will give me the specific week number for the date that i input.
datetime.date(2012,5,21).isocalendar()[1]
my problem is that i want to make the week 1 of my year to start in April 1st instead of Jan 1st. Which means, my calendar for every year starts from April 1 and ends at march 31. Not Jan 1 to dec 31. Is there any build in function i can use?

Use datetime
from datetime import date
end = date(2012, 5, 21)
start = date(2012, 4, 1)
Diff = end - start
print Diff.days / 7 # convert to number of weeks

Extract day of year and Julian day from a string date

I have a string "2012.11.07" in python. I need to convert it to date object and then get an integer value of day of year and also Julian day. Is it possible?

First, you can convert it to a datetime.datetime object like this:
>>> import datetime
>>> fmt = '%Y.%m.%d'
>>> s = '2012.11.07'
>>> dt = datetime.datetime.strptime(s, fmt)
>>> dt
datetime.datetime(2012, 11, 7, 0, 0)
Then you can use the methods on datetime to get what you want… except that datetime doesn't have the function you want directly, so you need to convert to a time tuple
>>> tt = dt.timetuple()
>>> tt.tm_yday
312
The term "Julian day" has a few different meanings. If you're looking for 2012312, you have to do that indirectly, e.g., one of the following.
>>> int('%d%03d' % (tt.tm_year, tt.tm_yday))
2012312
>>> tt.tm_year * 1000 + tt.tm_yday
2012312
If you're looking for a different meaning, you should be able to figure it out from here. For example, if you want the "days since 1 Jan 4713 BC" meaning, and you have a formula that requires Gregorian year and day in year, you've got those two values above to plug in. (If you have a formula that takes Gregorian year, month, and day, you don't even need the timetuple step.) If you can't work out where to go from there, ask for further details.
If you don't have a formula—and maybe even if you already do—your best bet is probably to look around PyPI and ActiveState for pre-existing modules. For example, a quick search turned up something called jdcal. I'd never seen it before, but a quick pip install jdcal and a brief skim of the readme, and I was able to do this:
>>> sum(jdcal.gcal2jd(dt.year, dt.month, dt.day))
2456238.5
That's the same result that the USN Julian date converter gave me.
If you want integral Julian day, instead of fractional Julian date, you have to decide which direction you want to round—toward 0, toward negative infinity, rounding noon up to the next day, rounding noon toward even days, etc. (Note that Julian date is defined as starting since noon on 1 Jan 4713BC, so half of 7 Nov 2012 is 2456238, the other half is 2456239, and only you know which one of those you want…) For example, to round toward 0:
>>> int(sum(jdcal.gcal2jd(dt.year, dt.month, dt.day)))
2456238

To get the Julian day, use the datetime.date.toordinal method and add a fixed offset.
The Julian day is the number of days since January 1, 4713 BC at 12:00 in the proleptic Julian calendar, or November 24, 4714 BC at 12:00 in the proleptic Gregorian calendar. Note that each Julian day starts at noon, not midnight.
The toordinal function returns the number of days since December 31, 1 BC at 00:00 in the proleptic Gregorian calendar (in other words, January 1, 1 AD at 00:00 is the start of day 1, not day 0). Note that 1 BC directly precedes 1 AD, there was no year 0 since the number zero wasn't invented until many centuries later.
import datetime
datetime.date(1,1,1).toordinal()
# 1
Simply add 1721424.5 to the result of toordinal to get the Julian day.
Another answer already explained how to parse the string you started with and turn it into a datetime.date object. So you can find the Julian day as follows:
import datetime
my_date = datetime.date(2012,11,7) # time = 00:00:00
my_date.toordinal() + 1721424.5
# 2456238.5

To simplify the initial steps of abarnert's answer:
from dateutil import parser
s = '2012.11.07'
dt = parser.parse(s)
then apply the rest of abanert's answer.

This functionality (conversion of date strings to Julian date/time) is also present in the astropy module. Please refer to their documentation for complete details. The astropy implementation is especially handy for easy conversions to Julian time, as opposed to just the Julian date.
Example solution for the original question:
>>> import astropy.time
>>> import dateutil.parser
>>> dt = dateutil.parser.parse('2012.11.07')
>>> time = astropy.time.Time(dt)
>>> time.jd
2456238.5
>>> int(time.jd)
2456238

For quick computations, you could find day of year and Julian day number using only stdlib datetime module:
#!/usr/bin/env python3
from datetime import datetime, timedelta
DAY = timedelta(1)
JULIAN_EPOCH = datetime(2000, 1, 1, 12) # noon (the epoch name is unrelated)
J2000_JD = timedelta(2451545) # julian epoch in julian dates
dt = datetime.strptime("2012.11.07", "%Y.%m.%d") # get datetime object
day_of_year = (dt - datetime(dt.year, 1, 1)) // DAY + 1 # Jan the 1st is day 1
julian_day = (dt.replace(hour=12) - JULIAN_EPOCH + J2000_JD) // DAY
print(day_of_year, julian_day)
# 312 2456239
Another way to get day_of_year:
import time
day_of_year = time.strptime("2012.11.07", "%Y.%m.%d").tm_yday
julian_day in the code above is "the Julian day number associated with the solar day -- the number assigned to a day in a continuous count of days beginning with the Julian day number 0 assigned to the day starting at Greenwich mean noon on 1 January 4713 BC, Julian proleptic calendar -4712".
The time module documentation uses the term "Julian day" differently:
Jn The Julian day n (1 <= n <= 365). Leap days are not counted, so in
all years February 28 is day 59 and March 1 is day 60.
n The
zero-based Julian day (0 <= n <= 365). Leap days are counted, and it
is possible to refer to February 29.
i.e., the zero-based Julian day is day_of_year - 1 here. And the first one (Jn) is day_of_year - (calendar.isleap(dt.year) and day_of_year > 60) -- the days starting with March 1 are shifted to exclude the leap day.
There is also a related term: Julian date.
Julian day number is an integer. Julian date is inherently fractional: "The Julian Date (JD) of any instant is the Julian day number for the preceding noon plus the fraction of the day since that instant."
In general, to avoid handling edge cases yourself, use a library to compute Julian day as suggested by #abarnert.

According to this article there is an unpublished one-line formula created by Fliegel and Van Flandern to calculate an Gregorian Date to an Julian Date:
JD = 367 * year - 7 * (year + (month + 9)/12)/4 - 3 * ((year + (month - 9)/7)/100 + 1)/4 + 275 * month/9 + day + 1721029
This was compacted by P. M. Muller and R. N. Wimberly of the Jet Propulsion Laboratory, Pasadena, California for dates after March of 1900 to:
JD = 367 * year - 7 * (year + (month + 9)/12)/4 + 275 * month/9 + day + 1721014
These formulas are off by 0.5, so just subtract 0.5 from the formulas.
Use some string manupulation to actually extract the data and you will be good
>>> year, month, day = map(int,"2018.11.02".split("."))
>>> 367 * year - 7 * (year + (month + 9)/12)/4 + 275 * month/9 + day + 1721014 - 0.5
2458424.5

I import datetime lib, and use strftime to extract 'julian day', year, month, day...
import datetime as dt
my_date = dt.datetime.strptime('2012.11.07', '%Y.%m.%d')
jld_str = my_date.strftime('%j') # '312'
jld_int = int(jld_str) # 312

From the above examples, here is the one liner (non-Julian):
import datetime
doy = datetime.datetime.strptime('2014-01-01', '%Y-%m-%d').timetuple().tm_yday

def JulianDate_to_date(y, jd):
month = 1
while jd - calendar.monthrange(y,month)[1] > 0 and month <= 12:
jd = jd - calendar.monthrange(y,month)[1]
month += 1
date = datetime.date(y,month,jd).strftime("%m/%d/%Y")
return date

While the answer of #FGol provides self-contained formulae, it should be noted that those formulae are only valid if division rounds towards zero (so-called "truncated division"), which is language-dependent.
Python, for example, implements rounding towards -infinity, which is quite different. To use the formulae given in Python, you can do something like this:
def trunc_div(a, b):
"""Implement 'truncated division' in Python."""
return (a // b) if a >= 0 else -(-a // b)
def formula1(year, month, day):
"""Convert Gregorian date to julian day number."""
return 367 * year - trunc_div(7 * (year + trunc_div(month + 9, 12)), 4) - trunc_div(3 * (trunc_div(year + trunc_div(month - 9, 7), 100) + 1), 4) + trunc_div(275 * month, 9) + day + 1721029
def formula2(year, month, day):
"""Convert Gregorian date to julian day number (simplified); only valid for dates from March 1900 and beyond."""
return 367 * year - trunc_div(7 * (year + trunc_div(month + 9, 12)), 4) + trunc_div(275 * month, 9) + day + 1721014