I need to iterate through all the days from a custom day to now. I need all the correct day not just the count of the days.
For example, I enter 10 for month and 2018 for year. I need to get:
2018-10-01
2018-10-02
...
2018-10-31
2018-11-01
2018-11-02
...
2019-05-21
How to increment a datetime by one day?
https://docs.python.org/3.7/library/datetime.html
date = datetime.datetime(2007,12,5)
while date != datetime.date.today():
date += datetime.timedelta(days=1)
print(date)
The sample uses for loop
. But, I try to use recursion.
from datetime import timedelta, date
def getdate(date):
print (date)
if date == date.today(): return
getdate(date+timedelta(1))
start_date = date(2018, 1, 1)
getdate(start_date)
If you want to get a dataframe from pandas , the date_range
function does all the work for you (doc) . The pd.datetime.now()
method gives you the current date (doc)
Here one example:
def get_df_days_to_now(year, month, day = 1):
date_start = "%4d/%d/%d" % (year, month, day)
date_end = pd.datetime.now().strftime("%d/%m/%Y")
return pd.date_range(start=date_start, end=date_end, freq="D")
print(get_df_days_to_now(2019, 5))
# DatetimeIndex(['2019-05-01', '2019-05-02', '2019-05-03', '2019-05-04',
# '2019-05-05', '2019-05-06', '2019-05-07', '2019-05-08',
# '2019-05-09', '2019-05-10', '2019-05-11', '2019-05-12',
# '2019-05-13', '2019-05-14', '2019-05-15', '2019-05-16',
# '2019-05-17', '2019-05-18', '2019-05-19', '2019-05-20',
# '2019-05-21', '2019-05-22'],
# dtype = 'datetime64[ns]', freq = 'D')
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