how to add one day in 'yyyy-mm-dd' format in python?

Question

I have this string '2019-01-11', in year-month-day, i want to add one day to this date, and convert it into date format in python.

i tried to convert it to date format, so that i can add one day to it by using 'timedelta' class.

date_1 = datetime.datetime.strptime('2019-01-11', "%y-%m-%d")
date_1 =date_1 + datetime.timedelta(days=1)

but i got this error line 1.

raise ValueError("time data %r does not match format %r" %
ValueError: time data '2019-01-11' does not match format '%y-%m-%d'

Answer 1

The format you're after is:

date_1 = datetime.strptime('2019-01-11', "%Y-%m-%d")

(upper case "Y")

Answer 2

A four-digit year is %Y . See the doc: https://docs.python.org/3/library/datetime.html#strftime-and-strptime-format-codes

>>> datetime.datetime.strptime('2019-01-11', "%Y-%m-%d") + datetime.timedelta(days=1)
datetime.datetime(2019, 1, 12, 0, 0)

Note also that you have a couple of typos in your original code -- the first line has a random semicolon at the end, and the second line references date1 instead of date_1 .

Answer 3

For the past 5 years, I've been using pandas Timestamp and Timedelta for everything date related. Once something is in Timestamp format, it can be exported as a datetime:

In[1]:  from pandas import Timestamp, Timedelta
In[2]:  x=Timestamp("2022-10-31") + Timedelta(1, unit="days"))
In[3]:  x=x.date()
Out[3]: datetime.date(2022, 11, 1)

And if it's kept in Timestamp format, it can be printed as a string:

In[3]:  x=Timestamp("2022-10-31") + Timedelta(1, unit="days")
In[4]:  f"{x:%Y-%m-%d}"
Out[4]: 2022-11-01