I know how to get the last Saturday date using date --date="sat. last week" +"%m%d%Y" However, I am unable to get the last Saturday date for a particular given date.
I am trying to find the date of last saturday date from a given date. For an instance, let us say that the given date is 20140605 (in YYYYMMDD format) or any weekday date from that particular week. This given date could be any weekday date (20140602-20140606) in the following week for that Saturday, I need to derive the Saturday date for the week before this date which would be (in my case)= 20140531. How can I achieve this?
I just barfed a bit on my keyboard, but this seems to work:
D="20140605"; date --date "$D $[($(date --date "$D" +%u) + 1) % 7] days ago" +"%Y%m%d"
I have a small Python snippet for you:
from datetime import datetime, timedelta
import sys
fmt = "%Y%m%d"
daystring = sys.argv[1]
day = datetime.strptime(daystring, fmt)
for d in range(7):
# Go back a few days, but not more than 6.
testdate = day - timedelta(days=d)
# If the test date is a saturday (0 is monday, 6 sunday), output that date.
if testdate.weekday() == 5:
print testdate.strftime(fmt)
Test:
$ python test.py 20140605
20140531
Related
input: week_number = 34 (or 2022-34)
expected output:
["2022-08-21","2022-08-22", "2022-08-23","2022-08-24","2022-08-25","2022-08-26","2022-08-27"]
First date should be of Sunday
the last date should be Saturday
work with both leap and non leap year
Try:
import datetime
week_number = 34
out = []
date = datetime.datetime.strptime(f"2022-{week_number}-0", "%Y-%U-%w")
for day in range(7):
out.append((date + datetime.timedelta(days=day)).strftime("%Y-%m-%d"))
print(out)
Prints:
[
"2022-08-21",
"2022-08-22",
"2022-08-23",
"2022-08-24",
"2022-08-25",
"2022-08-26",
"2022-08-27",
]
From the reference:
%U - Week number of the year (Sunday as the first day of the week) as
a zero-padded decimal number. All days in a new year preceding the
first Sunday are considered to be in week 0.
I have 2 columns with timestamp on them, I need the difference between them in seconds in a 3rd column , excluding weekends. How am I supposed to do this in Python/pandas?
I want it to exclude Saturday/Sunday from the timeline.
Ex -
1 . Starts at Thursday/Friday and ends at Monday/Tuesday - Calculate duration only for the time it lied between Thursday/Friday and then directly Monday/Tuesday.
2 . If it starts on Saturday and ends on Monday - Calculate only for Monday.
3 . If ex.Starts on Friday and ends on Sunday, Calculate only for Friday.
4 . If starts and ends on Saturday and Sunday - result is 0 seconds
First, Convert the timestamp in both the columns to DateTime if not it's already in this format.
Second, find if the day is a weekend or a weekday and use total_seconds method to find the diff in seconds in the following way:
def find_diff_in_secs(row):
day_of_week = row["start"].weekday()
if day_of_week<5:
# Find the diff in secs
diff_in_secs =(row["end"]-row["start"]).total_seconds()
return diff_in_secs
else:
return "NA"
df.apply(find_diff_in_secs)
I want to find the last business day for the month of June and Dec for 2019. my below code gives me the last business day for last month and the current month. Ideally i want a code which allows me to input a year and month and it will tell me the last business day for the month and year i input.
i want my output to be like this for June 2019 and Dec 2019
2906
3112
hope someone can help me here, thanks
from pandas.tseries.offsets import BMonthEnd
from datetime import date
d=date.today()
offset = BMonthEnd()
#Last day of current month
current_month = offset.rollforward(d)
print(current_month)
#Last day of previous month
last_month = offset.rollback(d)
print(last_month)
Here's a function that will find the last weekday in a given month and format it as a string in the format given
import datetime
def last_weekday(year, month):
# Add a month
if month == 12:
month = 1
year += 1
else:
month += 1
d = datetime.date(year, month, 1)
# Subtract a day
d -= datetime.timedelta(days=1)
# Subtract more days if we have a weekend
while d.weekday() >= 5:
d -= datetime.timedelta(days=1)
return '{:02d}{:02d}'.format(d.month, d.day)
# Examples:
last_weekday(2019, 6) # -> '0628'
last_weekday(2019, 12) # -> '1231'
Having a lot of trouble translating the logic below in pandas/python, so I do not even have sample code or a df to work with :x
I run a daily report, that essentially filters for data from Monday thru the day before what 'Today' is. I have a Date column [ in dt.strftime('%#m/%#d/%Y') format] . It will never be longer than a Monday-Sunday scope.
1) Recognize the day it is 'today' when running the report, and recognize what day the closet Monday prior was. Filter the "Date" Column for the Monday-day before today's date [ in dt.strftime('%#m/%#d/%Y') format ]
2) Once the df is filtered for that, take this group of rows that have dates in the logic above, have it check for dates in a new column "Date2". If any dates are before the Monday Date, in Date2, change all of those earlier dates in 'Date2' to the Monday date it the 'Date' column.
3) If 'Today' is a Monday, then filter the scope from the Prior Monday through - Sunday in the "Date" Column. While this is filtered, do the step above [step 2] but also, for any dates in the "Date2" column that are Saturday and Sunday Dates - changes those to the Friday date.
Does this make sense?
Here're the steps:
from datetime import datetime
today = pd.to_datetime(datetime.now().date())
day_of_week = today.dayofweek
last_monday = today - pd.to_timedelta(day_of_week, unit='d')
# if today is Monday, we need to step back another week
if day_of_week == 0:
last_monday -= pd.to_timedelta(7, unit='d')
# filter for last Monday
last_monday_flags = (df.Date == last_mon)
# filter for Date2 < last Monday
date2_flags = (df.Date2 < last_monday)
# update where both flags are true
flags = last_monday_flags & date2_flags
df.loc[flags, 'Date2'] = last_monday
# if today is Monday
if day_of_week == 0:
last_sunday = last_monday + pd.to_timedelta(6, unit='d')
last_sat = last_sunday - pd.to_timedelta(1, unit='d')
last_week_flags = (df.Date >= last_monday) & (df.Date <= next_sunday)
last_sat_flags = (df.Date2 == last_sat)
last_sun_flags = (df.Date2 == last_sun)
# I'm just too lazy and not sure how Sat and Sun relates to Fri
# but i guess just subtract 1 day or 2 depending on which day
...
What I want to do is figure out what X of the month this is, and returning the relative delta constant for it (Su Mo Tu...). I have found many examples of jumping to a specific day of the month (1). For instance today is the 3rd Tuesday of December and I can get to it by doing this: + relativedelta(month=12, day=1, weekday=TU(3))) but what I want to do is the opposite:
Put in today's date and subtract the first of the month and get something like TU(3) or if it were the 4th wednesday to get: WE(4)
My ultimate goal is to then be able to transfer this constant to a different month or timedelta object and find the equivalent 3rd Tuesday, or 4th Wednesday, etc...
This is a solution that I have come up with, maybe you'll find it less complicated.
It also seems to be about 4 times faster, which if you process a lot of dates can make a difference.
from datetime import *
from dateutil.relativedelta import *
def weekDayOfTheMonth(xdate):
daylist = [MO,TU,WE,TH,FR,SA,SU]
weekday = xdate.weekday()
firstDayOfTheMonth = datetime(xdate.year, xdate.month, 1)
interval = (weekday + 7 - firstDayOfTheMonth.weekday() ) % 7
firstOfThisWeekDay = datetime(xdate.year, xdate.month, 1 + interval)
n = ((xdate.day - firstOfThisWeekDay.day) / 7) + 1
return daylist[weekday](n)
print(weekDayOfTheMonth(datetime.today()))
print(weekDayOfTheMonth(datetime(2018,11,24)))
Basically what happens is that:
I find what day of the week is the first day of given month.
Based on that information I can easily calculate first day of any given weekday in given month.
Then I can even more easily calculate that for example 18th of December 2018 is third Tuesday of this month.
Ok I found a way using rrule to create a list of days in the month that share the current weekday up until today, then length of this list becomes the Nth. Than I use a list as a lookup table for the weekday constants. Not tested to see if this will work for every day of the month but this is a start.
from datetime import *; from dateutil.relativedelta import *
from dateutil.rrule import rrule, WEEKLY
import calendar
def dayofthemonth(xdate):
daylist = [MO,TU,WE,TH,FR,SA,SU]
thisweekday = daylist[xdate.weekday()]
thisdaylist = list(rrule(freq=WEEKLY, dtstart=xdate+relativedelta(day=1), until=xdate, byweekday=xdate.weekday()))
return thisweekday(len(thisdaylist))
print(dayofthemonth(datetime.today())) #correctly returns TU(+3) for 2018, 12, 18