Let's say I have 11 Sessions for myself to complete. I haven't set dates for these sessions but rather just weekdays where one session would take place. Let's say when scheduling these sessions, I chose MON, TUE and WED. This means that after today, I want the dates to 11 my sessions which would be 4 Mondays, 4 Tuesdays and 3 Wednesdays from now after which my sessions will be completed.
I want to automatically get the dates for these days until there are 11 dates in total.
I really hope this makes sense... Please help me. I've been scratching my head over this for 3 hours straight.
Thanks,
You can use pd.date_range and the CustomBusinessDay object to do this very easily.
You can use the CustomBusinessDay to specify your "business days" and create your date range from it:
import pandas
from datetime import date
session_days = pd.offset.CustomBusinessDay(weekmask="Mon Tue Wed")
dates = pd.date_range(date.today(), freq=session_days, periods=11)
I figured it out a while ago but my internet died. All it took was Dunhill and some rest.
import datetime
def get_dates():
#This is the max number of dates you want. In my case, sessions.
required_sessions = 11
#These are the weekdays you want these sessions to be
days = [1,2,3]
#An empty list to store the dates you get
dates = []
#Initialize a variable for the while loop
current_sessions = 0
#I will start counting from today but you can choose any date
now = datetime.datetime.now()
#For my use case, I don't want a session on the same day I run this function.
#I will start counting from the next day
if now.weekday() in days:
now = now + datetime.timedelta(days=1)
while current_sessions != required_sessions:
#Iterate over every day in your desired days
for day in days:
#Just a precautionary measure so the for loops breaks as soon as you have the max number of dates
#Or the while loop will run for ever
if current_sessions == required_sessions:
break
#If it's Saturday, you wanna hop onto the next week
if now.weekday() == 6:
#Check if Sunday is in the days, add it
if 0 in days:
date = now + datetime.timedelta(days=1)
dates.append(date)
current_sessions += 1
now = date
else:
#Explains itself.
if now.weekday() == day:
dates.append(now)
now = now + datetime.timedelta(days=1)
current_sessions += 1
#If the weekday today is greater than the day you're iterating over, this means you've iterated over all the days in a NUMERIC ORDER
#NOTE: This only works if the days in your "days" list are in a correct numeric order meaning 0 - 6. If it's random, you'll have trouble
elif not now.weekday() > day:
difference = day - now.weekday()
date = now + datetime.timedelta(days=difference)
dates.append(date)
now = date
current_sessions += 1
#Reset the cycle after the for loop is done so you can hop on to the next week.
reset_cycle_days = 6 - now.weekday()
if reset_cycle_days == 0:
original_now = now + datetime.timedelta(days=1)
now = original_now
else:
original_now = now + datetime.timedelta(days=reset_cycle_days)
now = original_now
for date in dates:(
print(date.strftime("%d/%m/%y"), date.weekday()))
Btw, I know this answer is pointless compared to #Daniel Geffen 's answer. If I were you, I would definitely choose his answer as it is very simple. This was just my contribution to my own question in case anyone would want to jump into the "technicalities" of how it's done by just using datetime. For me, this works best as I'm having issues with _bz2 in Python3.7 .
Thank you all for your help.
I'm quite lost and I'm in need of trying to format some code so it ends up having dashes in the date. I can get 3, 12, 28 but I can't get 3-12-28. I am a super new beginner so I'm quite lost at the moment.
year = 3
month = 12
day = 28
print(date)
Try
print("{0}-{1}-{2}".format(year,month,day))
You could use datetime to format the result
import datetime
year = 3
month = 12
day = 28
dt = (datetime.date(year, month, day))
print(dt)
the result will be 0003-12-28
if you want more examples of datetime you could take a look at https://docs.python.org/2/library/datetime.html#
As you say you are new to python you can concatenate the strings together.
year = 3
month = 12
day = 28
date = year + "-" + month + "-" + day
print(date)
Alternatively you can use format to set the variables in your required format.
print(f"{year}-{month}-{day}")
Another method is to use datetime if you are using todays date
import datetime
today = datetime.date.today()
print(today)
I have a pandas dataframe in which each cell of a column contains a timestamp, saved as a string:
>>>dataset['DateTime'][1]
'2018-03-14 00:34:46'
I would like to create a new column in which those dates are manipulated in the following way:
year += 1,
month += 2,
day += 3,
hour += 4,
minute += 5,
second += 6
(Important to this manipulation is that the initial date and the new date have a one-to-one relation, so that I can transform the date back later onwards)
In my case, the output I am looking for is as follows:
>>> dataset['newTimestamp'][1]
'2019-05-17 04:39:52'
To do so I am using the datetime library to create datetime objects (as a test, I have started with one variable first):
timestamp = dataset['DateTime'][1]
p = datetime.datetime.strptime(timestamp, "%Y-%m-%d %H:%M:%S")
Currently I am doing arithmetics on the single variables:
year = p.year + 1
if p.month < 12:
month = p.month + 1
else:
month = 1
year += 1
However, as with the months, there are exceptions when you can and when you cannnot add values so that the new timestamp is still a real timestamp (12 + 1 = 13, which is not an actual month).
I could program every rule explicitly, but that seems too much work and I expect there are better ways. How could I do this faster?
Use DateOffset.
Also, have a look at relativedelta module for this kind of manipulations:
dataset['newTimestamp'] = pd.to_datetime(dataset['DateTime']) + pd.DateOffset(years=1, months=2, days=3, hours=4, minutes=5, seconds=6)
You should try out beautiful-date library:
pip install beautiful-date
And use it like so:
from beautiful_date import *
...
dataset['DateTime'].apply(lambda d: d + 1 * years + 2 * months + ... + 6 * seconds)
should do the trick.
strptime() and strftime() are the functions you are looking for.
Just go ahead and google the two fuctions. surely enough,you will be abe to solve the stated problem.
these can be used to directly manipulate date-time quantities
I'm trying to calculate the nth weekday for a given date. For example, I should be able to calculate the 3rd wednesday in the month for a given date.
I have written 2 versions of a function that is supposed to do that:
from datetime import datetime, timedelta
### version 1
def nth_weekday(the_date, nth_week, week_day):
temp = the_date.replace(day=1)
adj = (nth_week-1)*7 + temp.weekday()-week_day
return temp + timedelta(days=adj)
### version 2
def nth_weekday(the_date, nth_week, week_day):
temp = the_date.replace(day=1)
adj = temp.weekday()-week_day
temp += timedelta(days=adj)
temp += timedelta(weeks=nth_week)
return temp
Console output
# Calculate the 3rd Friday for the date 2011-08-09
x=nth_weekday(datetime(year=2011,month=8,day=9),3,4)
print 'output:',x.strftime('%d%b%y')
# output: 11Aug11 (Expected: '19Aug11')
The logic in both functions is obviously wrong, but I can't seem to locate the bug - can anyone spot what is wrong with the code - and how do I fix it to return the correct value?
Your problem is here:
adj = temp.weekday()-week_day
First of all, you are subtracting things the wrong way: you need to subtract the actual day from the desired one, not the other way around.
Second, you need to ensure that the result of the subtraction is not negative - it should be put in the range 0-6 using % 7.
The result:
adj = (week_day - temp.weekday()) % 7
In addition, in your second version, you need to add nth_week-1 weeks like you do in your first version.
Complete example:
def nth_weekday(the_date, nth_week, week_day):
temp = the_date.replace(day=1)
adj = (week_day - temp.weekday()) % 7
temp += timedelta(days=adj)
temp += timedelta(weeks=nth_week-1)
return temp
>>> nth_weekday(datetime(2011,8,9), 3, 4)
datetime.datetime(2011, 8, 19, 0, 0)
one-liner
You can find the nth weekday with a one liner that uses calendar from the standard library.
import calendar
calendar.Calendar(x).monthdatescalendar(year, month)[n][0]
where:
x : the integer representing your weekday (0 is Monday)
n : the 'nth' part of your question
year, month : the integers year and month
This will return a datetime.date object.
broken down
It can be broken down this way:
calendar.Calendar(x)
creates a calendar object with weekdays starting on your required weekday.
.monthdatescalendar(year, month)
returns all the calendar days of that month.
[n][0]
returns the 0 indexed value of the nth week (the first day of that week, which starts on the xth day).
why it works
The reason for starting the week on your required weekday is that by default 0 (Monday) will be used as the first day of the week and if the month starts on a Wednesday, calendar will consider the first week to start on the first occurrence of Monday (ie. week 2) and you'll be a week behind.
example
If you were to need the third Saturday of September 2013 (that month's US stock option expiry day), you would use the following:
calendar.Calendar(5).monthdatescalendar(2013,9)[3][0]
The problem with the one-liner with the most votes is it doesn't work.
It can however be used as a basis for refinement:
You see this is what you get:
c = calendar.Calendar(calendar.SUNDAY).monthdatescalendar(2018, 7)
for c2 in c:
print(c2[0])
2018-07-01
2018-07-08
2018-07-15
2018-07-22
2018-07-29
c = calendar.Calendar(calendar.SUNDAY).monthdatescalendar(2018, 8)
for c2 in c:
print(c2[0])
2018-07-29
2018-08-05
2018-08-12
2018-08-19
2018-08-26
If you think about it it's trying to organise the calendars into nested lists to print a weeks worth of dates at a time. So stragglers from other months come into play. By using a new list of valid days that fall in the month - this does the trick.
Answer with appended list
import calendar
import datetime
def get_nth_DOW_for_YY_MM(dow, yy, mm, nth) -> datetime.date:
#dow - Python Cal - 6 Sun 0 Mon ... 5 Sat
#nth is 1 based... -1. is ok for last.
i = -1 if nth == -1 or nth == 5 else nth -1
valid_days = []
for d in calendar.Calendar(dow).monthdatescalendar(yy, mm):
if d[0].month == mm:
valid_days.append(d[0])
return valid_days[i]
So here's how it could be called:
firstSundayInJuly2018 = get_nth_DOW_for_YY_MM(calendar.SUNDAY, 2018, 7, 1)
firstSundayInAugust2018 = get_nth_DOW_for_YY_MM(calendar.SUNDAY, 2018, 8, 1)
print(firstSundayInJuly2018)
print(firstSundayInAugust2018)
And here is the output:
2018-07-01
2018-08-05
get_nth_DOW_for_YY_MM() can be refactored using lambda expressions like so:
Answer with lambda expression refactoring
import calendar
import datetime
def get_nth_DOW_for_YY_MM(dow, yy, mm, nth) -> datetime.date:
#dow - Python Cal - 6 Sun 0 Mon ... 5 Sat
#nth is 1 based... -1. is ok for last.
i = -1 if nth == -1 or nth == 5 else nth -1
return list(filter(lambda x: x.month == mm, \
list(map(lambda x: x[0], \
calendar.Calendar(dow).monthdatescalendar(yy, mm) \
)) \
))[i]
The one-liner answer does not seem to work if the target day falls on the first of the month. For instance, if you want the 2nd Friday of every month, then the one-liner approach
calendar.Calendar(4).monthdatescalendar(year, month)[2][0]
for March 2013 will return March 15th 2013 when it should be March 8th 2013. Perhaps add in a check like
if date(year, month, 1).weekday() == x:
delivery_date.append(calendar.Calendar(x).monthdatescalendar(year, month)[n-1][0])
else:
delivery_date.append(calendar.Calendar(x).monthdatescalendar(year, month)[n][0])
Alternatively this will work for Python 2, returns the occurance of weekday in the said month, i.e if 16 June 2018 is the input, then returns the occurance of the day on 16th June 2018
You may substitute the month/year/date integers to anything you might want - right now it's getting the input / date from the system via datetime
Omit out print statements or use pass where they're not needed
import calendar
import datetime
import pprint
month_number = int(datetime.datetime.now().strftime('%m'))
year_number = int(datetime.datetime.now().strftime('%Y'))
date_number = int(datetime.datetime.now().strftime('%d'))
day_ofweek = str(datetime.datetime.now().strftime('%A'))
def weekday_occurance():
print "\nFinding current date here\n"
for week in xrange(5):
try:
calendar.monthcalendar(year_number, month_number)[week].index(date_number)
occurance = week + 1
print "Date %s of month %s and year %s is %s #%s in this month." % (date_number,month_number,year_number,day_ofweek,occurance)
return occurance
break
except ValueError as e:
print "The date specified is %s which is week %s" % (e,week)
myocc = weekday_occurance()
print myocc
A little tweak would make the one-liner work correctly:
import calendar
calendar.Calendar((weekday+1)%7).monthdatescalendar(year, month)[n_th][-1]
Here n_th should be interpreted as c-style, e.g. 0 is the first index.
Example: to find 1st Sunday in July 2018 one could type:
>>> calendar.Calendar(0).monthdatescalendar(2018, 7)[0][-1]
datetime.date(2018, 7, 1)
People here seem to like one-liner, I will propose below.
import calendar
[cal[0] for cal in calendar.Calendar(x).monthdatescalendar(year, month) if cal[0].month == month][n]
The relativedelta module that's an extension from the Python dateutil package (pip install python-dateutil) does exactly what you want:
from dateutil import relativedelta
import datetime
def nth_weekday(the_date, nth_week, week_day):
return the_date.replace(day=1) + relativedelta.relativedelta(
weekday=week_day(nth_week)
)
print(nth_weekday(datetime.date.today(), 3, relativedelta.FR))
The key part here evaluates to weekday=relativedelta.FR(3): the third Friday of the month. Here are the relevant part of the docs for the weekday parameter,
weekday:
One of the weekday instances (MO, TU, etc) available in the
relativedelta module. These instances may receive a parameter N,
specifying the Nth weekday, which could be positive or negative
(like MO(+1) or MO(-2)).
How can I print the next year if the current year is given in python using the simplest code, possibly in one line using datetime module.
Both date and datetime objects have a year attribute, which is a number. Just add 1:
>>> from datetime import date
>>> print date.today().year + 1
2013
If you have the current year in a variable, just add 1 directly, no need to bother with the datetime module:
>>> year = 2012
>>> print year + 1
2013
If you have the date in a string, just select the 4 digits that represent the year and pass it to int:
>>> date = '2012-06-26'
>>> print int(date[:4]) + 1
2013
Year arithmetic is exceedingly simple, make it an integer and just add 1. It doesn't get much simpler than that.
If, however, you are working with a whole date, and you need the same date but one year later, use the components to create a new date object with the year incremented by one:
>>> today = date.today()
>>> print date(today.year + 1, today.month, today.day)
2013-06-26
or you can use the .replace function, which returns a copy with the field you specify changed:
>>> print today.replace(year=today.year + 1)
2013-06-26
Note that this can get a little tricky when today is February 29th in a leap year. The absolute, fail-safe correct way to work this one is thus:
def nextyear(dt):
try:
return dt.replace(year=dt.year+1)
except ValueError:
# February 29th in a leap year
# Add 365 days instead to arrive at March 1st
return dt + timedelta(days=365)
here is another simple way...
import datetime
x = datetime.datetime.now()
print(x.year+1)