For the function below, I am inputting a string like "6/29/2020" and "8/10/2010" and I want to get a numbers of days after Jan. 1, 2010. For example, if I input "1/29/2010", I want the integer 29 to be returned.
Currently, I have gotten "6/29/2020" to a string "2020-06-29". Now I just need help with converting that string into the days after Jan. 1, 2010.
I feel like I have posted everything needed for you to help, but if you need more information, let me know. Thank You for helping me with this problem.
def day_conversion(dates):
import datetime
i = 0
for day in dates:
day = day.split('/')
if len(day[0]) == 1:
day[0] = f"0{day[0]}"
if len(day[1]) == 1:
day[1] = f"0{day[1]}"
day = f"{day[2]}-{day[0]}-{day[1]}"
# day = date.format(day)
# from datetime import date
# day0 = date(2000, 1, 1)
# day = day - day0
dates[i] = day
i += 1
return dates
datetime has a function for parsing dates, and subtracting two datetime objects gives a timedelta object with a .days attribute:
from datetime import datetime
def days_since_jan1_2010(date):
dt = datetime.strptime(date, '%m/%d/%Y')
diff = dt - datetime(2010, 1, 1)
return diff.days
def day_conversion(dates):
return [days_since_jan1_2010(d) for d in dates]
print(day_conversion(['6/29/2020', '8/10/2010', '1/1/2010', '1/2/2010']))
Output:
[3832, 221, 0, 1]
Everything in the previous answer is correct, but just thought I'd point out that you were very nearly there if you include the commented out part in your code above except for the following points:
from datetime import date needs to come before you try to use date.
You want date.fromisoformat, not date.format.
Your code has Jan 1 2000 but you state in your question that you want the number of days from Jan 1 2010.
If you substitute the commented part of your original code for the following four lines you should get the result you are after.
from datetime import date
day = date.fromisoformat(day)
day0 = date(2010, 1, 1)
day = day - day0
Related
If I want to add a loop to constrain days as well, what is the easiest way to do it, considering different length of month, leap years etc.
This is the script with years and months:
yearStart = 2010
yearEnd = 2017
monthStart = 1
monthEnd = 12
for year in list(range(yearStart, yearEnd + 1)):
for month in list(range(monthStart, monthEnd + 1)):
startDate = '%04d%02d%02d' % (year, month, 1)
numberOfDays = calendar.monthrange(year, month)[1]
lastDate = '%04d%02d%02d' % (year, month, numberOfDays)
If you want only the days then this code, using the pendulum library, is probably the easiest.
>>> import pendulum
>>> first_date = pendulum.Pendulum(2010, 1, 1)
>>> end_date = pendulum.Pendulum(2018, 1, 1)
>>> for day in pendulum.period(first_date, end_date).range('days'):
... print (day)
... break
...
2010-01-01T00:00:00+00:00
pendulum has many other nice features. For one thing, it's a drop-in replacement for datetime. Therefore, many of the properties and methods that you are familiar with using for that class will also be available to you.
You may want to use datetime in addition to calendar library. I am exactly not sure on requirements. But it appears you want the first date and last date of a given month and year. And, then loop through those dates. The following function will give you the first day and last day of each month. Then, you can loop between those two dates in whichever way you want.
import datetime
import calendar
def get_first_last_day(month, year):
date = datetime.datetime(year=year, month=month, day=1)
first_day = date.replace(day = 1)
last_day = date.replace(day = calendar.monthrange(date.year, date.month)[1])
return first_day, last_day
Adding the logic for looping through 2 dates as well.
d = first_day
delta = datetime.timedelta(days=1)
while d <= last_day:
print d.strftime("%Y-%m-%d")
d += delta
I am making a program where I input start date to dataStart(example 21.10.2000) and then input int days dateEnd and I convert it to another date (example 3000 = 0008-02-20)... Now I need to count these dates together, but I didn't managed myself how to do that. Here is my code.
from datetime import date
start=str(input("type start date (DD.MM.YYYY)"))
end=int(input("how many days from it?"))
dataStart=start.split(".")
days=int(dataStart[0])
months=int(dataStart[1])
years=int(dataStart[2])
endYears=0
endMonths=0
endDays=0
dateStart = date(years, months, days)
while end>=365:
end-=365
endYears+=1
else:
while end>=30:
end-=30
endMonths+=1
else:
while end>=1:
end-=1
endDays+=1
dateEnd = date(endYears, endMonths, endDays)
For adding days into date, you need to user datetime.timedelta
start=str(input("type start date (DD.MM.YYYY)"))
end=int(input("how many days from it?"))
date = datetime.strptime(start, "%d.%m.%Y")
modified_date = date + timedelta(days=end)
print(datetime.strftime(modified_date, "%d.%m.%Y"))
You may use datetime.timedelta to add certain units of time to your datetime object.
See the answers here for code snippets: Adding 5 days to a date in Python
Alternatively, you may wish to use the third-party dateutil library if you need support for time additions in units larger than weeks. For example:
>>> from datetime import datetime
>>> from dateutil import relativedelta
>>> one_month_later = datetime(2017, 5, 1) + relativedelta.relativedelta(months=1)
>>> one_month_later
>>> datetime.datetime(2017, 6, 1, 0, 0)
It will be easier to convert to datetime using datetime.datetime.strptime and for the part about adding days just use datetime.timedelta.
Below is a small snippet on how to use it:
import datetime
start = "21.10.2000"
end = 8
dateStart = datetime.datetime.strptime(start, "%d.%m.%Y")
dateEnd = dateStart + datetime.timedelta(days=end)
dateEnd.date() # to get the date format of the endDate
If you have any doubts please look at the documentation python3/python2.
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.
In Python, given a date, how do I find the preceding weekday? (Weekdays are Mon to Fri. I don't care about holidays)
Simply subtract a day from the given date, then check if the date is a weekday. If not, subtract another, until you do have a weekday:
from datetime import date, timedelta
def prev_weekday(adate):
adate -= timedelta(days=1)
while adate.weekday() > 4: # Mon-Fri are 0-4
adate -= timedelta(days=1)
return adate
Demo:
>>> prev_weekday(date.today())
datetime.date(2012, 8, 20)
>>> prev_weekday(date(2012, 8, 20))
datetime.date(2012, 8, 17)
Alternatively, use an offset table; no need to make this a mapping, a tuple will do just fine:
_offsets = (3, 1, 1, 1, 1, 1, 2)
def prev_weekday(adate):
return adate - timedelta(days=_offsets[adate.weekday()])
This is an old question, but anyway, a simpler way to do that, which doesn't require loops
from datetime import datetime, timedelta
today = datetime.today() # Today date
weekday = 6
days = today.isoweekday() - weekday
if days<0:
days += 7
previous_date = today - timedelta(days=days)
print(previous_date)
See the datetime module, in particular the date() and weekday() function. For example:
import datetime
temp = datetime.date(2012,8,21) # Today's date: 2012/08/21
print temp.weekday()
This will output 1. 0 stands for Monday, 1 for Tuesday, etc., until 6 for Sunday. Finding the preceding weekday is easy from here.
in datetime module you can do something like this: a = date.today() - timedelta(days=1)
and then a.weekday(). Where monday is 0 and sunday is 6.
I need to get date objects for the first and last day in the current year.
Currently I'm using this code which works fine, but I'm curious if there's a nicer way to do it; e.g. without having to specify the month/day manually.
from datetime import date
a = date(date.today().year, 1, 1)
b = date(date.today().year, 12, 31)
The only real improvement that comes to mind is to give your variables more descriptive names than a and b.
from datetime import datetime
starting_day_of_current_year = datetime.now().date().replace(month=1, day=1)
ending_day_of_current_year = datetime.now().date().replace(month=12, day=31)
There is nothing in the python library but there are external libraries that wrap this functionality up. For example, pandas has a timeseries library, with which you can do:
from datetime import date
from pandas.tseries import offsets
a = date.today() - offsets.YearBegin()
b = date.today() + offsets.YearEnd()
Whilst pandas is overkill if all you want is year begin and year end functionality, it also has support for a lot of other high level concepts such as business days, holiday calendars, month/quarter/year offsets: http://pandas.pydata.org/pandas-docs/stable/timeseries.html#dateoffset-objects
You'll have some funky stuff going on if you happen to be running this late on New Year's Eve and the two calls to today() cross the year boundary. It's safer to do this:
from datetime import date
epoch_year = date.today().year
year_start = date(epoch_year, 1, 1)
year_end = date(epoch_year, 12, 31)
import datetime
year = 2016
first_day_of_year = datetime.date.min.replace(year = year)
last_day_of_year = datetime.date.max.replace(year = year)
print(first_day_of_year, last_day_of_year)
duh.
What if we want to get the exact time the year begins or ends? Same thing. Replace datetime.date with datetime.datetime, and there you go, you got the first last day of year in datetime.datetime format.
To make this even fancier, wrap the whole thing in a function:
import datetime
def year_range(year, datetime_o = datetime.date):
return (
datetime_o.min.replace(year = year),
datetime_o.max.replace(year = year)
)
print(year_range(2016, datetime.date))
print(year_range(2016, datetime.datetime))
Output:
(datetime.date(2016, 1, 1), datetime.date(2016, 12, 31))
(datetime.datetime(2016, 1, 1, 0, 0), datetime.datetime(2016, 12, 31, 23, 59, 59, 999999))
use relative timedelta and substract from date object
from dateutil.relativedelta import relativedelta
import datetime
date = datetime.date.today()
fistOfYear = date - relativedelta(years=0, month=1, day=1)
one genius way to find out first and last day of year is code below
this code works well even for leap years
first_day=datetime.date(year=i,month=1, day=1)
first_day_of_next_year=first_day.replace(year=first_day.year+1,month=1, day=1)
last_day=first_day_of_next_year-jdatetime.timedelta(days=1)