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

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)

There must be a more elegant way to do this, interested for educational purposes as much as anything else!

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 in order 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)