Get Yesterday's date as a string in YYYY-MM-DD format

As an input to an API request I need to get yesterday's date as a string in the format YYYY-MM-DD. My working version seems inelegant:

yesterday = datetime.date.fromordinal(datetime.date.today().toordinal()-1)
report_date = str(yesterday.year) + \
   ('-' if len(str(yesterday.month)) == 2 else '-0') + str(yesterday.month) + \
   ('-' if len(str(yesterday.day)) == 2 else '-0') + str(yesterday.day)

You just need to subtract one day from today's date. In Python datetime.timedelta object lets you create specific spans of time as a timedelta object.

datetime.timedelta(1) gives you the duration of "one day" and is subtractable from a datetime object. After you subtracted the objects you can use datetime.strftime to convert the result --which is a date object-- to string format based on your format of choice:

>>> from datetime import datetime, timedelta
>>> yesterday = datetime.now() - timedelta(1)

>>> type(yesterday)
>>> datetime.datetime

>>> datetime.strftime(yesterday, '%Y-%m-%d')
'2015-05-26'

Note that instead of calling the datetime.strftime function, you can also directly use strftime method of datetime objects:

>>> (datetime.now() - timedelta(1)).strftime('%Y-%m-%d')
'2015-05-26'

As a function:

from datetime import datetime, timedelta


def yesterday(frmt='%Y-%m-%d', string=True):
    yesterday = datetime.now() - timedelta(1)
    if string:
        return yesterday.strftime(frmt)
    return yesterday

example:

In [10]: yesterday()
Out[10]: '2022-05-13'

In [11]: yesterday(string=False)
Out[11]: datetime.datetime(2022, 5, 13, 12, 34, 31, 701270)