I want to create a list of dates, starting with today, and going back an arbitrary number of days, say, in my example 100 days. Is there a better way to do it than this?
import datetime
def date_generator():
cur = base = datetime.date.today()
end = base + datetime.timedelta(days=100)
delta = datetime.timedelta(days=1)
while(end>base):
base = base+delta
print base
date_generator()
from datetime import datetime, timedelta
from dateutil import parser
def getDateRange(begin, end):
""" """
beginDate = parser.parse(begin)
endDate = parser.parse(end)
delta = endDate-beginDate
numdays = delta.days + 1
dayList = [datetime.strftime(beginDate + timedelta(days=x), '%Y%m%d') for x in range(0, numdays)]
return dayList
From above answers i created this example for date generator
import datetime
date = datetime.datetime.now()
time = date.time()
def date_generator(date, delta):
counter =0
date = date - datetime.timedelta(days=delta)
while counter <= delta:
yield date
date = date + datetime.timedelta(days=1)
counter +=1
for date in date_generator(date, 30):
if date.date() != datetime.datetime.now().date():
start_date = datetime.datetime.combine(date, datetime.time())
end_date = datetime.datetime.combine(date, datetime.time.max)
else:
start_date = datetime.datetime.combine(date, datetime.time())
end_date = datetime.datetime.combine(date, time)
print('start_date---->',start_date,'end_date---->',end_date)
Another example that counts forwards or backwards, starting from Sandeep's answer.
from datetime import date, datetime, timedelta
from typing import Sequence
def range_of_dates(start_of_range: date, end_of_range: date) -> Sequence[date]:
if start_of_range <= end_of_range:
return [
start_of_range + timedelta(days=x)
for x in range(0, (end_of_range - start_of_range).days + 1)
]
return [
start_of_range - timedelta(days=x)
for x in range(0, (start_of_range - end_of_range).days + 1)
]
start_of_range = datetime.today().date()
end_of_range = start_of_range + timedelta(days=3)
date_range = range_of_dates(start_of_range, end_of_range)
print(date_range)
gives
[datetime.date(2019, 12, 20), datetime.date(2019, 12, 21), datetime.date(2019, 12, 22), datetime.date(2019, 12, 23)]
and
start_of_range = datetime.today().date()
end_of_range = start_of_range - timedelta(days=3)
date_range = range_of_dates(start_of_range, end_of_range)
print(date_range)
gives
[datetime.date(2019, 12, 20), datetime.date(2019, 12, 19), datetime.date(2019, 12, 18), datetime.date(2019, 12, 17)]
Note that the start date is included in the return, so if you want four total dates, use timedelta(days=3)
Pandas is great for time series in general, and has direct support for date ranges.
For example pd.date_range():
import pandas as pd
from datetime import datetime
datelist = pd.date_range(datetime.today(), periods=100).tolist()
It also has lots of options to make life easier. For example if you only wanted weekdays, you would just swap in bdate_range.
See date range documentation
In addition it fully supports pytz timezones and can smoothly span spring/autumn DST shifts.
EDIT by OP:
If you need actual python datetimes, as opposed to Pandas timestamps:
import pandas as pd
from datetime import datetime
pd.date_range(end = datetime.today(), periods = 100).to_pydatetime().tolist()
#OR
pd.date_range(start="2018-09-09",end="2020-02-02")
This uses the "end" parameter to match the original question, but if you want descending dates:
pd.date_range(datetime.today(), periods=100).to_pydatetime().tolist()
Here's a one liner for bash scripts to get a list of weekdays, this is python 3. Easily modified for whatever, the int at the end is the number of days in the past you want.
python -c "import sys,datetime; print('\n'.join([(datetime.datetime.today() - datetime.timedelta(days=x)).strftime(\"%Y/%m/%d\") for x in range(0,int(sys.argv[1])) if (datetime.datetime.today() - datetime.timedelta(days=x)).isoweekday()<6]))" 10
Here is a variant to provide a start (or rather, end) date
python -c "import sys,datetime; print('\n'.join([(datetime.datetime.strptime(sys.argv[1],\"%Y/%m/%d\") - datetime.timedelta(days=x)).strftime(\"%Y/%m/%d \") for x in range(0,int(sys.argv[2])) if (datetime.datetime.today() - datetime.timedelta(days=x)).isoweekday()<6]))" 2015/12/30 10
Here is a variant for arbitrary start and end dates. not that this isn't terribly efficient, but is good for putting in a for loop in a bash script:
python -c "import sys,datetime; print('\n'.join([(datetime.datetime.strptime(sys.argv[1],\"%Y/%m/%d\") + datetime.timedelta(days=x)).strftime(\"%Y/%m/%d\") for x in range(0,int((datetime.datetime.strptime(sys.argv[2], \"%Y/%m/%d\") - datetime.datetime.strptime(sys.argv[1], \"%Y/%m/%d\")).days)) if (datetime.datetime.strptime(sys.argv[1], \"%Y/%m/%d\") + datetime.timedelta(days=x)).isoweekday()<6]))" 2015/12/15 2015/12/30