For a private project I want to check in Python if one day is in a range of two dates. The tricky part is that I have to check it without the year.
The project is meant to calculate what star sign a specific person has and the year is not needed.
Example:
Aquarius 01-21 - 02-19 (21.01. - 19.02.)
All the tutorials I've seen are with the specific year but it seems too complicated for me to calculate that too.
I don't need the whole code, but a hint would be very much appreciated.
I would add a reference year to the dates, just for sake of comparing it. The only tricky part would then be the range December - January. In those cases you would have to change the year to the next year.
start_date = datetime.datetime(2000, 1, 21)
end_date = datetime.datetime(2000, 2, 19)
check_date = datetime.datetime(2000, 2, 1)
if start_date < check_date and check_date < end_date:
print('success!')
Use the datetime module of Python, it will handle the year part as well. Refer to the example code below. If anyone finds any mistake or has a better approach, please comment also.
import datetime
def datetime_format(date):
date_list = date.split('-')
date_list = list(map(int, date_list))
date = datetime.date(date_list[0], date_list[1], date_list[2])
return date
def compare_dates(first, last, date_):
first = datetime_format(first)
last = datetime_format(last)
date_ = datetime_format(date_)
if date_ <= last and date_ >= first:
return("Within range")
return("Out of range")
first = '2020-01-01'
last = '2022-12-31'
date_ = '2021-05-05'
print(compare_dates(first, last, date_))
Related
I am trying to print the working days between two dates, but excluding the latter one and the weekends, holding in there. I've tried the following code:
from datetime import timedelta, date, datetime
def print_working_dates(date1, date2):
excluded = (6,7)
for n in range(int((date2 - date1).days)+1):
if date1.isoweekday == excluded:
continue
else:
yield date1 + timedelta(n)
start_dt = datetime.now()
end_dt = datetime(2022, 4, 28)
for dt in print_working_dates(start_dt, end_dt):
print(dt.strftime("%Y-%m-%d"))
which prints every day before the last one, but it holds all weekends. Could anyone please anyone could give a piece of advice to define better this function? To my mind, the excluded object should be better detailed, but I cannot know how.
Firstly, you can check if a value is in an iterable (such as a tuple or list) using the in keyword. isoweekday is a function, so you need to call it with isoweekday(). Finally, you need to work out the new date, then check it's weekday, otherwise it just checks if the start date is a weekend for every date.
def print_working_dates(date1, date2):
excluded = (6,7)
for n in range(int((date2 - date1).days)+1):
new = date1 + timedelta(n)
if new.isoweekday() not in excluded:
yield new
I've negated the if statement with not to make it a bit more compact.
First off, you are not checking the correct time to be a weekday or not you have to check if (date1 + timedelta(n)) is a weekday, not date1
And secondly you have to you have to check if isoweekday's return value is in the excluded
Here's my solution to your code:
from datetime import timedelta, datetime
def print_working_dates(date1, date2):
excluded = (6,7)
for n in range(int((date2 - date1).days)+1):
if (date1 + timedelta(n)).isoweekday() in excluded:
continue
else:
yield date1 + timedelta(n)
start_dt = datetime.now()
end_dt = datetime(2022, 4, 28)
for dt in print_working_dates(start_dt, end_dt):
print(dt.strftime("%Y-%m-%d"))
I'm not a Python developer but have to fix an existing code.
In this code, a method (extract) is called providing an interval of dates:
extract(start_date, end_date)
The parameters can have for exemple the values:
start_date : 2020-10-01
end_date : 2022-01-03
The problem
The issue with this call is that the extract method only support a 1 year max interval of dates. If greater, the interval must be split, for exemple as follow:
extract('2020-10-01', '2020-12-31')
extract('2021-01-01', '2021-12-31')
extract('2022-01-01', '2022-01-03')
So I'm trying to create loop where the start_date and end_date are computed dynamically. But being new to Python, I have no ideas for now how this can be done. Any help would be greatly appreciated.
EDIT
Answer to some comments here
Tried so far finding a solution starting from code like this so far:
from datetime import datetime
from dateutil import relativedelta
from datetime import datetime
from_date = datetime.strptime('2020-10-01', "%Y-%m-%d")
end_date = datetime.strptime('2022-01-03', "%Y-%m-%d")
# Get the interval between the two dates
diff = relativedelta.relativedelta(end_date, from_date)
Then I thought iterating accross the years using diff.years and adding some logic to build the start_date and end_date from there, but I thought there might be a much simplier approach.
Also saw others possibilities like here but still no final simple result found at the moment.
from_str = '2020-10-01'
end_str = '2022-01-03'
from_year = int(from_str[:4])
end_year = int(end_str[:4])
if from_year != end_year:
# from_date to end of first year
extract(from_str, f"{from_year}-12-31")
# full years
for y in range(from_year + 1, end_year):
extract(f"{y}-01-01", f"{y}-12-31")
# rest
extract(f"{end_year}-01-01", end_str)
else:
extract(from_str, end_str)
As mentioned in the comments, you can either use the datetime library or you can also use pandas if you want. The pandas version is the following (admittively not the most pretty, but it does the job):
import pandas as pd
import datetime
start = datetime.datetime(2020,10,1)
end = datetime.datetime(2022,1,3)
def extract(from_dt, to_dt):
print(f'Extracting from {from_dt} to {to_dt}')
prev_end = pd.to_datetime(start)
for next_end in pd.date_range(datetime.datetime(start.year, 12, 31), end, freq='y'):
if next_end < end:
extract(prev_end.strftime('%Y-%m-%d'), next_end.strftime('%Y-%m-%d'))
else:
extract(prev_end.strftime('%Y-%m-%d'), end.strftime('%Y-%m-%d'))
prev_end = next_end + datetime.timedelta(days=1)
if prev_end < end:
extract(prev_end.strftime('%Y-%m-%d'), end.strftime('%Y-%m-%d'))
If you need to parse the original dates from strings, check out datetime.strptime
This kind of problems are nice ones to resolve by recursion:
from datetime import datetime
start_date = '2020-10-01'
end_date = '2022-01-03'
def intervalcalc(datestart,dateend):
newdate=dateend[:4] + '-01-01'
startd = datetime.strptime(datestart, "%Y-%m-%d")
endd = datetime.strptime(newdate, "%Y-%m-%d")
if endd < startd:
print(datestart, dateend)
return True
else:
print(newdate, dateend)
previousyear=str(int(newdate[:4])-1) + '-12-31'
intervalcalc(datestart,previousyear)
intervalcalc(start_date, end_date)
output:
2022-01-01 2022-01-03
2021-01-01 2021-12-31
2020-10-01 2020-12-31
You just need to change the prints by calls to extract function.
As mentioned by #Wups the conversion to date is not really necessary, it could be an string compare as they are YYYYMMDD dates.
Also, this can be done the other way around and calculate from the start date year + '-12-31' and then compare dateend>end_date to determine the anchor for the recursion.
I want tot create a list of dates starting from 10/09/2020 with increments of 182 days until reaching 05/03/2020.
my code is this :
start_date="10/09/2020"
last_date="05/03/2020"
start_date=datetimedatetime.strptime(date,"%d/%m/%Y").date()
last_date=datetimedatetime.strptime(date,"%d/%m/%Y").date()
dates=[]
while dates[-1] != last_date:
i=star_date+timedelta(days=182)
dates.append(i)
dates[i]=i+timedelta(days=pago_cupon)
I don't know what's your problem.
You can try this code
from datetime import date, timedelta
start_day = date(year=2020, month=9, day=10)
end_day = date(year=2021, month=3, day=5)
one_data_delta = timedelta(days=1)
res = []
while end_day != start_day:
start_day += one_data_delta
res.append(start_day)
print(res)
A few problems:
You had a spelling mistake in your for loop star_date instead of start_date.
Also, you may want to change the comparison from != to less than equals or more than equals.
I am checking against (last_date - timedelta(days=182)) so that we won't exceed the last_date.
Your original start date is after your original end date.
As an example, I have adjusted the end day to be a couple years in the future.
I'm appending the dates as text.
from datetime import datetime, timedelta
start="10/09/2020"
last="05/03/2022"
start_date=datetime.strptime(start,"%d/%m/%Y")
last_date=datetime.strptime(last,"%d/%m/%Y")
dates=[]
while start_date <= (last_date - timedelta(days=182)):
start_date += timedelta(days=182)
dates.append(start_date.strftime("%d/%m/%Y"))
# not quite sure what you are trying to do here:
#dates[i]=i+timedelta(days=pago_cupon)
print(dates)
Output:
['11/03/2021', '09/09/2021']
In pandas I have a variable defined as
start_date = date (2020,7,1)
How do I update it to be the next day? I have a dataframe and I am filtering on individual days but I want to iterate through a full time range. I suppose I could have a for loop like so
for x < 10:
start_date = date (2020,7,x)
x +=1
But is there another way? I couldn't find any other stack exchange questions for python dates.
Assuming date is the regular python one you can add a day as follows:
from datetime import date, timedelta
start_date = date(2020, 7, 1)
next_date = start_date + timedelta(days=1)
I have a function that removes a file after a certain amount of time. The problem is that it works at later parts of the month, but when I try and remove 7 days from the start of the month it will not substract into the previous month. Does anyone know how to get this to work? The code is below that works out the date and removes the days.
today = datetime.date.today() # Today's date Binary
todaystr = datetime.date.today().isoformat() # Todays date as a string
minus_seven = today.replace(day=today.day-7).isoformat() # Removes 7 days
Thanks for any help.
minus_seven = today - datetime.timedelta(days = 7)
The reason this breaks is that today is a datetime.date; and as the docs say, that means that today.day is:
Between 1 and the number of days in the given month of the given year.
You can see why this works later in the month; but for the first few days of the month you end up with a negative value.
The docs immediately go on to document the correct way to do what you're trying to do:
date2 = date1 - timedelta Computes date2 such that date2 + timedelta == date1.